Group Theory
Massoud Malek
One of the amazing features of the twentieth century mathematics has been the recognition
of the power of the abstract approach. This has given rise to a large body of new results,
and has, in fact, led us to open up whole new mathematics whose very existence had not
even suspected.
♣ Modular Arithmetic. The foundations of modular arithmetic were introduced in the third
century BCE, by Euclid, in the 7th book of his Elements.
Modular arithmetic is a system of arithmetic for integers, where numbers ”wrap around” upon reaching
a certain value, the modulus.
A familiar use of modular arithmetic is in the 12-hour clock, in which the day is divided into two
12-hour periods. If the time is 9:00 now, then 7 hours later it will be 4:00.
When we divide two integers a and b, we will have the equation: a = b q + r, where a is the dividend, b
is the divisor, q is the quotient, and r is the remainder, with 0 ≤ r < b.
Sometimes, we are only interested in what the remainder is when we divide a by b. For these cases
there is an operator called the modulo operator (abbreviated as mod). For example by dividing 16 by
12, the remainder is 4. Thus instead of saying 16 o’clock, we say 4 o’clock.
The notion of modular arithmetic is related to that of the remainder in Euclidean division. The
operation of finding the remainder is sometimes referred to as the modulo operation, and denoted
with ”mod” used as an infix operator. For example, the remainder of the division of 16 by 12 is
denoted by 16 mod 12; as this remainder is 4, we have 16 mod 12 = 4.
The congruence, indicated by ”≡” followed by ”(mod)” means:
Given integers n , A and B, the expression A ≡ B (mod n) means that A and B
have the same remainders, when they are divided by n .
The fundamental property of multiplication in modular arithmetic may thus be written
(a mod n) (b mod n) ≡ a b (mod n),
or, equivalently,
(a mod n) (b mod n) mod n = (a b) mod n .
♥ Examples.
1. Consider the modular equation x +12 x ≡ 2 mod 12.
Then 2 x ≡ 2 , 14 , 26 , 38 , . . . mod 12 or x ≡ 1 , 7 , 13 , 19 , . . . mod 12.
Since 13 > 12, we conclude that x ≡ 1mod 12 and x ≡ 7 mod 12 .
2. Consider the modular equation 3 x +8 5 ≡ 7 mod 8.
Then 3 x ≡ 2 , 10 , 18 , 26 , . . . mod 8. The only non-negative integer less than 8
is 18/3 = 6. Hence x = 6 .
3. Consider the modular equation 4 x +8 5 ≡ 7 mod 8. Then 4 x ≡ 2 , 10 , 18 , 26 , 34 , . . . mod 8.
But none of the numbers in the sequence are divisible by 4. Thus the equation has no solution.
Group Theory
Groups
2
♣ Groups. A group is a pair ( G, ∗) where G is a non-empty set and ∗ is a binary operation defined
on G satisfying the following conditions:
Axiom 1. Closure: ∀ a, b ∈ G, a ∗ b ∈ G.
Axiom 2. Associative: ∀ x, y, z ∈ G, x ∗ (y ∗ z) = (x ∗ y) ∗ z .
Axiom 3. Identity: There exists a unique Neutral or identity element e ∈ G (because the German
word for identity is ”Einheit”), such that ∀ a ∈ G, a ∗ e = e ∗ a = a.
Lemma 1. Uniqueness of the Identity in the group G.
Proof. Suppose both e and e0 are identities of G. Then,
(i) a e = a for all a ∈ G, and
(ii) e0 a = a for all a ∈ G.
By choosing a = e0 in (i) and a = e in (ii), we get e0 e = e0 and e0 e = e . Thus, e and e0 are both equal
to e0 e and so are equal to each other.
Axiom 4. Inverse: ∀ a ∈ G, there is a unique element a−1 ∈ G such that a ∗ a−1 = a−1 ∗ a = e .
Note that (a ∗ b) ∗ (b−1 ∗ a−1 ) = a ∗ (b ∗ b−1 ) ∗ a−1 = a ∗ e ∗ a−1 = a ∗ a−1 = e which implies that
(a ∗ b)−1 = b−1 ∗ a−1 .
Lemma 2. (Uniqueness of Inverses). ∀ a ∈ G, there is a unique element b ∈ G such that a ∗ b =
b ∗ a = e.
Proof. Suppose b and c are both inverses of a. Then a ∗ b = e and a ∗ c = e; thus a ∗ b = a ∗ c.
Canceling the a on both sides gives us b = c.
Lemma 3. (Cancellation). In a group G, the right and left cancellation laws hold; that is, b ∗ a = c ∗ a
implies b = c and a ∗ b = a ∗ c implies b = c.
Proof. Suppose b ∗ a = c ∗ a. Let a0 be an inverse of a. Then, multiplying on the right by a0 we obtain
(b ∗ a) ∗ a0 = (c ∗ a) ∗ a0 . Associativity yields b ∗ (a ∗ a0 ) = c ∗ (a ∗ a0 ). Then, b ∗ e = b and c ∗ e = c,
therefore, b = c. Similarly, one may prove a ∗ b = a ∗ c implies b = c, by a left multiplying of a0 .
Axiom 5. Commutativity: ∀ x, y ∈ G x ∗ y = y ∗ x.
A group G satisfying Axiom 5 is often called an abelian group.1 To prove that ( G, ∗) is a group, one
must verify first the closure, before other axioms. Clearly the fourth axiom must always be verified
after finding an identity. If the commutativity is verified right after the closure; then it is not necessary
to prove that the right identity is the same as the left identity or the right inverse is equal to the left
inverse.
Lemma 4. In a group ( G, ∗), if ∀ a, b ∈ G, (a ∗ b)−1 = a−1 ∗ b−1 ; then ( G, ∗) is an abelian group.
Proof. We only need to show that a ∗ b and b ∗ a have exactly the same inverse. Clearly
e = (a ∗ b) ∗ (a ∗ b)−1 and
(b ∗ a) ∗ (a ∗ b)−1 = (b ∗ a) ∗ a−1 ∗ b−1 = b ∗ (a ∗ a−1 ) ∗ b−1 = b ∗ (e) ∗ b−1 = b ∗ b−1 = e .
It is common practice to talk of the group G rather than ( G, ∗), if there is no confusion.
1
In recognition of the fact that there is no Nobel Prize for mathematics, in 2002 Norway established the Abel Prize
as the “Nobel Prize in mathematics” in honor of its native son. At approximately the $1,000,000 level, the Abel Prize is
now seen as an award equivalent to the Nobel Prize.
Niels Henrik Abel (1802-1829) suffered from tuberculosis and was unable to find a teaching position. Two days after his
death at 26, a letter came in which he was offered a teaching post at the University of Berlin.
Group Theory
Groups
3
Definition 1. If a group G contains a finite number of elements, we say that G is a finite group.
Otherwise we call G an infinite group. The order of a finite group G with n elements is denoted by
| G | = n.
The order of an element g in a group G is the smallest positive integer n such that g n = e. (In
additive notation, this would be n g = 0). If no such integer exists, we say that g has infinite order.
The order of an element g is denoted by | g |.
Lemma 5. ∀ a ∈ G, (a−1 )−1 = a.
Proof. let b = (a−1 )−1 , then b ∗ a−1 = e, but a ∗ a−1 is also equal to e. Thus b = (a−1 )−1 must be
a.
Lemma 6. For any elements a and b from a group and any integer n , (a−1 ∗ b ∗ a)n = a−1 ∗ bn ∗ a.
Proof. We have (a−1 ∗ b ∗ a)n = (a−1 ∗ b ∗ a) ∗ (a−1 ∗ b ∗ a) ∗ · · · ∗ (a−1 ∗ b ∗ a). The proof then follows
from the associativity and the facts that a ∗ a−1 = e and b ∗ e ∗ b ∗ e ∗ · · · ∗ e ∗ b = bn .
To facilitate future computations, we construct an operation table or Cayley table
Cayley table for ( Z5 , +5 ) and ( Z∗5 , ×5 ) are as follows:
+5
0̄
1̄
2̄
3̄
4̄
0̄
1̄
2̄
3̄
4̄
0̄
1̄
2̄
3̄
4̄
1̄
2̄
3̄
4̄
0̄
2̄
3̄
4̄
0̄
1̄
3̄
4̄
0̄
1̄
2̄
4̄
0̄
1̄
2̄
3̄
×5
1̄
2̄
3̄
4̄
1̄
2̄
3̄
4̄
1̄
2̄
3̄
4̄
2̄
4̄
1̄
3̄
3̄
1̄
4̄
2̄
4̄
3̄
2̄
1̄
2
For example the
♥ Examples.
1. (N, +) (N∗ , .) are not groups, since Axiom 4 is not satisfied.
2. (Z, +)) is an abelian group but (Z∗ , .)) is not a group, because of Axiom 4.
3. (Q, +)), (Q∗ , .)), (R, +)) (R∗ , .) are all abelian groups.
4. (Zn , +) and (Z∗p , .), where p is a prime number, are finite abelian groups. Note that (Z∗8 , .) is not a
group since 2.4 = 0 is not in Z∗8 .
5. Let U[n] be the set of all positive integers relatively prime to n , then (U[n], .) is an abelian group.
Clearly U[p] = Z∗p , whenever p is prime.
Recall that a b mod n is the unique integer r with the property a b = n q + r , where 0 ≤ r < n and
a b is the ordinary multiplication.
6. Let C = z = eiθ = (cos θ + i sin θ) : 0 ≤ θ ≤ 2 π be the set of all complex points on the unit circle.
Then C is an infinite abelian group.
7. The n-th roots of unity: Cn={z ∈ C : z n=1}, where z = cos 2 nkπ +i sin 2 nkπ , k = 0, 1, 2 . . . , n .
8. The set of all m × n matrices is an abelian group for addition but it is not a group under multiplication, since either it is not closed, when m 6= n or for m = n, not every square matrix has an inverse.
2
The English mathematician, Arthur Cayley (1821-1895) who took great pleasure in paintings and traveling, published
967 papers covering nearly every aspect of modern mathematics. He was fluentin French, German, Greek and Italian.
In 1854 Cayley wrote two papers which are remarkable for the insight they have of abstract groups. He defined an abstract
group and gave a table to display the group multiplication. He realized that matrices and quaternions were groups.
Group Theory
Groups
4
9. Let F be one of the following set: Z2 , Q, R, or C. The set GLn (F) of n × n invertible matrices,
together with the operation of ordinary matrix multiplication is called a linear group of degree n .
GLn (F) is a group, because the product of two invertible matrices is again invertible, and the inverse
of an invertible matrix is invertible.
10. The set of all 3 × 3 permutation matrices

 
 
 
 
 


0 1 0
1 0 0
0 0 1
0 1 0
0 0 1 
 1 0 0


 
 
 
 
 

P3 = 0 1 0, 1 0 0, 0 0 1, 0 1 0, 0 0 1, 1 0 0.




0 0 1
0 0 1
0 1 0
1 0 0
1 0 0
0 1 0
is a non-abelian group for multiplication.
A summary of group examples:
Group
Operation
Identity
Element
Inverse
abelian
Z
Addition
0
k
-k
Yes
Zn
Addition mod n
0
k
n-k
Yes
Multiplication
1
x
1/x
Yes
A−1
No
Q∗ , R∗
"
GL2 (F)
SL2 (F))
Matrix
multiplication
"
1 0
0 1
#
Matrix
multiplication
"
1 0
0 1
#
Matrix
multiplication
"
1 0
0 1
#
Multiplication
z=1
a b
A=
c d
ad − bc 6= 0
"
#
a b
c d
ad − bc = 1
"
D2 (R)
C
#
a 0
0 b
"
#
d −b
−c a
No
"
#
1/a 0
0 1/b
Yes
cos α−i sin α
Yes
#
ab 6= 0
cos α+i sin α
Exercise.
(I) If a and b are in the group G and a b = b a, we say that a and b commute. Assuming that a and b
commute, prove the following:
1. a−1 and b−1 commute.
2. a and b−1 commute.
3. a2 and b2 commute.
Solution. a2 b2 = a (a b) b = a (b a) b = (a b) (a b) = (b a) (b a) = b (a b) a = b (b a) a = b2 a2
Group Theory
Groups
5
4. a and a b commute.
5. For any x ∈ G, x a x−1 and x b x−1 commute.
Solution. We have (x a x−1 ) (x b x−1 ) = (x a) (x−1 x) (b x−1 ) = (x a) (b x−1 ) = x (a b) x−1 . (i)
and (x b x−1 ) (x a x−1 ) = (x b) (x−1 x) (a x−1 ) = (x b) (a x−1 ) = x (a b) x−1 . (ii)
Since a b = b a, we conclude that expressions in (i) and (ii) are equal.
6. Prove that a b = b a, if and only if a b a−1 b−1 = e.
(II) Let a, b, and b be elements of the group G and let e be the neutral element of G. Then prove the
following:
1. If a b = e, then b a = e.
2. If a b c = e, then c a b = e and b c a = e .
3. If for some x and y in G, x a y = a−1 , then y a x = a−1 .
4. a = a−1 , if and only if a2 = e.
5. Let c = c−1 , then a b = c, if and only if a b c = e.
Solution. If a b = c, then (a b) c = c c = c = c−1 = e.
Now if a b c = e, then c = e c = (a b c) c = (a b) (c c) = (a b) e = a b .
