Solution of Some Homework Problems
Section 2.2 Permutations
(2.1) Find sgn(α) and α−1 , where

α=
1 2 3 4 5 6 7 8 9
9 8 7 6 5 4 3 2 1

.
Solution: Since α = (19)(28)(37)(46)(5), sgn(α)=(−1)9−5 =1 and that α−1 = (5)(46)(37)(28)(19).
(2.2) If α ∈ Sn , prove that sgn(α−1 )=sgn(α).
Proof: Since αα−1 = e, we have sign(αα−1 ) = sign(e). That is sign(α)sign(α−1 ) = 1 and
so sign(α) = sign(α−1 ) = 1 or -1.
(2.8) Define f : {0, 1, 2, · · · , 10} 7→ {0, 1, 2, · · · , 10} by
f (n) = the raminder afer dividing4n2 − 3n7 by 11.
(i) Show that f is a permutation.
(ii) compute the parity of f .
(iii) Computer the inverse of f .
Proof: (i) Easy to get f (0) = 0, f (1) = 1 and f (10) ≡ f (−1) ≡ 4 + 3 ≡ 7 (mod 11). We
perform the following computation in Z11 .
f (2) = 4(22 ) − 3(27 ) = 16(1 − 3 · 8) ≡ 5(−1) ≡ 6 (mod 11)
f (9) ≡ f (−2) ≡ 16(1 + 24) ≡ 5 · 3 ≡ 15 ≡ 4 (mod 11)
f (3) = 4(32 ) − 38 = 9(4 − 36 ) ≡ 9(4 − (−2)3 ) ≡ 9(12) ≡ 9 (mod 11)
f (8) ≡ f (−3) ≡ (−3)2 (4 + (−3)6 ) ≡ 9(4 + (−2)3 ) ≡ (−2)(−4) ≡ 8 (mod 11)
f (4) = 4(42 ) − 3(47 ) = 43 (1 − 3(44 )) ≡ 9(1 − 3 · 25) ≡ 9(1 − 3(3)) ≡ 9(−8) ≡ 5 (mod 11)
f (7) ≡ f (−4) ≡ 43 + 3(47 ) = 43 (1 + 3(44 )) ≡ 9(1 + 3 · 25) ≡ 9(1 + 9)) ≡ 2 (mod 11)
f (5) = 4(52 ) − 3(57 ) = 52 (4 − 15(54 )) ≡ 3(4 − 4 · 9) ≡ 12(1 − 9) ≡ −8 ≡ 3 (mod 11)
f (6) ≡ f (−5) ≡ 4(52 ) + 3(57 ) = 52 (4 + 15(54 )) ≡ 3(4 + 4 · 9) ≡ 12(1 + 9) ≡ 10 (mod 11)
Thus f = (2 6 10 7)(3 9 4 5)(0)(1)(8) = (2 6 10 7)(3 9 4 5).
(ii) sgn(f ) = (−1)11−5 = 1, and so f is an even permutation.
1
(iii) f −1 = (3 9 4 5)−1 (2 6 10 7)−1 = (3 5 4 9)(2 7 10 6).
(2.10) (i) Prove that if α and β are (not necessary disjoint) permutations that commute,
then (αβ)k = αk β k for all k ≥ 1.
(ii) Give an example of two permutations and for which (αβ)2 6= α2 β 2 .
Proof: (i) First we prove, by induction, the statement L(k) that for any k ≥ 1, αβ k = β k α.
Since αβ = βα, L(1) holds.
Suppose that when k = n ≥ 1, we have αβ n = β n α.
We consider the case k = n + 1, αβ (n+1) = α(ββ n ) = (αβ)β n = (βα)β n = β(αβ n ) =
β(β n α) = (ββ n )α = β (n+1) α. Thus by induction L(k) holds for all integers k ≥ 1.
Now we use induction to prove the statement S(k) that (αβ)k = αk β k , for any integer
k ≥ 1.
Since S(1) = L(1), S(1) holds.
Suppose that when k = n ≥ 1, we have (αβ)n = αn β n .
We consider the case k = n + 1, (αβ)(n+1) = (αβ)n (αβ) = (αn β n )(αβ) = αn (β n α)β =
αn (αβ n )β = (αn α)(β n β) = α(n+1) β (n+1) . Thus, S(k) holds for all integers k ≥ 1.
Next, we show that for any integer k, (αβ)k = αk β k . This has been proved for k
being a positive integer. If k = 0, then
(αβ)0 = 1 = 1 · 1 = α0 β 0 .
It remains to show that this holds also when k < 0 is a negative integer. Let k = −n for
some positive integer n > 0. Note that as αβ = βα,
β −1 α−1 = (αβ)−1 = (βα)−1 = α−1 β −1 .
Therefore, by what we have proved above, (α−1 β −1 )n = (α−1 )n (β −1 )n . It follows
(αβ)k = (αβ)−n = ((αβ)−1 )n = (α−1 )n (β −1 )n = α−n β −n = αk β k .
Thus the statement holds also for all integer k.
An alternative proof for Part (i): We apply induction on n ≥ 0 for the statement:
S(n): (αβ)n = αn β n .
S(0) holds trivially and S(1) is in the hypothesis. Therefore, we assume that S(n) holds
for n ≤ k ≥ 1 to prove that S(k + 1) is also true.
(αβ)k+1 = (αβ)n−1 αβ
associative law
2
= αk β k αβ
induction hypothesis
= αk β(α1−k αk−1 )β k−1 αβ
multiplying by an identity
= αk βα1−k (αk−1 β k−1 )αβ
associative law
= αk βα1−k (αβ)k−1 (αβ)
= αk βα1−k (αβ)k
induction hypothesis & asso. law
law of exponents
= αk βα1−k αk β k
induction hypothesis
= αk β(α1−k αk )β k
associative law
= αk βαβ k
law of exponents
= αk αββ k
S(1) is true
= αk+1 β k+1
This proves S(+1). The rest of the proof is the same as above.
(ii) Let α = (123), β = (13). We have (αβ)2 = (132)2 = (123) and α2 β 2 = (12)2 (13)2 = 1.
(2.13) (i) How many permutations in S5 commute with (123), and how many even permutations in S5 commute with (123)?
(ii) Same questions for (12)(34).
Solution: (i) Let σ ∈ S5 be a permutation that commute with (123). Then by
σ(123)(4)(5)σ −1 = (σ(1) σ(2) σ(3))(σ(4))(σ(5))
we have σ ∈ {(1), (123), (132), (45), (123)(45), (132)(45)}, which are all permutations in S5
commuting with (123). Among them (1), (132), (123) are even.
(ii) With the same method, in S4 , a permutation σ ∈ S4 commutes with (12)(34) if and