6. Let a = a−1 and b = b−1 , then b a is the inverse of a b.
7. Let a, b, and c each be equal to its own inverse. If a b = c, then b c = a and c a = b.
8. If a b c is its own inverse, then b c a is its own inverse and c a b is its own inverse.
9. Which of the following Cayley tables defined on the set G = { a , b , c , d } form a group? Support
your answer in each case.
a
(a) b
c
d
a
b
c
d
a
b
c
d
c
b
d
a
d
c
a
b
a
d
b
c
a
(b) b
c
d
a
b
c
d
a
b
c
d
b
a
d
c
c
d
a
b
d
c
b
a
a
(c) b
c
d
a
b
c
d
a
b
c
d
b
a
b
d
c
c
a
b
d
d
d
c
Definition 2. If there is an element x ∈ G such that a = x2 , then we say that a has a square root in
G. Similarly, if a = y 3 for some y ∈ G, then we say a has cubic root in G. In general a has an n-th
rooth in G, if an = z, for some z ∈ G.
Exercise.
1. If x a x = b, then a b has a square root in G.
2. If a3 = e, then a has a square root in G.
Solution. From a3 = e we have a (a2 ) = e. This implies that a is the inverse of a2 . Thus a =
(a a)−1 = a−1 a−1 = (a−1 )2 . This clearly complete the proof.
3. If a2 = e, then a has a cubic root in G.
4. If a−1 has a cubic root in G, then so does a.
♣ Subgroups. Let H be a non-empty subset of a group G. If H forms a group with its own elements,
then we call H a subgroup of G and we denote by H < G. Note that the associativity property is
independent of the subset of the group. Thus to prove that a subset of a group is a subgroup, one
does not need to verify the associativity.
Group Theory
Subgroups
6
Remark 1. An important point to be noted is this: if H < G, the operation of H is the same as the
operation of G. In other words, if a and b are in H, the product of a b computed in H is precisely the
product of a b computed in G.
For example, it would be meaningless to say that (Q∗ , . ) is a subgroup of (R, + ); for although Q∗ is a
subset of R, but the operations on these two sets are different.
Theorem 1. Let H ⊂ G such that ∀ a , b ∈ H, a ∗ b−1 ∈ H , then H < G .
Proof. By choosing a for b, we show that e ∈ H. Now if a = e, then the inverse of b must also be in
H. Finally, since (b−1 )−1 = b, we conclude that H is a closed set.
Corollary 1. Let H be a closed subset of G such that ∀ a ∈ H, a−1 ∈ H , then H < G .
Proof. Let a , b ∈ H. Since b ∈ H, then b−1 ∈ H. The fact that H is closed, the proof then follows
from the previous theorem.
Corollary 2. Let H be a nonempty finite subset of a group G. If H is closed under the operation of
G, then H is a subgroup of G.
Proof. In view of Corollary 1, we need only to prove that a−1 ∈ H whenever a ∈ H. If a = e, then
a−1 = a and we are done. If a 6= e, then consider the sequence a , a2 , . . .. By closure, all of these
elements belong to H. Since H is finite, not all of these elements are distinct. Say ai = aj , i > j.
Then, ai−j = e; and since a 6= e, i − j > 1. Thus, a ai−j−1 = ai−j = e, and, therefore, ai−j−1 = a−1 .
But, i − j − 1 ≥ 1 implies ai−j−1 = a−1 ∈ H and we are done.
Notation. For a ∈ G, the set . . . , a−3 , a−2 , a−1 , a0 = e , a , a2 , a3 , . . . = { an , where n ∈ Z }
is denoted by < a > . Note that (an )−1 = a−n .
Corollary 3. Let G be a group and a ∈ G, then < a > is a subgroup of G.
Proof. Clearly < a > is not empty, since a ∈ < a > . Now let an , am ∈ < a >, then an (am )−1 =
an−m ∈ < a >, so by Theorem 1, < a > is a subgroup of G.
Definition 3. The center, Z( G), of a group G is the subset of elements in G that commute with every
element of G. In symbols,
Z( G) = { a ∈ G : ∀ x ∈ G, x a = a x } .
The notation Z( G) comes from the fact that the German word Zentrum for center.
Theorem 2. Z( G) < G.
Proof. Clearly e ∈ Z( G), so Z( G) is non-empty. Suppose a , b ∈ Z( G). Then
(ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab) ∀ x ∈ G .
Let a ∈ Z( G), then ∀ x ∈ G,
a−1 x = a−1(x e) = a−1x(a a−1 ) = a−1(x a)a−1= a−1(a x)a−1= (a−1 a)(x a−1) = e(x a−1 ) = x a−1.
Definition 4. The centralizer of an element a of a group G is the set of elements of the group G
which commute with a,
C(a) = {x ∈ G : x a = a x} .
Group Theory
Subgroups
7
♥ Examples.
1. The singleton {e} is a subgroup of G which is called the trivial subgroup.
2. Clearly G is a subgroup of itself. Any subgroup of G that is different from the trivial subgroup
and G is called a proper subgroup. If we don’t know if the non-trivial subgroup H of G is proper or
not, then we denote by H ≤ G.
3. The group (Z8 , +) contains two proper subgroups: H1 = {0̄ , 4̄} and H2 = {0̄ , 2̄ , 4̄ , 6̄}.
4. The proper subgroups of (U[30], .) are as follows:
H1 = {1, 11}, H2 = {1, 19}, H3 = {1, 29}, H4 = {1, 7 , 13, 19}, H5 = {1 , 17 , 19, 23}.
5.
of
6.
of
The set H1 = z = ei kπ/n : k = 0, 1, 2, . . . and H2 = z = ei k : k = 0, 1, 2, . . . are both subgroup
C. Although H1 is a finite subgroup, but H2 has infinitely many elements.
The set H, of all n × n invertible diagonal matrices is an abelian subgroup under multiplication
the non-abelian group of all n × n invertible matrices. G.
Exercise.
1. Prove that the subset C(a) is a subgroup of G.
2. Prove that the subset C(a) = C(a−1 )
Solution. Let x ∈ C(a), then using the fat that the operation is associative, we have
(x a)−1 = (a x)−1 ⇒ a−1 x−1= x−1 a−1 ⇒ x a−1 x−1 x = x x−1 a−1 x ⇒ x a−1 = a−1 x.
3. Consider the group G = ( R, + ). Prove that H = {log a : a ∈ Q+ } is a subgroup of G.
4. Let H = { z = a + b i : a , b ∈ R, a b ≥ 0 }. Prove or disprove that the subset H is a subgroup of
C under addition.
Solution. Clearly 0 = 0+0 i is in H. If z = a+b i ∈ H, then a b ≥ 0 which implies that (−a) (−b) = a b
is greater than or equal zero. Hence −z ∈ H. Corollary 1 implies that H is a subgroup of C.
5. Let K = z = a + b i : a , b ∈ R, a2 + b2 = 1 . Prove or disprove that the set K is a subgroup of
C under multiplication.
6. Let F be the set of all real functions and let H = {f ∈ F : f (1) = 1 } . Prove or disprove that H
is a subgroup of F.
7. Let K = {f ∈ F : f (1) = 0 } . Prove or disprove that K is a subgroup of F.
8. Let H and K be subgroups of a group G. Show that H ∩ K ≤ G.
9. Let H and K be subgroups of a group G. Show that H ∪ K ≤ G, if and only if H ≤ K or K ≤ H.
Solution. Assume without loss of generality that H ≤ K, so H ⊆ K, which implies that H ∪ K =
K ≤ G.
Conversely, assume that H and K are not subgroups of each other, then H 6⊂ K and K 6⊂ H, which
implies that H ∪ K 6= H and H ∪ K 6= K. Then there exist an h ∈ H − K (this means that h ∈ H
but h is not in K) and a k ∈ K − H such that h k ∈ H ∪ K.
If h k ∈ H, then since h−1 ∈ H, we conclude that h−1 h k = e k = k ∈ H, which is a contradiction.
Similarly we can obtain a contradiction by assuming that h k ∈ K.
10. Let H and K be subgroups of an abelian group G. Show that H K = { x y : x ∈ H, y ∈ K} is a
subgroup of G.
11. Let H be a subgroup of a group G. Prove that ∀ g ∈ G, g H g −1 = {g h g −1 : h ∈ H } is a
subgroup of G.
Group Theory
Cyclic Groups
8
Solution. Let a , b ∈ g H g −1 , then a = g h1 g −1 and b = g h2 g −1 , for some h1 , h2 ∈ H. The fact
that (g −1 )−1 = g, implies that b−1 = b ; hence
a b−1 = (g h1 g −1 ) (g h2 g −1 ) = g h1 (g −1 g) h2 g −1 = g (h1 h2 ) g −1 .
Thus by Theorem 1, g H g −1 is a subgroup of G.
♣ Cyclic Groups. A group G is called cyclic if there exists an element g ∈ G such that
G = < g >= {g n : n ∈ Z} .
Such an element g is called a generator of G.
Definition 5. Let n be a positive integer. The number of divisors of n is denoted by τ (n). Also the
sum of divisors of n is denoted by σ(n) For example, σ(8) = 1 + 2 + 4 + 8 = 15 and τ (8) = 4.
Theorem 3. A group G is a union of proper subgroups if and only if G is not cyclic.
Proof. Suppose G = < g > is cyclic and the union of proper subgroups, then any subgroup of G
containing g contains G, hence a contradiction.
Conversely, say G is not cyclic. Then for every element g ∈ G, the subgroup < g > is a proper
subgroup. Clearly G is the union, taken over all g ∈ G, of the subgroups < g > .
It is easy to show that in a finite cyclic group G of order n , for any g ∈ G, | g | divides n .
Theorem 4. The number of proper subgroups of ( Zn , +n ) is equal to τ (n) − 2 .
Proof. If n is prime, then 1 and n are the only divisors of n , so τ (n) = 2, and Zn has no proper
subgroup; we also have τ (n) − 2 = 0.
If n is not prime, then the order of any proper subgroup of Zn generated by a member a ∈ Zn must
divide n . Thus there are exactly τ (n) = 2 proper subgroups in Zn .
Here is the list of all subgroups of the cyclic group ( Z8 , +8 ), where the last two subgroups are proper:
<
<
<
<
0̄
1̄
2̄
4̄
> = {0̄} ,
> = < 3̄ > = < 5̄ > = < 7̄ > = Z8 ,
> = < 6̄ > = { 2̄ , 4̄ , 6̄ , 0̄ } ,
> = { 4̄ , 0̄ } .
Notice that the cyclic group Z8 = { 0̄ , 1̄ , 2̄ , 3̄ , 4̄ , 5̄ , 6̄ , 7̄ } is not the union of the proper subgroups
{ 4̄ , 0̄ } and { 2̄ , 4̄ , 6̄ , 0̄ }.
Theorem 5. Let G be a cyclic group of order n and suppose that a ∈ G is a generator of the group.
Then ak = e if and only if n divides k.
Proof. Suppose that ak = e. By the division algorithm, k = n q + r, where 0 ≤ r < n; hence,
e = ak = an q+r = an q ar = e ar = ar .
Since the smallest positive integer m such that am = e is n , then r = 0 .
Conversely, suppose n divides k, then k = n s for some positive integer s. Consequently,
ak = an s = (an )s = es = e.
Group Theory
Cyclic Groups
9
Corollary 4. Let G be a cyclic group of order n and suppose that a ∈ G is a generator of the group.
If b = ak , then the order of b is n/d, where d = gcd(k, n).
Proof. We wish to find the smallest integer m such that e = bm = ak m . By Theorem 5, this is
the smallest integer m such that n divides k m or, equivalently, n/d divides m (k/d). Since d is the
greatest common divisor of n and k, then n/d and k/d are relatively prime. Hence, for n/d to divide
m (k/d), it must divide m. The smallest such m is n/d.
An immediate result of this Corollary is the following result concerning ( Zn , ).
Corollary 5. The generators of the cyclic group ( Zn , ) are the integers k such that 1 ≤ k < n and
gcd(k, n) = 1 .
Let us examine the group ( Z8 , +8 ). The numbers { 1̄ 3̄ 5̄ 7̄} are the elements of Z8 that are relatively
prime to 8. Each of these elements generates Z8 . For example
1 × 5̄ = 5̄ 2 × 5̄ = 2̄ 3 × 5̄ = 7̄ 4 × 5̄ = 4̄
5 × 5̄ = 1̄ 6 × 5̄ = 6̄ 7 × 5̄ = 3̄ 8 × 5̄ = 0̄
Note that by 2 × 5̄ = 5̄ + 5̄, 3 × 5̄ = 5̄ + 5̄ + 5̄, and so on.
♦ Basic Properties.
1. All cyclic groups are abelian.
2. Every subgroup of a finite cyclic group is cyclic. But the converse is not true.
Consider the finite group ( U[30], . ) = { 1 , 7 , 11 , 13 , 17 , 19 , 23 , 29 } . Then
< 1 > = { 1 },
< 7 > = < 13 > = { 7 , 19 , 13 , 1 } and < 17 > = < 23 > = { 17 , 19 , 23 , 1 },
< 11 > = { 11 , 1 }, < 19 > = { 19 , 1 }, and < 29 > = { 29 , 1 }.
Notice that ( U[30], · ) is not cyclic, but all of its subgroups are cyclic.
3. The order of any subgroup of a cyclic group of order n is a divisor of n .
4. A group of order n is cyclic if and only if for every divisor d of n , the group has exactly one
subgroup of order d.
The group ( U[30], · ) is not cyclic, because both subgroups { 7 , 19 , 13 , 1 } and { 17 , 19 , 23 , 1 } are
of order 4.
5. The twelfth roots of unity C12 = ei kπ/6 : k = 0, 1, 2, . . . , 11 , which are points on the unit circle,
form a multiplicative cyclic group.
6. An infinite cyclic group always has countable elements (its cardinality is ℵ0 ). An example of an
infinite cyclic group is ( Z, + ). Its subgroups are n Z. The set of complex points on the unit circle
C = {| z | = 1 : z ∈ C } is not a cyclic group; it is called a circle group.
7. In a non-finite cyclic group, there are exactly two elements that each generate the group. In the
cyclic group n Z, the integers n and −n are both (and only ones) generators of n Z.
Exercise.
1. Prove that a finite group of prime order is a cyclic group with no proper subgroups.
Solution. The order of any proper subgroup must divide the order of the group. But a prime number
p has no divisor greater than one and less than p.
Group Theory
Permutation Groups
10
2. Which of the following Cayley tables defined on the set G = { a , b , c , d } form a cyclic group?
Support your answer in each case.
a
(a) b
c
d
a
b
c
d
c
a
d
b
a
b
c
d
d
c
b
a
b
d
a
b
a
(b) b
c
d
a
b
c
d
a
b
c
d
b
a
d
c
c
d
a
b
d
c
b
a
a
(c) b
c
d
a
b
c
d
a
b
c
d
b
c
d
a
c
d
a
b
d
a
b
c
3. Prove that ( Q, + ) is not a cyclic group.
Solution. Suppose Q is cyclic, then for some relatively prime integers p , q, such that
p
p
p
p
p
Q = < > . Since
∈ Q, there exists an n ∈ Z∗ , such that
= n . By canceling on both side,
q
2q
2q
q
q
1
we conclude that ∈ Z, which is a contradiction. Thus, Q is not cyclic.
2
4. Use one of the properties of cyclic groups to show that ( R, + ) is not cyclic.
5. Let G be a cyclic group such that | G | = m n, where m > 1 , n > 1. Show that G has a non-trivial
subgroup.
6. Find all generators of ( Z10 , + ).
7. Let G be a cyclic group of order 30. Find the number of elements of order 6 in G and also find the
number of elements of order 5 in G.
8. Give an example of a non-cyclic abelian group all of whose proper subgroups are cyclic.
9. Consider the group ( Z∗7 , ×7 ) with the Cayley table:
×7
1̄
2̄
3̄
4̄
5̄
6̄
1̄
2̄
3̄
4̄
5̄
6̄
1̄
2̄
3̄
4̄
5̄
6̄
2̄
4̄
6̄
1̄
3̄
5̄
3̄
6̄
2̄
5̄
1̄
4̄
4̄
1̄
5̄
2̄
6̄
3̄
5̄
3̄
1̄
6̄
4̄
2̄
6̄
5̄
4̄
3̄
2̄
1̄
Is the group cyclic, if yes then find all its generators. List all its proper subgroups.
♣ Permutation Groups. A permutation on a non-empty set X is a function σ : X 7→ X such that
σ is one-to-one and onto. The set of all permutations on X denoted by S(X), is a non-abelian group
under the composition of functions. To prove this, we state that the composition of two bijections on
X is also a bijection on X; the composition is associative, preserving the injectivity and surjectivity
properties; the function that assigns each element of X to itself is the identity element i(x) = x; and
finally, since a permutation is a bijection, its inverse is also a permutation with
σ −1 ◦ σ (x) = σ ◦ σ −1 (x) = i(x) = x .
If |X| = n, then there are exactly n! permutations on X. It is common practice to chose
X = N∗n = {1, 2, 3, . . . , n}, whenever |X| = n; in this case, we denote by Sn the set of all permutations
Group Theory
Permutation Groups
11
on X; and a permutation is denoted by
σ=
1 2 3 ···
k1 k2 k3 · · ·
n−1 n
kn−1 kn
!
.
The group Sn is called a Symmetric group and contains exactly n ! elements.
!
!
1 2 3 4 5 6
1 2 3 4 5 6
Let α =
and β =
.
3 4 2 1 6 5
5 3 6 4 2 1
Then we obtain α ◦ β and α−1 as follows:
1 2 3 4 5 6
5 3 6 4 2 1
6 2 5 1 4 3
1 2 3 4 5 6
3 4 2 1 6 5
1 2 3 4 5 6
α◦β =
!
1 2 3 4 5 6
,
6 2 5 1 4 3
α−1 =
!
1 2 3 4 5 6
.
4 3 1 2 6 5
There is another notation commonly used to specify permutations. It is called cycle notation and was
first introduced by the great French mathematician Cauchy in 1815.3 A permutation cycle in Sn is a
subset of a Sn whose elements trade places with one another. For example, in the permutation
σ1 =
!
1 2 3 4
:1→4→3→2→1
4 1 2 3
and in the permutation
σ2 =
!
1 2 3 4
: 1 → 4 → 3 → 1 and 2 → 2.
4 2 1 3
So we say that σ1 is a 4-cycle and write σ1 = ( 1 4 3 2 ) and σ2 is the product of a 3-cycles and a
1-cycle and write σ2 = ( 1 4 3 ) ( 2 ) or simply σ = ( 1 4 3 ) by omitting the 1-cycle. ( 1 3 4 2 ) is called
the orbit of σ1 . A 2-cycle is also called a transposition or a two-element swap.
When X is a finite set of at least two elements, the permutations of X fall into two classes of equal
size: the even permutations and the odd permutations. There are many ways to write a permutation
as the product of transpositions, and they can vary in length, but those products will have either an
odd or an even number of factors, never both.
!
1 2 3 4 5
♥ Example. Consider the permutation σ =
. It can be obtained by three transpo3 4 5 2 1
sitions:
First exchange the places of 1 and 3, then exchange the places of 2 and 4, and finally exchange the
3
An arrogant royalist, self-righteous and narrow-minded bigot, once a week appeared before the Academy to present
a new paper. The Academy, largely on his account, was obliged to introduce a rule restricting the number of articles a
member could request to be published in a year. This great French mathematician known as a smug hypocrite by his fellow
scientists is known to us by the name of Augustine Louis Cauchy (1789-1857).
Numerous terms in mathematics bear Cauchy’s name. He produced 789 mathematics papers, an incredible achievement.
His collected works were published in 27 volumes.
Group Theory
Permutation Groups
12
places of 1 and 5. This shows that the given permutation σ is odd. So
σ=
!
1 2 3 4 5
= 1 3 5 2 4 = 1 5 1 3 2 4 .
3 4 5 2 1
There are many other ways of writing σ as a composition of transpositions, for instance σ may be
expressed as:
σ=
12345
13245
!
12345
21345
!
12345
14325
!
12345
12543
!
12345
12354
!
= ( 2 3 ) ( 1 2 ) ( 2 4 ) ( 3 5 ) ( 4 5 ),
but it is impossible to write it as a product of an even number of transpositions.
There are exactly n!/2 of even (odd) permutations in Sn . The even permutations of Sn form a subgroup
called Alternating group denoted by An .
The identity permutation is of length zero which makes it even; hence the set of odd permutations can
not be a subgroup of Sn .The group of even permutations is denoted by An and is called the alternating
n!
group of degree n. For n > 1, the order of An is
2.
The alternating groups are among the most important examples of groups. Many molecules with
chemical formulas of the form AB4, such as methane (CH4) and carbon tetrachloride (CCl4), have A4
as their symmetry group.
Theorem 6. For n > 1, the order of An is
n!
2.
Proof. For each odd permutation α, the permutation ( 1 2 ) α is even and ( 1 2 ) α 6= ( 1 2 ) β, when
α 6= β. Thus, there are at least as many even permutations as there are odd ones.
On the other hand, for each even permutation α, the permutation ( 1 2 ) α is odd and ( 1 2 ) α 6= ( 1 2 ) β
when α 6= β. Thus, there are at least as many odd permutations as there are even ones. It follows
n!
that there are equal numbers of even and odd permutations. Since | Sn | = n!, we have | A | =
2.
Remark 2. The fact that the set of odd permutations and even permutations of Sn are disjoint, we
conclude that if α is an even permutation and since An is a subgroup of Sn , then α−1 is also even.
This clearly implies that the inverse of an odd permutation is also odd.
The order of a permutation σ is the smallest natural number n such that σ n = ( 1 ). For example,
!
!
1 2 3 4 5 6
1
2
3
4
5
6
if σ =
, then σ 4 =
= ( 1 ).
3 4 2 1 6 5
1 2 3 4 5 6
Here is how:
1
3
2
4
1
2
4
1
3
2
3
2
4
1
3
4
1
3
2
4
5
6
5
6
5
6
5
6
5
6
Group Theory
Rubik’s Cube
13
♦ Basic Properties.
1. Any permutation can be expressed as a cycle or a product of disjoint cycles (more precisely: cycles
with disjoint orbits); such cycles commute with each other.
2. A cycle of k objects can be written as a product of (k − 1) transpositions.
3. If σ = C1 (k1 ) C2 (k2 ) · · · Cr (kr ), where Cj (kj )0 s are kj -cycles, then |σ| = lcm(k1 , k2 , . . . , kr ).
Exercise.
1. Let α =
!
1 2 3 4 5 6 7
,β =
2 1 5 4 6 7 3
!
1 2 3 4 5 6 7
, and δ = 1 2 5 7 3 4 6 .
7 6 1 2 3 4 5
Then find
(a.) α ◦ β , β ◦ α , α ◦ δ , β ◦ δ , δ ◦ α , δ ◦ β , α−1 , β −1 , and δ −1 .
(b.) Express α , β , and δ as a product of disjoint cycles.
(c.) Express α , β , and δ as a product of disjoint transpositions.
(d.) Find the order of α , β , and δ.
2. Determine whether the following permutations are even or odd.
a. ( 1 3 5 6 ) ;
b. ( 1 3 5 6 7) ;
c. ( 1 2 ) ( 1 3 4 ) ( 1 5 2 ).
3. Express σ = ( 1 2 3 4 5 ) as a product of (i) four transposition and (ii) six transpositions.
Solution.
(i) σ = ( 1 2 3 4 5 ) = ( 5 4) ( 5 3) ( 5 2) ( 5 1) ;
(ii) σ = ( 1 2 3 4 5 ) = ( 5 4) ( 5 2) ( 2 1) ( 2 5) ( 2 3) ( 1 3) .
4. Let α and β belong to Sn . Prove that α β is even if and only if α and β are both even or both odd.
!
1 2 3
5. How do you express α =
as a product of disjoint transpositions?
3 1 2
The Rubik’s Cube consists of 6 faces, each with 9 colored squares called facets, for a
total of 54 facets. A solved cube has all of the facets on each face having the same color.
A cube move rotates one of the 6 faces 90◦ , 180◦ or −90◦ . A center facet rotates about its axis but
otherwise stays in the same position.
There are 8! = 40, 320 ways to arrange the corner cubes. Seven can be oriented independently, and
the orientation of the eighth depends on the preceding seven, giving 37 = 2, 187 possibilities.
There are 12!/2 = 239, 500, 800 ways to arrange the edges, since an even permutation of the corners
implies an even permutation of the edges as well. Eleven edges can be flipped independently, with the
flip of the twelfth depending on the preceding ones, giving 211 = 2, 048 possibilities.
Therefore there are
8! × 37 × (12!/2) × 211 = 43, 252, 003, 274, 489, 856, 000
moves which is approximately 43 quintillion.
Group Theory
Homomorphism
14
♣ The Rubik’s Cube group. This is a group (G, ∗) that corresponds to the set G of all cube
moves on the Rubik’s Cube. The group operation will be defined like this: if M1 and M2 are two
moves, then M1 ∗ M2 is the move where you first do M1 and then do M2 .
• The closure: If M1 and M2 are moves, then M1 ∗ M2 is a move as well.
• Associativity: If M1 , M2 , and M3 are moves, then by moving M1 and M2 first and then M3 , you
obtain the same result if you move M1 , then M2 and M3 after that.
• The identity: If we let e be the “empty” move (that is, a move which does not change the configuration of the Rubik?s cube at all), then M ∗ e means ?first do M , then do nothing.? This is certainly
the same as just doing M ; so M ∗ e = e ∗ M.
• The inverse. If M is a move, we can reverse the steps of the move to get a move M 0 . Then, the
move M ∗ M 0 means “first do M , then reverse all the steps of M .” This is the same as doing nothing,
so M ∗ M 0 = e, so M 0 is the inverse of M .
We can identify each of the six face rotations as elements in the symmetric group on the set of noncenter facets. More concretely, we can label the non-center facets by the numbers 1 through 48, and
then identify the six face rotations as elements of the symmetric group S48 according to how each
move permutes the various facets. The Rubik’s Cube group, G, is then defined to be the subgroup of
S48 generated by the 6-face rotations,
{ F , B , U , D , L , R }, { F 2 , B 2 , U 2 , D2 , L2 , R2 }, and { F −1 , B −1 , U −1 , D−1 , L−1 , R−1 }, where
F, B, U, D, L, R : turning ”Front, Back, Up, Down, Left and Right sides” 90◦ clockwise;
F 2 , B 2 , U 2 , D2 , L2 , R2 : are turning ”Front, Back, Up, Down, Left and Right sides” 180◦ clockwise ;
F −1 , B −1 , U −1 , D−1 , L−1 , R−1 : turning ”Front, Back, Up, Down, Left and Right sides” 90◦ counterclockwise.