only if σ ∈ {(1), (12), (34), (12)(34), (13)(24), (14)(23), (1324), (1432)}. Among them, (1),
(12)(34), (13)(24),(14)(23) are even.
(2.14) Give an example of α, β, γ ∈ S5 , with α 6= (1), such that αβ = βα, αγ = γα
and βγ 6= γβ.
Solution:
Let α = (12), β = (34), γ = (45). Then we have αβ = βα, αγ = γα and
βγ 6= γβ.
(2.15) If n ≥ 3, show that if α ∈ Sn commute with every β ∈ Sn , then α = (1).
Proof:
As α can be expressed as a product of disjoint cycles, we may assume that
3
(1 2 · · · k) is a longest cycle in the factorization of α into disjoint cycles.
First, we claim that k ≤ 2. By contradiction, we assume that k ≥ 3. Pick β = (12).
Then we must have βαβ −1 = α. However, βαβ −1 (1) = 3 6= 2 = α(1). Therefore, we must
have k ≤ 2.
Nest, we claim that α cannot have more than one disjoint transpositions. If not, we may
assume that α = (12)(34) · · ·. Let β = (13). Then we must have βαβ −1 = α. However,
βαβ −1 (1) = 3 6= 2 = α(1). Therefore, the claim must hold.
Thus if α is not the identity, it can only be a 2-cycle. We may assume that α = (12).
Now we pick β = (13) again. Then as βαβ −1 (1) = 3 6= 2 = α(1), we have a contradiction
again. This implies that α must be the identity.
Section 2.3 Groups
(2.19) (i) Compute the order, inverse, and parity of
α = (12)(43)(13542)(15)(13)(23).
(ii) What are the respective orders of the permutations in Exercise 2.1 on page 49 and 2.8
on page 50?
Solution: (i) Since
α = (12)(43)(13542)(15)(13)(23)
= (43)(12)(12)(14)(15)(13)(15)(13)(23)
= (43)(14)(135)(135)(23)
= (43)(14)(153)(23)
= (14)(13)(13)(15)(23)
= (14)(15)(23)
= (154)(23),
the order of (154) is 3 and the order of (23) is 2. Thus the order of α is lcm(2, 3) = 6.
Direct computation yields that α−1 = (23)(145) and sgn(α) = (−1)5−2 = −1. Therefore α
is odd.
(ii) For Exercise 2.1 on page 49, α = (19)(28)(37)(46) is a product of 4 disjoint 2-cycles,
and so the order is lcm(2, 2, 2, 2) = 2.
For Exercise 2.1 on page 49, α = (2 6 10 7)(3 9 4 5) is a product of two disjoint 4-cycles,
and so the order is lcm(4, 4) = 4.
4
(2.21) If G is a group, prove that the only element g ∈ G with g 2 = g is 1.
Proof: If g 2 = g, then g −1 g 2 = g −1 g and so g = 1.
(2.22) This exercise gives a shorter list of axioms defining a group. Let H be a set containing
an element e, and assume that there is an associative binary operation ? on H satisfying
the following properties:
(1). e ? x = x for all x ∈ H;
(2). for every x ∈ H, there is x0 ∈ H with x0 ? x = e.
(i) Prove that if h ∈ H satisfies h ? h = h, then h = e.
(ii) For all x ∈ H, prove that x ? x0 = e.
(iii) For all x ∈ H, prove that x ? e = x.
(iv) Prove that if e0 ∈ H satisfies e0 ? x = x for all x ∈ H, then e0 = e.
(iv) Prove that H is a group.
(v) Let x ∈ H. Prove that if x00 ∈ H satisfies x00 x = e, then x00 = x0 .
Proof:
(i) Let h ∈ H satisfies h ? h = h. By (2), ∀h, ∃h0 ∈ H with h0 ? h = e. We have
h0 ? (h ? h) = h0 ? h and then (h0 ? h) ? h = h0 ? h, that is e ? h = h. Thus, h = e.
(ii) Since (x ? x0 )2 = (x ? x0 )(x ? x0 ) = x ? (x0 x?)x0 = x ? x0 , by (i), x ? x0 = e.
(iii) x ? e = x ? (x0 ? x) = (x ? x0 ) ? x = e ? x = x.
(iv) Since (e0 )2 = e0 e0 = e0 , by (i), e0 = e.
(v) Since x0 ? x ? x00 = e ? x00 = x00 and x0 ? x ? x00 = x0 ? x ? x00 ? (x ? x0 ) = x0 ? x ? (x00 ? x) ? x0 =
x0 ? x ? x0 = x0 , we have x00 = x0 .
(vi) From above, we have the following observations:
(A) ? is an associative binary operation;
(B) there is a unique identity e in H;
(C) for any element x in H, there is a unique x0 in H such that xx0 = x0 x = e.
By the group axioms, H is a group.
(2.24) Let G be a group and let a ∈ G have order k. If p is a prime divisor of k, and
if there is x ∈ G with xp = a, prove that x has order pk .
Proof: Let the order of x be n. Since xpk = (xp )k = ak = 1, n|pk. Since p is a prime, we
have either (n, p) = 1 or (n, p) = p.
Suppose first assume that (n, p) = 1. Since n|pk and since (n, p) = 1, by the Euclid
Lemma, n|k. Since 1 = xnp = (xp )n = an , k|n and so n = k. But p|k and so (n, p) = p, a
contradiction.
5
Thus we must have (n, p) = p. Then p|n and so for some integer n1 , n = pn1 . It follows
that 1 = xn = (xp )n1 = an1 . Thus k|n1 , and so for some integer n2 , n1 = kn2 . Multiplying both sides by p to get n = pn1 = pkn2 . This implies that (pk)|n, and so we have n = pk.
(2.25) Let G = GL(2, R), and let