Note that if X is the move of a face, then X 4 = X. This means that the order of any element of G is
less that or equal to 4. Also turning a face 270◦ is the same as turning the face 90◦ counter-clockwise.
Therefore one may solve the puzzle in much less than 43 quintillion moves.
♣ Homomorphism. Given two groups G 1 and G 2 with binary operations ∗1 and ∗2 respectively;
a group homomorphism from ( G 1 , ∗1 ) to ( G 2 , ∗2 ); is a mapping ϕ from G 1 to G 2 , such that for all
x1 , y 1 ∈ G 1 ,
ϕ (x1 ∗1 y1 ) = ϕ (x1 ) ∗2 ϕ (y1 ) .
The set of all homomorphisms from G 1 to G 2 is denoted by Hom( G 1 , G 2 ).
♥ Example. Let ( G, · ) be a group and g ∈ G . Define a map ϕ : Z 7→ G by ϕ (n) = g n . Then ϕ is a
group homomorphism, since
ϕ (m + n) = g m+n = g m g n = ϕ (m) ϕ (n) .
Definition 6.
1. A homomorphism from G into itself is called an endomorphism.
2. A injective homomorphism is called an monomorphism.
3. A surjective homomorphism is called an epimorphism.
4. A bijective homomorphism is called an isomorphism.
5. An isomorphism from G into itself is called an automorphism.
Group Theory
Homomorphism
15
The isomorphism is denoted by the symbol ∼
=.
• ϕ maps the identity element e1 of G 1 into the identity element e2 of G 2 .
We have ϕ (x) = ϕ (e1 ∗1 x) = ϕ (e1 ) ∗2 ϕ (x). Thus ϕ (e1 ) = e2 .
• ϕ(x−1 ) = ϕ(x)−1 .
From the facts that x ∗1 x−1 = e1 , and ϕ(e1 ) = ϕ(x ∗1 x−1 ) = ϕ(x) ∗2 ϕ(x−1 ) = e2 , we conclude that
ϕ(x−1 ) = ϕ(x)−1 .
• If G1 is commutative, then G2 is also commutative.
The proof follows from ϕ (x1 ∗1 y1 ) = ϕ (x1 ) ∗2 ϕ (y1 ) and ϕ (y1 ∗1 x1 ) = ϕ (y1 ) ∗2 ϕ (x1 ) and the fact
that x1 ∗1 y1 = y1 ∗1 x1 .
♥ Example. Consider the Cayley tables of Z4 , U[8] , U[10] , and U[12] :
+4
0̄
1̄
2̄
3̄
×8
1
3
5
7
×10
1
3
7
9
×12
1
5
7
11
0̄
1̄
2̄
3̄
0̄
1̄
2̄
3̄
1̄
2̄
3̄
0̄
2̄
3̄
0̄
1̄
3̄
0̄
1̄
2̄
1
3
5
7
1
3
5
7
3
1
7
5
5
7
1
3
7
5
3
1
1
3
7
9
1
3
7
9
3
9
1
7
7
1
9
3
9
7
3
1
1
5
7
11
1
5
7
11
5
1
1
7
7
11
1
5
11
7
3
1
In Z4 : 2̄ +4 2̄ = 0̄ ; and 1̄ and 3̄ are the inverse of each other.
In U[10] : 9 ×10 9 = 1 ; and 3 and 7 are the inverse of each other.
So there are two ways to define isomorphisms between these two groups:
(a.) 0̄ 7→ 1 , 2̄ 7→ 9 , 1̄ 7→ 3 , and 3̄ 7→ 7 .
(b.) 0̄ 7→ 1 , 2̄ 7→ 9 , 1̄ 7→ 7 , and 3̄ 7→ 3 .
In both U[8] and U[12] , each member is the inverse of itself, so they have the same structure, which
means they are isomorphic to each other. There are actually six different isomorphism between these
two groups. Could you find these isomorphisms? Using the fact that e1 7→ e2 .
♣ Kernel and Image. The kernel of a homomorphism ϕ, denoted by Ker(ϕ), is the set of elements
in G 1 which are mapped to the identity e2 ∈ G 2 .
Ker (ϕ) = {x ∈ G 1 : ϕ (x) = e2 } .
The image of ϕ, denoted by Im (ϕ) is to be
Im(ϕ) = {ϕ (x) : x ∈ G 1 } .
♥ Example. Define the function ϕ : ( Z, + ) 7→ ( Zn , +n ) by ϕ (a) = ā, ∀ a ∈ Z. Let a , b ∈ Z . Then
ϕ (a + b) = a + b = ā +n b̄ = ϕ (a) +n ϕ (b).
Note that Ker (ϕ) = n Z, Also the finite group Zn is the image of the infinite group Z under this
epimomorphism.
Theorem 7. Ker (ϕ) < G 1 and Im (ϕ) < G 2 .
Proof. Let x1 , y1 ∈ Ker (ϕ). Then ϕ (x1 ∗1 y1−1 ) = ϕ (x1 ) ∗2 ϕ (y1−1 = e2 ∗2 e2 = e2 . Thus
Ker (ϕ) < G 1 .
Group Theory
Homomorphism
16
∀ x2 , y2 ∈ Im(ϕ), there exist x1 , y1 ∈ G 1 and a z1 ∈ G 1 such that ϕ (x1 ) = x2 , ϕ (y1−1 ) = y2−1 , and
z1 = x1 ∗1 y1−1 . We have :
x2 ∗2 y2−1 = ϕ (x1 ) ∗2 ϕ (y1−1 ) = ϕ (x1 ∗1 y1−1 ) = ϕ (z1 ) .
By Theorem 1, we conclude that Im (ϕ) < G 2 .
The kernel of the epimomorphism ϕ : ( Z, + ) 7→ ( Zn , +n ) by ϕ (a) = ā, for all a ∈ Z is:
Ker (ϕ) = {a ∈ Z : ϕ (a) = 0̄ }
= { a ∈ Z : ā = 0̄ }
= { a ∈ Z : a is divisible by n}
= { a ∈ Z : a = q n for some q ∈ Z}
= { n q : q ∈ Z }.
Theorem 8. Let ϕ be a homomorphism of a group G 1 into a group G 2 . Then ϕ is one-to-one if and
only if Ker (ϕ) = {e1 }.
Proof. Suppose ϕ is one-one. Let a ∈ Ker (ϕ) = {e1 }. Then ϕ (a) = e2 = ϕ (e1 ). This implies that
a must be equal to e1 . Hence Ker (ϕ) = {e1 } .
Conversely, suppose that Ker (ϕ) = {e1 }. Let a , b ∈ G 1 and suppose varphi (a) = ϕ (b). Then
ϕ (a b−1 ) = ϕ (a) ϕ (b−1 ) = ϕ (b) ϕ (b−1 ) = ϕ (b b−1 ) = ϕ (e1 ) = e2 .
Thus, a b−1 ∈ Ker (ϕ) = {e1 } and so a b−1 = e1 i.e., a = b. This proves that f is one-one.
Theorem 9. If H1 < G 1 , then ϕ( H1 ) < G 2 .
Proof. ∀ x2 , y2 ∈ ϕ( H1 ), there exist x1 , y1 ∈ H1 and a z1 ∈ H1 such that ϕ (x1 ) = x2 , ϕ (y1−1 ) =
y2−1 , and z1 = x1 ∗1 y1−1 . We have :
x2 ∗2 y2−1 = ϕ (x1 ) ∗2 ϕ (y1−1 ) = ϕ (x1 ∗1 y1−1 ) = ϕ (z1 ) .
By Theorem 1, we conclude that varphi( H1 ) < G 2 .
If ϕ , ψ ∈ Hom( G 1 , G 2 ), the sum of ϕ and ψ is defined as:
∀ u ∈ G 1,
(ϕ + ψ)(u) = ϕ(u) + ψ(u) .
Theorem 10. ( Hom( G 1 , G 2 ), + ) is a group. Moreover, If G 2 is abelian, then Hom( G 1 , G 2 ) is also
abelian.
Proof. Hom( G 1 , G 2 ) is closed by definition. It is not difficult to show that the operation + is
associative; the map that takes every element of the group G 1 and changes it into e2 is the neutral
element; finally if ϕ ∈ Hom( G 1 , G 2 ), then − ϕ is its inverse. Thus Hom( G 1 , G 2 ) is a group.
If the group G 2 is abelian, then for any ϕ and ψ, we have:
(ϕ + ψ)(u) = ϕ(u) + ψ(u) = ψ(u) + ϕ(u) = (ψ + ϕ)(u)
∀ u ∈ G1 .
Group Theory
Homomorphism
17
Thus (Hom( G 1 , G 2 ), +) also becomes abelian. Note that if G 1 is not abelian, the commutativity of
∗2 in G 2 makes ( Hom( G 1 , G 2 ), +) an abelian group.
Remark 3. We proved previously that the image of an abelian group must be abelian but according to
the above theorem, a non-commutative group may have a commutative image.
♥ Examples.
A cyclic group G of order n is isomorphic to the cyclic group (Zn ,+). The next three examples
illustrate this fact.
1. Consider the group (Z3 , + ) and the group of permutations G = { ( 1 ) , ( 1 , 2 , 3 ) , ( 1 , 3 , 2 ) } .
The mapping ϕ : Z3 7→ G is an isomorphism, where
ϕ (0̄) = ( 1 ),
ϕ (1̄) = ( 1 , 2 , 3 ),
and ϕ (2̄) = ( 1 , 3 , 2 ) .
2. Consider the groups (Z4 , + ) and < i >= { 1 , i , −1 , −i } . The mapping ϕ : Z4 7→ < i > is an
isomorphism, where
ϕ (0̄) = 1,
ϕ (1̄) = i,
ϕ (2̄) = −1,
and ϕ (3̄) = −i .
3. Consider the groups ( Z6 , + ), where
< 0̂ >= {0̂}, < 1̂ >=< 5̂ >= Z6 , < 2̂ > = < 4̂ >= {2̂ , 4̂ , 0̂}, and < 3̂ >= {3̂, 0̂};
and (Z∗7 , .), where
< 1 > = {1}, < 2 > = < 4 > = {2 , 4 , 1}, < 3 > = < 5 > = Z∗7 , and < 6 > = {6 , 1}.
The mapping
ϕ : Z6 −→ Z∗7
with
ϕ(0̂) = 1,
ϕ(1̂) = 3,
ϕ(2̂) = 2,
ϕ(3̂) = 6,
ϕ(4̂) = 4,
and ϕ(5̂) = 5
is an isomorphism.
4. Consider the abelian groups (Z8 , +) and (U[30], ×30 ) , where
Z8 = {0̂ , 1̂ , 2̂, 3̂ , 4̂ , 5̂ , 6̂ , 7̂} and U[30] = {1 , 7 , 11 , 13 , 17 , 19 , 23 , 29}; with
< 0̂ > = {0̂}, < 2̂ > = < 6̂ > = {2̂ , 4̂ , 6̂ , 0̂}, < 4̂ > = {4̂ , 0̂}, < 1̂ > = < 3̂ > = < 5̂ > = < 7̂ > = Z8
< 1 > = {1}, < 7 > = < 13 > = {7 , 13 , 19, 1}, < 11 > = {11 , 1}, < 17 > = < 23 > = {17 , 19 , 23 , 1},
< 19 > = {19 , 1}, < 29 > = {29 , 1}.
Notice that Z8 is a cyclic group but U[30] is not. Also, 4̂ is the only member of Z8 which is self inverse,
−1
−1
−1
but in U[30], we have 11 = 11, 19 = 19 and 29 = 29 . Although both groups have the same
order, but based on the above facts, there is no way one could define an isomorphism between these
two groups.
The only homomorphism from Z8 to U[30] (resp. U[30] to Z8 ) is the one that maps all members of Z8
to 1 (resp. U[30] to 0̂).
Group Theory
Homomorphism
18
5. Consider the abelian group U[15] = 1, 2, 4, 7, 8, 11, 13, 14 with
1
−1
= 1, 2
−1
= 8, 4
−1
= 4, 7
−1
= 13, 11
−1
= 11, 13
−1
= 7, and 14
−1
= 14 ;
and the mapping ϕ : U[15] −→ U[15] with ϕ(x) = x3. We have
ϕ(1) = 1, ϕ(2) = 8, ϕ(4) = 4, ϕ(7) = 13, ϕ(8) = 2, ϕ(11) = 11, ϕ(13) = 7, and ϕ(14) = 14 .
ϕ is bijective with ϕ(1) = 1, and ∀ x, y ∈ U[15]:
ϕ(x)−1 = ϕ(x−1 ) and ϕ(x y) = (x y)3 = x3 y 3 = ϕ(x) ϕ(y) .
Hence ϕ is an automorphism. Also note that ϕ2 (x) = ϕ ◦ ϕ(x) = x, ∀ x ∈ U[15].
Now define the mapping ψ : U[15] 7−→ U[15] with ψ(x) = x2. We have
ψ(1) = 1, ψ(2) = 4, ψ(4) = 1, ψ(7) = 4, ψ(8) = 4, ψ(11) = 1, ψ(13) = 4, and ψ(14) = 1 .
Note that ∀ x, y ∈ U[15]:
ψ(x)−1 = ψ(x−1 ) and ψ(x y) = (x y)2 = x2 y 2 = ψ(x) ψ(y) ;
and Ker(ψ) = 1, 4, 11, 14 . Thus ψ is just an endomorphism.
6. Let G be a group. For any g ∈ G, define the function σg : G 7→ G by σg (a) = g a, where a ∈ G.
To show that σg is one-to-one, suppose that σg (a) = σg (b). Then we have
g a = σg (a) = σg (b) = g b
⇒ a = b.
To show that σg is onto, for each b ∈ G, we choose a = b g −1 ∈ G. Then σg (a) = g (g −1 b) = b. Hence
σg is onto. The isomorphism σg (a) = g a is known as the left regular representation of G.
Remark 4. Note that the isomorphism σg permutes the members of the group G, therefore it is a
permutation group. We shall prove that every group is isomorphic to a permutation group by assigning
to every g in a group, its left representation σg .
Lemma 7. The set Gb = {σg : g ∈ G} is a group under composition of functions.
Proof. We have closure under composition of functions since
σg ◦ σh (a) = σg (h a) = g h a = σgh .
It is well-known that the composition of functions is associative.
b
Also σe (a) = e a = a , which makes σe the neutral element of G.
−1
Finally ( σg−1 (a) σg (a) ) = g (g a) = a = σe (a).
Lemma 8. The mapping φ : G 7→ Gb with a 7→ σg (a) is a group isomorphism.
Proof. The group operation is preserved since ∀ g , h ∈ G, we have
φ (g h) = σg h = σg σh = φ (g) φ (h) .
Group Theory
Homomorphism
19
φ is one-to-one, because if φ (g) = φ (h), then for any a ∈ G,
g a = σg (a) = φ (g) (a) = φ (h) (a) = σh (a) = h a .
Hence g = h .
φ is onto follows from the facts that σg = φ (g) and σg ∈ Gb is define by using g ∈ G.
The English mathematician, Arthur Cayley proved that every group is isomorphic to a permutation
group. The proof follows from combining lemmas 8 and 9.
♦ Basic Properties.
1. Given an homomorphism from G1 into G2 , If G1 is abelian, then G2 must be abelian. But nonabelian groups may have abelian homomorphic images.
2. If ϕ is an isomorphism, then ϕ−1 is also an isomorphism.
3. If ϕ is a homomorphism, then ϕ (g n ) = [ϕ (g)], for all n ∈ Z.
4. The mapping ϕ : G1 7→ G2 defined by ϕ(a−1 ) = (ϕ(a))−1 , for every a ∈ G1 is an isomorphism.
Exercise.
1. Prove that the groups (U[7], ×7 ) and (U[9], ×9 ). are cyclic and isomorphic.
2. Find m > 7 and n > 9 such that (U[m], ×m ) and (U[n], ×n ) are cyclic and isomorphic.
3. Find m and n (m 6= n) such that (U[m], ×m ) ∼
= (U[n], ×n ) ∼
= ( Z4 , +4 ) .
4. Find an isomorphism ϕ : ( R, + ) 7→ ( R+ , × ) .
Solution. Let ϕ (x) = 2 x . Clearly ϕ is a function from R into R+ .
To prove that ϕ is one-to-one, suppose that 2 x = 2 y . Then log2 x2 = log2 y 2 , and therefore x = y.
To show that ϕ is onto, we must find some real number x such that ϕ (x) = 2x = y. Taking log2 of
both side of the equation 2x = y, we obtain x = log2 y.
Finally,
ϕ (x + y) = 2 x+y = 2 x 2 y = ϕ (x) ϕ (y) .
5. Find an isomorphism ϕ : ( R+ , × ) 7→ ( R, + ) .

0 , if σ is even
6. Prove that the mapping ϕ : Sn 7→ Z2 defined by ϕ(σ) =
is an epimorphism.
1 , if σ is odd
7. Is the function ϕ : ( R, + ) 7→ ( R, + ) defined by ϕ (a) = a2 , a homomorphism?
8. Is the function φ : ( R∗ , × ) 7→ ( R∗ , , × ) defined by φ (a) = a2 , a homomorphism? If yes, then find
Ker (φ).
9. Is the function φ : ( R∗ , × ) 7→ ( R∗ , × ) defined by φ (a) = | a |, a homomorphism? If yes, then find
Ker (φ).
10. Prove that there is no isomorphism from ( Q, + ) onto ( Q∗ , × )
Solution. If φ were such a mapping, there would be a rational number a with φ (a) = −1 . But then
−1 = φ (a) = φ
1
1
a+ a
2
2
=φ
2
1
1
1
a ×φ
a = φ
a
.
2
2
2
However, the square of a rational number can’t be −1 .
11. Let (G, · ) be the group of 2 × 2 real matrices with determinant 1 and let M be a fixed real matrix
Group Theory
Homomorphism
20
in G. Prove that ψ : G 7→ G, defined by ψ (A) = M A M −1 , for all A ∈ G is an automorphism.
Solution. We have ψ (A B) = M A B M −1 = M A M −1 M B M −1 = M A M −1 = ψ (A) ψ (B) .
If ψ (A) = ψ (B) , then M A M −1 = M B M −1 . By multiplying the left side by M −1 and the right
side by M , we conclude that A = B. Thus ψ is one-to-one.
To show that ψ is onto, for any B ∈ G, we choose A to be M −1 B M , then
ψ (A) = M A M −1 = M M −1 B M M −1 = B.
12. Show that the function ϕ : ( R + ) 7→ ( C, · ), defined by ϕ (θ) = cos θ + i sin θ is an epimorphism.
♣ Normal Subgroups. A subgroup n of a group G, is called a normal subgroup and denoted by
N C G , if
∀ g ∈ G and ∀ n ∈ N, g n g −1 ∈ N .
Normal subgroups are also known as invariant subgroups or self-conjugate subgroups.
For any subgroup, the following conditions are equivalent to normality. Therefore any one of them
may be taken as the definition:
∀ g ∈ G, g N g −1 ⊆ N.
(1)
∀ g ∈ G, g N g −1 = N.
(2)
∀ g ∈ G, g N = N g.
(3)
Clearly, any subgroup of an abelian group is normal. The trivial subgroup {e} is a normal subgroup.
The center of a group is also a normal subgroup.
Évariste Galois
4
was the first to realize the importance of the existence of normal subgroups.
♥ Example. The group SL2 [R] of 2 × 2 real matrices with determinant 1 is a normal subgroup of
GL2 [R] of 2 × 2 real matrices with non-zero determinant.
To verify this, we choose a matrix GL2 [R] and a matrix A ∈ SL2 [R] and note that
det M A M −1 = det(M ) det(A) det M −1 = det(M ) det M −1 det(A) = 1 det(A) = 1.
So M A M −1 ∈ SL2 [R].
4
Although Évariste Galois Galois (1811-1832) had mastered the works of Legendre and Lagrange at age 15, but he
failed twice at his entrance examination to l’ Ecole Polytechnique. He did not know some basic mathematics, and he did
mathematics almost entirely in his head, to the annoyance of the examiner.
While still in his teens, he was able to determine a necessary and sufficient condition for a polynomial to be solvable by
radicals, thereby solving a problem standing for 350 years. His work laid the foundations for Galois theory and group
theory.
At 18, Galois wrote his important research on the theory of equations and submitted it to the French Academy of Sciences
for publication. The paper was given to Cauchy for refereeing. Cauchy, impressed by the paper, agreed to present it to
the academy, but he never did.
At the age of 19, Galois sumitted a paper of the highest quality in the competition for the Grand Prize in Mathematics,
given by the French Academy of Sciences. The paper was given to Fourier, who died shortly thereafter. Galois’s paper
was never seen again.
Galois at seventeen was making discoveries of epochal significance in the theory of equations, discoveries whose consequences are not yet exhausted after for two centuries.
Galois spent most of the last year and a half of his life in prison for revolutionary political offenses. While in prison, he
attempted suicide and prophesied that he would die in a duel. On May 30, 1832, Galois was shot in a duel and died the
next day at the age of 20.
Galois’s entire collected works fill only 60 pages.
Group Theory
Normal Groups
21
Theorem 11. If ϕ ∈ Hom( G 1 , G 2 ), then Ker(ϕ) / G 1 .
Proof. For all g ∈ G 1 and n ∈ Ker(ϕ), we have:
ϕ(g n g −1 ) = ϕ(g)ϕ(n)ϕ(g −1 ) = ϕ(g) e2 ϕ(g −1 ) = ϕ(g)ϕ(g −1 ) = ϕ(g g −1 ) = ϕ(e1 ) = e2 .
Thus g n g −1 ∈ Ker(ϕ) .
Remark 5. Let ϕ be an endomorphism on S3 with ϕ(σ) = σ 2 .
(
Im(ϕ) =
σ0 =
!
1 2 3
, σ1 =
1 2 3
!
1 2 3
Now let α =
, then α ◦ σ1 =
2 1 3
is not a normal subgroup of S3 .
!
1 2 3
6=
3 2 1
Then
!
1 2 3
, σ2 =
3 1 2
1 2 3
2 3 1
!)
.
!
1 2 3
= σ1 ◦ α; so we conclude that Im(ϕ)
1 3 2
Theorem 12. If H and K are normal subgroups of G with H ∩ K = {e} , then every element of H
commutes with every element of K
Proof. Let h ∈ H and k ∈ K, then let u = (h−1 k −1 )(h k) = h−1 (k −1 h k) = (h−1 k −1 h)k. Since H is
normal, k −1 h k ∈ H, hence u = h−1 (k −1 h k) ∈ H. Similarly, we use the facts that K is normal and
k ∈ K, to show that u = (h−1 k −1 h)k ∈ K; thus u = e. According to Lemma 1, we have h k = k h .
Remark 6. 1. The image of a normal subgroup is a normal subgroup, provided that the group map
is surjective.
2. A subgroup H of a group G occurs as the kernel of a group homomorphism if and only if,H it is a
normal subgroup of G.
3. Unlike cyclic groups which are abelian; permutation groups are not. Also most subgroups of permutation groups are not normal.
4. If H is a normal subgroup of G and K is a normal subgroup of H, it does not necessarily imply that
K is a normal subgroup of G.
5. In the Rubik’s Cube group, the subgroup consisting of operations which only affect the corner pieces
is normal. By contrast, the subgroup consisting of turns of the top face only is not normal.
Exercise.
1. Is H = {( 1 ) , ( 1 , 3 ) } a normal subgroup of S3 ?
2. Is H = {( 1 ) , ( 1 , 3 ) ( 3 , 4 ) } a normal subgroup of A4 ?
(
!
)
a b
3. Let H =
: a c 6= 0 . Is H a normal subgroup of GL2 [R]?
0 c
4. Let N be a normal subgroup of G. Prove that if a b ∈ N , then b a ∈ N .
5. Let H be a subgroup of G and suppose that if a b ∈ H, then b a ∈ H. Prove that H is a normal
subgroup of G .
6. Prove that the intersection of two normal subgroups of a group is also a normal subgroup.
7. Let ϕ : G1 7→ G2 be an epimorphism and N C G . Prove that ϕ (N ) C G2 .
Solution. We know that ϕ (N ) < G2 . Let y ∈ ϕ (N ) and g2 ∈ G2 , then since ϕ is an epimormism, we
Group Theory
Quaternion Group
22
have y = ϕ (x) and g2 = ϕ (g1 ), for some x ∈ N and g1 ∈ G1 .
The fact that N C G1 implies that g1 x g1−1 ∈ N . But
ϕ g1 x g1−1 = ϕ (g1 ) ϕ (x) ϕ (g1−1 ) = g2 y g2−1 ∈ ϕ(N ).
Hence ϕ (N ) C G2 .
8. Find a non-normal subgroup of S3 .
♣ Quaternion Group.5 The set Q8 = {±1, ±i, ±j, ±k}, where
i2 = j 2 = k 2 = − 1,
i j = k,
j k = i,
k i = j,
j i = − k,
k j = − i,
i k = − j.
is called the Quaternion group. The group is a non-abelian of order eight; and is one of the two
non-abelian groups of the five total finite groups of order 8.
Since i2 = j 2 = k 2 = −1, the elements i, j and k are known as the imaginary units, by analogy with
i ∈ C. Any pair of the imaginary units generate the group. Thus
Q8 = < i , j > = < i , k > = < j , k > .
Note that i, j, and k all have order four in Q8 and any two of them generate the entire group. Another
presentation of Q8 demonstrating this is :
x , y | x 4 = 1 , x 2 = y 2 , y −1 x y = x −1 .
One may take, for instance, i = x, j = y and k = x y .
Remark 7. A Dedekind group is a group G such that every subgroup of G is normal. All abelian
groups are Dedekind groups. A non-abelian Dedekind group is called a Hamiltonian group. The most
familiar (and smallest) example of a Hamiltonian group is the quaternion group Q8 .
5
On October 16, 1843, the above equations occurred at once to the Irish astronomer, mathematician, and physicist,
Sir William Rowan Hamilton (1805-1885) as he was crossing the Brougham Bridge in Dublin. He carved them into the
stone of the bridge (the original carving is gone but a plaque celebrates this famous act of “mathematical vandalism”).
On the following day, Hamilton wrote a letter to his friend and fellow mathematician, John T. Graves, describing the
train of thought that led to his discovery.
And here there dawned on me the notion that we must admit, in some sense, a fourth dimension of space
for the purpose of calculating with triples ...