0 −1

1
0


 and 
0
1
−1 1

.

Show that A4 = I = B 6 , but that (AB)n 6= I for all n > 0, where I = 
1
0

 is the
−1 1
2 × 2 identity matrix. Conclude that AB can have infinite order even though both factors
A and B have finite order.
Proof:

A4 = 

0
B6 = 
1
−1 1
6
4
0 −1
1

 =

 =
0
−1 1
−1 0
2
−1
0
0
−1

0 −1

−1 1
1 −1

−1 0
n

3
 =
 =I

=I
Arguing by induction on n, we can prove that

(AB)n = 
0 −1

1

0
0
1
−1 1
n

 = 
0 −1
0
1
 = 
1 −n
0
1

 6= I.
Section 2.4 Lagrange’s Theorem
(2.30) (i) Define the special linear group by SL(2, R) = {A ∈ GL(2, R); det(A) = 1}. Prove
that SL(2, R) is a subgroup of GL(2, R).
(ii) Prove that GL(2, Q) is a subgroup of GL(2, R).

1 0

 ∈ SL(2, R), so SL(2, R) 6= ∅.
0 1
(2) For any A, B ∈ SL(2, R), det(AB) = det(A) det(B) = 1, so AB ∈ SL(2, R).
Proof: (i) (1) I = 
(3) For any A ∈ SL(2, R), as AA−1 = I, det(A) det(A−1 ) = det(I) = 1. It follows that
det(A−1 ) =
1
det(A)
= 1 and so A−1 ∈ SL(2, R).
Combing (1),(2),(3),