Group Theory
Quaternion Group
23
Q8 = { ± i , ± j , ± k , ± 1 }
I = { i , −1 , −i , 1 }
J = { j , −1 , −j , 1 } K = { k , −1 , −k , 1 }
{ −1 , 1 }
{1}
Figure 1: Laeeice of Subgroups of Q8
Notice that all the proper subgroups of the non-abelian group Q8 are cyclic.
Here is the table for the Quaternion group and how Red i and Blue j generating Q8 :
Q8
1
i
j
k
−1
−i
−j −k
1
1
i
j
k −1
−i
−j −k
i
i
−1
j
j
−k
k
k
j
k −j
−i
1
−k
j
i −j
k
1
−i
−i −1 −k −j
i
1
−1
−1 −1 −i
−j
−k
1
i
j
k
−i
−i
1
−k
j
−i
−1
k
−j
−j
−j
k
−1
−i
j
−k −k −j
−i
1 −k
−k −1 −i
j
−i −1
The Quaternion group can be represented as a subgroup of the general linear group GL2 (C).
A representation of Q8 is given by:
b
1 7→
!
1 0 b
, i→
7
0 1
!
i 0
, b
j→
7
0 −i
!
0 1
, b
k→
7
−1 0
!
0 i
.
i 0
Thus Q8 is isomorphic to the following group :
(
!
!
!
!
!
!
!
!)
1 0
i 0
0 1
0 i
−1 0
−i 0
0 −1
0 −i
,
,
,
,
,
,
,
.
0 1
0 −i
−1 0
i 0
0 −1
0 i
1 0
−i 0
♣ Biquaternions. are a number system that extends the complex numbers. A feature of biquaternions is that multiplication of two biquaternions is non-commutative.
Biquaternions find uses in both theoretical and applied mathematics, in particular for calculations
involving three-dimensional rotations such as in three-dimensional computer graphics. The unit quasi-
Group Theory
Quaternion Group
24
sphere of the biquaternions provides a presentation of the Lorentz group, which is the foundation of
special relativity.
A biquaternion is in the form of
q = a + b i + c j + d k,
where a, b, c, d are real numbers and i , j , k are elements of the Quaternion group.
To distinguish the square root of minus one in the biquaternions, Hamilton used the convention of
representing the square root of minus one in the scalar field C by h , since there is an i in the
quaternion group. Then
h i = i h, h j = j h, and h k = k h .
Biquaternions are added component-wise and multiplied using the “foil method”. For addition
(a1 + b1 i + c1 j + d1 k) + (a2 + b2 i + c2 j + d2 k) = (a1 + a2 ) + (b1 + b2 ) i + (c1 + c2 ) j + (d1 + d2 ) k
. To do the multiplication, you expand out the product
(a1 + b1 i + c1 j + d1 k) × (a2 + b2 i + c2 j + d2 k)
as you would for a polynomial and then simplify all the terms involving i j, i k, etc., using the rules
for the products of i j and k. For instance
(3 i + j) (7 j + 2 k) = 21 i j + 6 i k + 7 j 2 + 2 j k = 21 k − 6 j − 7 + 2 i = −7 + 2 i − 6 j + 21 k.
Given a biquaternion q = a + b i + c j + d k, its conjugate is q̄ = a − b i − c j − d k.
√
We have q q̄ = a2 + b2 + c2 + d2 and its magnitude or norm is defined as | q | = q q̄. q is called a unit
biquaternion if | q | = 1. In other words q q̄ = q̄ q = 1 or q̄ = 1/q.
Consider the products
( a1 + b1 i + c1 j + d1 k ) ( a2 + b2 i + c2 j + d2 k ) and ( 7 + 2 i − 3 j + 4 k ) ( 4 + 5 i + 2 j + 3 k ).
One may use the following method to obtain the result :
step
step
step
step
Thus
and
1
2
3
4
a2
a1
b2
b1
c2
c1
d2
d1
− d1 d21
− c1 c22
− b1 b23
a1 a24
d1 c22
c1 d21
b1 a24
a1 b23
d1 b23
c1 a24
− b1 d21
a1 c22
d1 a24
− c1 b13
b1 c22
a1 d21
S1
S2
S3
S4
step
step
step
step
1
2
3
4
4
7
5
2
2
−3
− 12 1
62
− 10 3
28 4
−8 2
−9 1
84
35 3
20 3
− 12 4
−6 1
14 2
12
26
16
( a1 + b1 i + c1 j + d1 k ) ( a2 + b2 i + c2 j + d2 k ) = S1 + S2 i + S3 j + S4 k
( 7 + 2 i − 3 j + 4 k ) ( 4 + 5 i + 2 j + 3 k ) = 12 + 26 i + 16 j + 56 k .
† The powers indicate the orders of the products.
3
4
16 4
15 3
42
21 1
56
Group Theory
Dihedral Groups
25
The biquaternion u = a + b bi + c b
j + db
k may be represented by the matrix
U=
!
a + bi c + di
.
−c + di a − bi
To find the product of two biquaternions, we may use the product of matrices and take advantage
of the fact that the (2, 2) entry is the conjugate of the (1, 1) entry and the (2, 1) entry is minus the
conjugate of (1, 2) entry.
Remark. One could construct a four-dimensional vector space with basis { 1 , i , j , k }. This vector
space is used in computer graphics.
Exercise.
1. Prove that all proper subgroups of Q8 are normal.
2. Find the subgroup of GL2 (C) generated by
!
i 0
and
0 −i
!
0 1
.
−1 0
3. Is Q8 isomorphic to U [ 30 ] ? Explain.
4. Let p = 2 + i − 3 j + 4 k and q = 3 + 4 i − j + 2 k . Then
a. find p q and q p without using matrix multiplication ;
b. find p q and q p using matrix multiplication .
♣ Dihedral Groups. A Dihedral group is the group of symmetries of a regular polygon, including
both rotations and reflections. In geometry the group is denoted by Dn , while in algebra the same
group is denoted by D2n , to indicate its order. Dihedral groups are non-abelian generated by two
elements, R and L, where R represents the rotation and L , the reflection, and I , the identity.
Dn = h R, L | R n = I, L 2 = I, and L R L = R −1 i .
A Video Explaining Dihedral Groups
The dihedral group D4 is a non-abelian group. It is sometimes called the octic group. It is generated
as follows :
D4 = h R , L | R 4 = L 2 = I , L R L = R −1 i .
Geometric Definition. The elements of D4 may be represented by the following eight squares:
Rotation :
identity
1··· · 2
..
..
.
.
4··· · 3
R0 = I = (1)
reflect 45 ◦
Ref lection :
3··· · 2
..
.
. & ..
4··· · 1
L0 = (1, 3)
rotate 90 ◦
2←3
↓
↑
1 7→ 4
R1 = (1, 4, 3, 2)
reflect 90 ◦
2 ↔ 1
..
..
.
.
3 ↔ 4
L1 = (1, 2)(3, 4)
rotate 180 ◦
3←4
↓
↑
2 7→ 1
R2 = (1, 3)(2.4)
reflect 135 ◦
1··· · 4
..
.
. . ..
2··· · 4
L2 = (2, 4)
rotate 270 ◦
4←1
↓
↑
3 7→ 2
R3 = (1, 2, 3, 4)
reflect 180 ◦
4··· · 3
↓
↑
1··· · 2
L3 = (1, 4)(2, 3)
Group Theory
Dihedral Groups
26
Matrix Representation. If we center the regular polygon at the origin, then elements of the Dihedral
group act as linear transformations of the plane. This lets us represent elements of Dn as matrices,
with composition being matrix multiplication. The matrices for elements of Dn are:
ck =
R
cos(2kπ/n) − sin(2kπ/n)
sin(2kπ/n) cos(2kπ/n)
!
ck =
and L
!
cos(2kπ/n) sin(2kπ/n)
,
sin(2kπ/n) − cos(2kπ/n)
ck is a rotation matrix, expressing a counterclockwise rotation through an angle of 2π k/n and
where R
ck is a reflection across a line that makes an angle of π k/n with the x-axis.
L
For example, the elements of the group D4 can be represented by the following eight matrices:
!
1 0
c1 =
, R
0 1
c0 = I =
R
c0 =
L
!
1 0
c1 =
, L
0 −1
!
0 −1
c2
, R
1 0
!
0 1
c2
, L
1 0
!
−1 0
c3
, R
0 −1
!
−1 0
c3
, L
0 1
!
0 1
,
−1 0
!
0 −1
.
−1 0
The Cayley table for the dihedral group D4
Definition as a Permutation Group. The elements of the group D4 are :
(
D4 =
ρ 0 = (1), ρ 1 = (1, 4, 3, 2), ρ 2 = (1, 3)(2, 4), ρ 3 = (1, 2, 3, 4),
σ 0 = (1, 3), σ 1 = (1, 2) (3, 4), σ 2 = (2, 4), σ 3 = (1, 4) (2, 3).
)
⊂ S4
Theorem 13. The number of proper subgroups of a Dihedral group Dn is : τ (n) + σ(n) − 2 .
According to this theorem, D4 has τ (4) + σ (4) − 2 = 3 + 7 − 2 = 8 proper subgroups. They are :
Order two :
N1 = { R 0 , R 2 } , H 2 = { R 0 , L 0 } , H 3 = { R 0 , L 1 } , H 4 = { R 0 , L 2 } , H 5 =
R0 , L3 .
Order four :
N6 = { R 0 , R 0 , R 2 , R 3 } , N7 = { R 0 , R 2 , L 0 , L 2 } , N8 = { R 0 , R 2 , L 1 , L 3 } .
• The center of D4 is N1 .
Group Theory
Cosets
27
• All subgroups of order two are cyclic and isomorphic to Z2 .
• N6 is cyclic and isomorphic to Z4 .
• N1 and all subgroups of order four are normal .
The dihedral group D4 and quaternion group are the only two non-abelian groups of the five groups
total of order eight.
Remark 8. the quaternion group Q8 is generated as follows :
h x , y | x 4 = 1 , x 2 = y 2 , y −1 x y = x −1 i
and the dihedral group D4 is generated as follows :
h x , y | x 4 = 1 , y 2 = 1 , y −1 x y = x −1 i.
Although they have similar definitions, but they are very different. For example all proper subgroups
of Q8 are normal and cyclic, but among eight proper subgroups of D4 , only four are normal and only
six are cyclic.
The following pictures show the effect of the sixteen elements of D8 on a stop sign:
The effect of eight rotations :
The effect of eight reflections :
Notice that a rotation followed by a reflection doesn’t produce the same picture as if the same reflection
is followed by the same rotation. This proves that the group is not abelian.
Exercise.
1. Prove that N7 is a subgroup of D4 .
2. Prove that N8 is a subgroup of D4 .
3. Prove that N7 is normal .
4. Use permutations to prove that N8 is a normal subgroup of D4 .
5. Draw the lattice of subgroups for D4 .
♣ Cosets. The notion of a coset which is a powerful tool for analyzing a group was introduced by
Galois in 1830.
For a subgroup H of a group G and an element g of G , define
g H = {g h : h ∈ H} and Hg = {h g : h ∈ H} .
The subset g H is called a left coset of H and H g is called a right coset of H . It is not difficult to
Group Theory
Cosets
28
show that any two left cosets (resp. right cosets) of H have the same cardinality, and in particular,
every coset (right or left) of H has the same cardinality as H.
Although derived from a subgroup, cosets are not usually themselves subgroups of G , only subsets.
The number of cosets associated with a subgroup H is called the index of H and denoted by [ G : H ] .
For abelian groups, left cosets and right cosets are always the same. Only when H is normal will the
set of right cosets and the set of left cosets of H coincide, which is one definition of normality of a
subgroup.
If H is not normal in G , then its left cosets are different from its right cosets. That is, there is an
a ∈ G such that no element b ∈ G satisfies a H = H b . This means that the partition of G into the
left cosets of H is a different partition than the partition of G into the right cosets of H . (Some
cosets may coincide. For example, if a is in the center Z( G ) , then a H = H a .) Also the center
Z( G ) is normal, since its left cosets are equal to its right cosets.
On the other hand, the subgroup N is normal if and only if g N = N g for all g ∈ G . In this case,
the set of all cosets form a group called the quotient group G / N with the operation ∗ defined by
( a N ∗ b N = a b N ) . Since every right coset is a left coset, there is no need to distinguish ”left
cosets” from ”right cosets”. Given a group G and H < G , the relation
∼L on G defined by a ∼L b , if and only if b−1 a ∈ H is called left congruence modulo H;
similarly the relation
∼R on G defined by a ∼R b , if and only if a b−1 ∈ H is called Right congruence modulo H .
Clearly these two relations coincide if G is abelian.
The equivalence classes of these two left and right equivalence relations are exactly the left cosets or
right cosets of H . Thus the left cosets (resp. right cosets) of H form a partition of G .
Lemma 9. Let H be a subgroup of G and let a , b be two members of G , such that a ∈
/ b H . Then
a H and b H are disjoint (resp. H a and H b are disjoint).
Proof. Suppose a H and b H are not disjoint. Then there exist h1 , h2 ∈ H such that a h1 = b h2 .
−1
By multiplying both sides of the equality by h−1
1 , we obtain a = b h2 h1 . This implies that a ∈ b H
which is a contradiction.
Similarly we may show for the right cosets
Lemma 10. Let H be a subgroup of G . Then all left and right cosets of H have the same order
(number of elements, or cardinality in the case of an infinite H ), equal to the order of H .
Proof. H is itself a coset since e H = H = H e .
For the left cosets, define the map ϕ : a H 7→ b H by setting ϕ (x) = b a−1 x . This map is bijective
because its inverse is given by ϕ−1 (y) = a b−1 y . Hence all left cosets have the same order.
Similarly we can show that the map ψ : H a 7→ H b by setting ψ (x) = x a b−1 is a bijection which
implies that right cosets have the same order.
Theorem 14. (Lagrange)
6
If G is a finite group, then
| G| = | H | × [ G : H ] .
6
In character, the Italian-born French mathematician and astronomer, Joseph-Louis Lagrange (1736-18130) was nervous and timid, he detested controversy, and to avoid it willingly allowed others to take the credit for what he had himself
Group Theory
Cosets
29
Proof. By Lemma 10, the order of each coset of H is precisely | H |. By Lemma 9, cosets are either
equal or disjoint. Thus the number of cosets of G times the order of H must be equal to | G | .
A immediate consequence of the theorem is that the order of any element a of a finite group divides
the order of that group, since the order of a is equal to the order of the cyclic subgroup generated by
a. If the index of a subgroup H is 2, then it must be normal, because the left coset g H equal the
right cost H g.
Remark 9. Lagrange’s theorem raises the converse question as to whether every divisor of the order
of a group is the order of some subgroup. This does not hold in general: given a finite group G and a
divisor d of | G | , there does not necessarily exist a subgroup of G with order d . The smallest example
is the alternating group G = A4 , which has 12 elements but no subgroup of order 6 .
Other consequences of the Lagrange Theorem are the following corollaries :
Corollary 6. Let G be a finite group, and let a ∈ G . Then a | G | = e .
Proof. Let a ∈ G , then | G | = | a | k , for some positive integer k . Thus
k
a|G | = a|a|k = a|a|
= ek = e.
Corollary 7. A finite group of prime order is cyclic.
Proof. . Let a 6= e be an element of G , then the order of the subgroup h a i must divide divide the
order of the group. Since the order is prime, h a i = G .
Corollary 8. All proper subgroups of the Quaternion group Q8 are normal.
The proof follows from the fact that the subgroup { 1 , −1 } is the center of Q8 and other proper
8
subgroups have order 4 and index = 2 .
4
For the same reason, the center N1 = { R0 , R2 } and all three subgroups of order four of the dihedral
group D4 are also normal.
The Sudoku puzzles are 9 × 9 matrices divided into nine 3 × 3 sub-matrices. Digits 1 through 9 appear
in some of the entries. Other entries are blank. The goal is to fill the blank entries with the digits 1
through 9 in such a way that each digit appears exactly once in each row and in each column, and in
each block.
A question naturally leaps to mind: “When and how can a Cayley table be arranged in such a way as
to partition the group into cosets.”
The following modified Cayley table of the abelian Z9 shows how the group is partitioned by the
subgroup H = { 0 , 3 , 6 } . There are three different cosets, colored Red, Green, and Blue.
done.
He always thought out the subject of his papers before he began to compose them, and wrote them straight off without a
single erasure or correction.
Lagrange is one of the founders of the calculus of variations and laid some of the foundations of group theory, anticipating
Galois. In number theory, he proved that an integer is either a square or the sum of two, three, or four squares. He also
showed that if n is a prime, then (n − 1)! + 1 is a multiple of n.
Group Theory
Cosets
30
Z9
0
3
6
1
4
7
2
5
8
0
3
6
1
4
7
2
5
8
0
3
0
1
4
7
2
2
5
3
0
3
4
7
1
5
5
2
6
4
2
7
1
4
8
3
0
1
4
7
2
5
8
3
6
0
4
7
1
5
8
2
6
0
3
7
1
4
8
2
5
0
3
6
2
5
8
3
6
0
4
7
1
5
8
2
6
0
3
7
1
4
8
2
5
0
3
6
1
4
7
Fermat’s Little Theorem. Let p be a prime number not dividing the integer a , then
a p−1 ≡ 1
(mod p)
⇐⇒
a p ≡ a (mod p) .
Note. If p divides a , then clearly p divides a p , so both a p and a are divisible by p . This means
that a p ≡ a (mod p) . For showing a p−1 ≡ 1 (mod p) we must assume that p does not divide a .
There are several proofs using different techniques to prove this theorem.
Proof 1 (Induction). For a = 1 , obviously 1 p ≡ 1 (mod p) is true.
Suppose p divides a p − a . Then examine a p+1 − (a + 1) .
From the binomial theorem,
p
p
p
p−1
p−2
(a + 1) = a +
a
+
a
+ ··· +
a + 1.
1
2
p−1
p
p
Rewriting,
p
p
p
p−1
p−2
(a + 1) − a − 1 =
a
+
a
+ ··· +
a.
1
2
p−1
p
p
But p divides the right side, so it also divides the left side. Combining with the induction hypothesis
gives that p divides the sum
[ ( a + 1 ) p − a p − 1 ] + (a p − a) = (a + 1) p − (a + 1),
as assumed, so the hypothesis is true for any a .
Proof 2 (Using Coset). By the division algorithm, a = p m + r , where 0 ≤ r < p . Thus, a ≡ r
(mod p) , or a p ≡ r p (mod p) . Hence it suffices to show that r p ≡ r (mod p) .
If r = 0 , the result is trivial, so we may assume that r ∈ U [ p ] . The proof then follows from Corollary
6 and the fact that the order of U [ p ] is p − 1 .
Proof 3. Let S = { 1 , 2 , 3 , · · · , p − 1 } . Then define
Sb = { 1 a , 2 a , · · · , (p − 1) a } (mod p) .
Clearly none of the i a for 1 ≤ i ≤ p − 1 are divisible by p . Suppose that a i ≡ a j (mod p) for i 6= j .
Since gcd (a, p) = 1 , by the cancellation rule, that reduces to i ≡ j (mod p) , which is a contradiction.
Thus all of the elements in Sb are distinct. So they must be congruent to 1 , 2 , 3 , · · · , p − 1 in some
Group Theory
Cosets
31
order. Multiply all these congruences together, we find
1 a · 2 a · 3 a · · · · · (p − 1) a ≡ 1 · 2 · · · (p − 1)
(mod p) .
By dividing both sides by (p − 1) !, we are left with the statement a p−1 ≡ 1 (mod p).
Remark 10. The Chinese hypothesis is a disproven conjecture stating that an integer p is prime if
and only if it satisfies the condition that 2 p ≡ 2 (mod p) is divisible by p . It is true that if p is
prime, then 2 p ≡ 2 (mod p) (it is derived from Fermat’s Little Theorem.by multiplying both sides of
the congruence by 2 ) . However, the converse is false, and therefore the hypothesis as a whole is false.
The smallest counter example is p = 341 = 11 × 31 .
Corollary 9. The order of U [ n ] is even if n > 2 .
Proof. We have (n − 1)2 = n2 − 2 n + 1 ≡ 1 mod n . Thus the order of the subgroup generated by
n − 1 is 2 . By Lagrange Theorem, the order of U [ n ] is divisible by 2 , therefore even.
♥ Examples.
1. Calculate 2 345 (mod 11), using Fermat’s Little Theorem .
Solution. The number 2 is not divisible by the prime 1 1, so 2 10 ≡ 1 (mod 11) by Fermat’s Little
Theorem . By the division algorithm, 345 = 34 · 10 + 5 and 2 345 = 2 10 · 34+5 = ( 2 10 ) 34 · 2 5 . Then
2 345 ≡ 1 34 · 2 5 ≡ 1 · 32 ≡ 10
(mod 11) .
2. Show that x5 ≡ 3 (mod 11) has no solutions .
Solution. Assume x5 ≡ 3 (mod 11) has a solution. Then any solution must satisfy gcd ( x , 11 ) = 1 ,
so Fermat’s Little Theorem gives x 10 ≡ 1 (mod 11) . Since 5 divides 10 , we square both sides of the
congruence to get
1 ≡ x 10 ≡ ( x 5 ) 2 ≡ 3 2 ≡ 9 (mod 11) .
This is not true and so the assumption is false; thus the congruence has no solution .
Properties of Cosets. Let H be a subgroup of G and let a and b belong to G . Then,
1. a ∈ a H .
a = ae ∈ aH.
2. a H = H , if and only if a ∈ H .
The proof follows from property 1 and the fact that H is closed .
3. a H = b H , if and only if a ∈ b H .
If a H = b H , then a = a e ∈ b H . Conversely, if a ∈ b H, then a = b h for some h ∈ H . Hence
aH = bH.
T
4. a H = b H or a H
bH = ∅.
T
This follows from property 3, for if there is an element c in a H
b H , then c H = a H and
cH = bH).
5. a H = b H , if and only if a −1 b ∈ H .
This property follows from property 2 and the fact that a H = b H , if and only if H = a −1 b H .
6. | a H | = | b H | .
Group Theory
Quotient Groups
32
Define the mappings ϕ : a H 7→ b H , where a h :7→ b h and ψ : b H 7→ a H , where b h :7→ a h .
The proof then follows from the fact that both ϕ and ψ are one-to-one.
7. H a = a H , if and only if H = a H a −1 .
The proof follows from multiplying the right sides of both equalities by a −1 .
8. a H is a subgroup of G if and only if a ∈ H .
A coset to be a subgroup of G is must contain e . If a H is a subgroup, then a −1 is in H which
makes a a member of H .
Conversely, if a ∈ H , then a a −1 = e ; hence a H is a subgroup of G .
♥ Example. Let G = S3 and H = {( 1 ) , ( 1 , 3 ) } . Then the left cosets of H in G are
(1)H = (1, 3)H = H;
( 1 , 2 ) H = {(1 , 2) , ( 1 2 ) , ( 1 , 3 ) } = { ( 1 , 2 ) , ( 1 , 3 , 2 ) } = ( 1 , 3 , 2 ) H ;
( 2 , 3 ) H = { ( 2 , 3 ) , ( 2 , 3 ) ( 1 , 3 ) } = { ( 2 , 3 ) , (1 , 2 , 3 ) } = ( 1 , 2 , 3 ) H .
Exercise.
1. First find all the proper subgroups of S3 and then find their right and left cosets in S3 .
2. Draw the modified versions of the Cayley table partitioned by its subgroups of order two.
3. Let H = { 0 , 4 , 8 } be a subgroup of ( Z 12 , + 12 ) . Then find all of its cosets.
4. First find all the proper subgroups of G = ( U [ 30 ] , · ) and then find their cosets in G .
5. Find all of the cosets of { 1 , 11 } in (U [ 30 ] , · )
Remark 11.
• Coset leaders are used in decoding received data in linear error-correcting codes.
• With about 35 CPU-years of idle computer time donated by Google, a team of researchers used Cosets
of the Rubik’s Cube group showed that every position of Rubik’s Cube can be solved in twenty moves
or less.
• It was only in the middle of the 20th century that Fermat’s Little Theorem became useful. At this
point, relatively primitive computers were being used. In the 1970’s, a trio of mathematicians at MIT,
were able to use Fermat’s?s Little Theorem to come up with the public-key encryption scheme known
as the RSA algorithm, which is the standard encryption method used for online transactions today.
♣ Quotient Groups. This is the study of groups of subsets generated by a normal subgroup.
Let N be a normal subgroup of a group G. The quotient group (also known as factor group) of N in G,
written G/N and read “ G modulo N ,” is the set of all cosets of N in G, i.e., G/N = {a N : a ∈ G}.
The group operation on G/N is the product of subsets defined as follows: For each a N and b N in
G/N :
(a N ) (b N ) = a (N b ) N = a (b N ) N = (a b) N N = (a b) N .
The normality of N is used in this equation.
Theorem 15. The set of cosets of N in G under the product of subsets is a group.
Proof. The product of subsets is closed, because (a N ) (b N ) really is a coset. The coset N is the
identity coset. The inverse of the coset a N is a−1 N .
Group Theory
Quotient Groups
33
To show associativity, let a , b , c ∈ G . Then
(N a N b) N c = N a b N c = N ( a b ) c = N a ( b c ) = N a N bc = N a ( N b N c ).
This proves that G/N is a group.
Theorem 16. Let N be a normal subgroup of a group G . Then G/N is abelian if and only if
a b a −1 b −1 ∈ N for all a , b ∈ G .
Proof. The quotient group G/N is abelian if and only if
N a b = N a N b = N b N a = N b a for all a , b ∈ G .
The subgroup N serves as the identity element of the quotient group G/N , since N = N e . We have
seen that if ( a b ) ( b a ) −1 = e , then a b = b a . Since N is the identity element of the quotient group,
we conclude that N a b = N b a if and only if ( a b ) ( b a ) −1 ∈ N Therefore, G/N is abelian if and
only if a b a −1 b −1 ∈ N , for all a , b ∈ G .
♥ Examples.
1. The subgroup n Z is a normal subgroup of Z. The group Zn = Z/n Z is a quotient group.
2. The set of positive real numbers R + is a subgroup of the multiplicative group R ∗ . We have
R ∗ = R + ∪ (−1) R + = R + ∪ R −
which make the quotient group R ∗ /R + a group of order 2, isomorphic to (Z2 , +) .
3. Let G be the set of n × n invertible real matrices, and N = { A ∈ G | det(A) = 1 }.
The mapping
ϕ : ( G, . ) −→ (R ∗ , . ) with ϕ(A) = det(A)
is a group homomorphism with Ker(ϕ) = N . This makes N a normal subgroup of G. The cosets of N
are the sets of matrices with a given determinant, and hence G/N is isomorphic to the multiplicative
group of non-zero real numbers R ∗ .
4. Let N = { 0̄ , 4̄ , 8̄ } be the normal subgroup of the additive group Z 12 . These four cosets of N
contain every element of Z 12 :
N + 0̄ = { 0̄ , 4̄ , 8̄ } = N ;
N + 1̄ = { 1̄ , 5̄ , 9̄ } ;
N + 2̄ = { 2̄ , 6̄ , 1̄0 } ;
N + 3̄ = { 3̄ , 7̄ , 1̄1 } .
Hence, every coset is one of these four. For instance, 5 is in N + 1̄ and 5 is also in N + 5̄ (Why?). So
the two cosets are not disjoint. Hence, N + 1̄ = N + 5̄ . Similarly, N + 4̄ = N + 0̄ and N + 6̄ = N + 2̄ .
The following Cayley tables are for the quotient group Z 12 /N and the group Z 4 :
N
N
N
N
+ 0̄
+ 1̄
+ 2̄
+ 3̄
N + 0̄
N + 1̄
N + 2̄
N + 3̄
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
+ 0̄
+ 1̄
+ 2̄
+ 3̄
+ 1̄
+ 2̄
+ 3̄
+ 0̄
+ 2̄
+ 3̄
+ 0̄
+ 1̄
+ 3̄
+ 0̄
+ 1̄
+ 2̄
0̄
1̄
2̄
3̄
0̄
1̄
2̄
3̄
0̄
1̄
2̄
3̄
1̄
2̄
3̄
0̄
2̄
3̄
0̄
1̄
3̄
0̄
1̄
2̄
Group Theory
Quotient Groups
34
We see that both Cayley tables are similar; therefore we assert that Z 12 /N is isomorphic to Z 4 .
In the additive group Z , the subgroup generate by 4 :
h 4 i = {0 , ± 4 , ± 8 , ± 12 , . . . , ± 4 n , . . . } = 4 Z
is normal and it partitions the group Z into four different cosets. By dividing Z by the subgroup
4 Z , we construct the quotient group Z/4 Z which we will denote by Z4 .
5. The normal subgroup C4 = e0π/6 , e3iπ/6 , e6iπ/6 , e9iπ/6 of the circular group
C12 =
n
ek iπ/6 : k = 0, 1, 2, . . . , 11
o
splits the group C12 into three cosets, shown in red, green, and blue :
C4 = e0π/6 , e3iπ/6 , e6iπ/6 , e9tπ/6
eiπ/6 C4 = eiπ/6 , e4iπ/6 , e7iπ/6 , e10iπ/6
e2iπ/6 C4 = e2iπ/6 , e5iπ/6 , e8iπ/6 , e11iπ/6
The quotient group C12 /C4 is the group of the above three cosets.
♣ Direct Product of Groups. One way of building a new group from two old groups, say G 1 and
G 2 is to take the set G 1 × G 2 of ordered pairs of elements, the first from G 1 and the second from G 2 ,
using the operation of G 1 for the first component and the operation of G 2 for the second component.
This operation is analogue of the Cartesian product of sets.
The direct product, also known as the external direct product is an operation that takes n ≥ 2 groups
G 1 , G 2 , G 3 , . . . , G n and constructs a new group, denoted
G = G1 × G2 × G3 × · · · × Gn .
In the context of abelian groups, the direct product is sometimes referred to as the direct sum, and is
denoted
G1 ⊕ G2 ⊕ G3 ⊕ · · · ⊕ Gn .
The binary operations on G are defined component-wise:
(x1 , x2 , x3 , . . . , xn ) . (y1 , y2 , y3 , . . . , yn ) = (x1 . y1 , x2 . y2 , x3 . y3 , . . . , yn . yn ).
This operation makes the set G a group, where the identity
e = (e1 , e2 , e3 , . . . , en )
♦ Basic Properties.
and (g1 , g2 , g3 , . . . , gn )−1 = g1−1 , g2−1 , g3−1 , . . . , gn−1 .
Group Theory
Direct Product of Groups
35
1. The order of G = G 1 × G 2 × G 3 × . . . × G n is the product of the orders of G 1 , G 2 , G 3 , . . . , G n :
| G | = | G 1 × G2 × G3 × . . . × Gn| = | G1| · | G 2| · | G 3| · · · · · · | Gn| .
This follows from the formula for the cardinality of the cartesian product of sets.
2. The order of each element (g1 , g2 , g3 , . . . , gn ) is the least common multiple of the orders of
g1 , g2 , g3 , . . . , gn :
| (g1 , g2 , g3 , . . . , gn ) | = lcm( | g1 |, | g2 | , | g3 | , . . . , | gn |) .
In particular, if | g1 |, | g2 | , | g3 | , . . . , and | gn | are relatively prime, then
| (g1 , g2 , g3 , . . . , gn ) | =
n
Y
|gi | .
i=1
As a consequence, if G k ’s are cyclic groups whose orders are relatively prime, then the group
G 1 × G 2 × G 3 × . . . × G n is cyclic as well. For example, the group (Z2 × Z3 × Z5 , + ) is isomorphic to
the cyclic group (Z30 , +30 ).
3. Let G = G 1 × G 2 with the subgroups H1 = G 1 × {e2 } and H2 = {e1 } × G 2 . Then every element
of H1 commutes with every element of H2 . It is not difficult to prove that G 1 is isomorphic to H1
and G 2 is isomorphic to H2 . Moreover both H1 and H2 are normal subgroups of G .
Remark 12. Let H and K be subgroups of G ; then define the sum of H and K as:
H + K = { x = h + k : h ∈ H, k ∈ K } .
It can be shown that H + K < G. Now if H ∩ K = {e} , then H + K is denoted as H ⊕ K and is
called the direct sum of H and K . This notation is a little confusing, since we previously defined the
symbol ⊕ as a direct product of abelian groups.
Definition 7. Given N 1 , N 2 , N 3 , . . . , N n , normal subgroups of G, the internal product of these
subgroups is
N 1 · N 2 · N 3 · · · · · N n = h1 · h2 · h3 · · · · · hn : h0k s ∈ N k .
♥ Examples.
1. A group of order 4 is isomorphic to either Z4 or Z2 × Z2 . To verify this, let G = {e, a b, a b}.
Because G has only 4 elements, a b must be equal to b a, so G is abelian. If G is cyclic, then it is
isomorphic to Z4 , if not, then according to Lagrange’s Theorem, |a| = |b| = |ab| = 2. The mapping
ϕ : G −→ Z2 × Z2 with
ϕ(e) = (0, 0), ϕ(a) = (0, 1), ϕ(b) = (1, 0), ϕ(ab) = (1, 1)
is an isomorphism.
2. Klein four-group (or just Klein group ) symbolized by the letter V and may be constructed as
U[8] = 1, 3, 5, 7 or Z2 × Z2 . This group is the smallest non-cyclic group.
The Klein group is a Dihedral group of order 2, its elements can be represented by the following four
Group Theory
Direct Product of Groups
matrices:
!
1 0
,
0 1
R0 =
!
−1 0
,
0 −1
R1
S0 =
!
1 0
,
0 −1
36
S1
!
−1 0
.
0 1
Here are the tables for the Klein group:
U8
1
3
5
7
Z2 × Z2
(0̂, 0̂)
(1̂, 0̂)
(0̂, 1̂)
(1̂, 1̂)
1
1
3
5
7
(0̂, 0̂)
(0̂, 0̂)
(1̂, 0̂)
(0̂, 1̂)
(1̂, 1̂)
3
3
1
7
5
(1̂, 0̂)
(1̂, 0̂)
(1̂.1̂)
(1̂, 1̂)
(0̂, 1̂)
5
5
7
1
4
(0̂, 1̂)
(0̂, 1̂)
(1̂, 1̂)
(0̂, 0̂)
(1̂, 0̂)
7
7
5
3
1
(1̂, 1̂)
(1̂, 1̂)
(0̂, 1̂)
(1̂, 0̂)
(0̂, 0̂)
3. The groups isomorphic to Z8 , Z2 × Z4 , and Z2 × Z2 × Z2 are the only abelian groups of order 8.
The Quaternion group Q and the Dihedral group D4 are the only non-abelian groups of order 8.
4. The group U[30] is isomorphic to Z2 × Z4 where
1 7→ (0, 0), 7 7→ (0, 1), 11 7→ (1, 0), 13 7→ (0, 3), 17 7→ (1, 3), 19 7→ (0, 2), 23 7→ (1, 1), 29 7→ (1, 2).
5. The group of nonzero complex numbers C∗ under multiplication is isomorphic to the direct product
C × R+ , where C is the circle group of unit complex numbers z = eiθ = (cos θ + i sin θ) and R+ is the
multiplicative group of positive real numbers.
Definition 8. An abelian group G is called elementary, if the order of each element is 2 . It is denoted
by En , where n is its order. For example the group Z2 × Z2 × Z2 is elementary. Every elementary
group of order must be isomorphic to Z2 × Z2 × Z2 .
6. The abelian group of 3 × 3 real diagonal matrices






d1 0 0



E8 = D =  0 d2 0  : | dk | = 1, k = 1, 2, 3




0 0 d3

is an elementary group and is isomorphic to the direct product group Z2 × Z2 × Z2 .
Application to Data Security. Cryptography is the science of sending and deciphering secret
messages.
A binary string of length n can (Z = {0 , 1 } be thought of as an element of
Z2 ⊕ Z2 ⊕ · · · Z2
( n copies ) .
Thus the binary string 1 1 0 1 0 1 0 1 corresponds to the element
( 1 , 1 , 0 , 1 , 0 , 1 , , , 0 , 1 } ∈ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 .
The fact that the sum of two binary sequences a1 , a2 , . . . , an + b1 , b2 , . . . , an = 0 , 0 , . . . , 0 , if
and only if the sequences are identical is the basis for a data security system used to protect internet
transactions.
Group Theory
Direct Product of Groups
37
Suppose that you want to purchase a book from Amazon, but you are concerned that a hacker will
intercept your credit-card number during the transaction. As you might expect, your credit-card
number is sent to Amazon in a way that protects the data. here is how the credit-card numbers is
send over the Web securely :
Let CC be the binary string corresponding to your credit-card number. When you place an order
with Amazon the company sends your computer a randomly generated string LU of 0’s and 1’s of
the same length as CC . the two strings are added. This process is called “locking the data”. The
resulting sum is then transmitted to Amazon. Amazon in turn adds LU to the received string which
then produces the original string CC . This process is called “unlocking the data”.
Application to Public-Key Cryptography. It refers to a method of encryption in which for each
person a pair of keys (digital codes) are included; one called the public key, which is disseminated
widely, and one called the private key, which is known only to the intended recipient. Even though the
method used to scramble the message is known publicly, only the person for whom it is intended will
be able to unscramble the message. The idea is based on the fact that there exist efficient methods
for finding very large prime numbers (say about 100 digits long) and for multiplying large numbers,
but no one knows an efficient algorithm for factoring large integers (say about 200 digits long).
An effort by several researchers concluded in 2009, factoring a 232-digit number (RSA-768), utilizing
hundreds of machines took two years and the researchers estimated that a 1024-bit RSA modulus
would take about a thousand times as long. However, it has not been proven that no efficient algorithm exists. The presumed difficulty of this problem is at the heart of widely used algorithms in
cryptography such as RSA.
Application to Digital Signatures. With so many financial transactions now taking place electronically, the problem of authenticity is paramount. How is a stockbroker to know that an electronic
message he receives that tells him to sell one stock and buy another actually came from his client?
The technique used in public key cryptography allows for digital signatures as well.
Application to Genetics. A DNA molecule is composed of two long strands in the form of a double
helix. Each strand is made up of strings of the four nitrogen bases adenine (A), thymine (T), guanine
(G), and cytosine (C). Each base on one strand binds to a complementary base on the other strand.
To model this process, we identify A with 0, T with 2, G with 1, and C with 3. Hence the genetic
code can be conveniently modeled using elements of
Z4 ⊕ Z4 ⊕ · · · Z4
( n copies ) .
Application to Electric Circuits. Many homes have light fixtures that are operated by a pair of
switches. They are wired so that when either switch is thrown the light changes its status (from on
to off or vice versa). Suppose the wiring is done so that the light is on when both switches are in the
up position. We can conveniently think of the states of the two switches as being matched with the
elements of
Z2 ⊕ Z2 = { ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) } .
The subgroup H = { ( 0 , 0 ) , ( 1 , 1 ) } (two switches in the up position corresponding to ( 0 , 0 ) and
the two switches in the down position corresponding to ( 1 , 1 ) ) turns the light on.̇ Each time a switch
is thrown, we add one to the corresponding component in the group Z2 ⊕ Z2 . We then see that the
Group Theory
Direct Product of Groups
38
lights are on when the switches correspond to the elements of the subgroup H = h ( 1 , 1 ) i and are
off when the switches correspond to the elements in the coset ( 1 , 0 ) + H = ( 0 , 1 ) + H .
A similar analysis applies in the case of three switches with the subgroup
H = { ( 0 , 0 , 0 ) , ( 1 , 1 , 0 ) , ( 0 , 1 , 1) , ( 1 , 0 , 1 ) }
corresponding to the lights-on situation.
Groups II