 we conclude that SL(2, R) is a subgroup of GL(2, R).
(ii) (1) I = 
1 0
0 1
 ∈ GL(2, Q), so GL(2, Q) 6= ∅.
6
(2) For any A, B ∈ GL(2, Q), (det)(AB) = (det)(A)(det)(B) 6= 0 and by the multiplication, (AB)ij =
P
k
aik bkj is a rational number, so AB ∈ GL(2, Q).
(3) For any A ∈ GL(2, Q),A−1 =
1
∗
det(A) A ,
(det)(A)(det)(A−1 ) = (det)(I) = 1. By
the knowledge of linear algebra, every element of matrix of A∗ is rational and (det)(A) is
rational. So A−1 ∈ GL(2, Q).
Combing (1),(2),(3), we get GL(2, Q) is a subgroup of GL(2, R).
(2.31) (i) Give an example of two subgroups H and K of a group G whose union H ∪ K is
not a subgroup of G.
(ii) Prove that the union H ∪ K of two subgroups is itself a subgroup if and only if either
H is a subset of K or K is a subset of H.
Proof: (i) Let G = Z, H = 2Z, K = 3Z. Then as 2 ∈ 2Z and 3 ∈ 3Z but 2+3 = 5 6∈ H ∪K,
H ∪ K 6≤ G.
(ii) If H ⊆ K or K ⊆ H, then H ∪ K is either H or K, and so a subgroup of G.
Conversely, suppose, to the contrary, that there is a ∈ H − K and b ∈ K − H such that
ab ∈ H ∪ K. If ab ∈ H, then b = a−1 (ab) ∈ H, a contradiction. If ab ∈ K, then
a = (ab)b−1 ∈ H, a contradiction. Thus if H ∪ K ≤ G, we must have either H − K = ∅ or
K − H = ∅.
(2.32) Let G be a finite group with subgroups H and K. If H ≤ K, prove that [G :
H] = [G : K][K : H].
Proof: By Lagrange, [G : H] =
|G|
|H| ,
[G : K] =
|G|
|K|
and [K : H] =
|K|
|H| .
Thus [G : H] = [G :
K][K : H].
(2.34) Prove that every subgroup S of a cyclic group G = hai is itself cyclic.
Proof: Let ak be the element of S such that k = min{i > 0 : ai ∈ S}. Since G = hai, any
b ∈ S can be expressed as b = al for some integer l. Write l = ks + t, 0 ≤ t < k. If t > 0,
then al = aks+t = aks at and so at = al a−ks ∈ H, contradicting the choice of k. Thus t = 0
and k|l. Hence S = hak i.
(2.37) If H is a subgroup of a group G, prove that the number of left closers of H in
G is equal to the number of right closets of H in G.
Proof: Let L and R denote the set of left cosets and right cosets of H in G, respectively.
Define a function f : L 7→ R by f (aH) = Ha−1 .
Since L is a set of equivalence classes, we must show that f is a well-defined map. In
7
other words, the value of f (aH) does not depend on the choice of representatives of the left
coset aH. To see that, let a, a1 ∈ aH be two representatives of the left coset aH. Then
−1 = Ha−1 and so f (aH) = f (a H).
aH = a1 H, and so aa−1
1
1 ∈ H. It follows that Ha
1
Hence f is well-defined.
We shall show that f : L 7→ R is injective. Suppose that f (a1 H) = f (a2 H). Then
Ha−1
1
−1
= Ha−2
2 , and so a1 a2 ∈ H. This implies that a1 H = a2 H. Hence f is injective.
It remains to show that f : L 7→ R is surjective. For any Ha ∈ R, a−1 ∈ G, and so
f (a−1 H) = H(a−1 )−1 = Ha. Hence f is surjective.
Summing up, f : L 7→ R is a bijective map.
Section 2.5 Homomorphisms
(2.40) (i) Show that the composite of homomorphisms is itself a homomorphism.
(ii) Show that the inverse of an isomorphism is an isomorphism.
(iii) Show that two groups that are isomorphic to a third group are isomorphic to each
other.
(iv) Prove that isomorphism is an equivalence relation on any set of groups.
Proof:
(i) Let f : G1 7→ G2 , h : G2 7→ G3 be homomorphisms. For any g1 , g2 ∈
G1 , f (g1 g2 ) = f (g1 )f (g2 ). Since f (g1 ), f (g2 ), f (g1 )f (g2 ) ∈ G2 and since h is a group homomorphism, h(f (g1 )f (g2 )) = h(f (g1 ))h(f (g2 )) = (hf )(g1 )(hf )(g2 ). That is (hf )(g1 g2 ) =
(hf )(g1 )(hf )(g2 ), and so hf : G1 7→ G3 is also a group homomorphism.
(ii) Let f be an isomorphism from G1 to G2 . Then f is a bijection and for any g1 , g2 ∈ G1 ,
f (g1 g2 ) = f (g1 )f (g2 ). And so f −1 is a bijection from G2 to G1 and for any h1 , h2 ∈ G2 ,
there are g1 , g2 ∈ G1 such that f (g1 ) = h1 , f (g2 ) = h2 . It follows that f −1 (h1 h2 ) =
f −1 (f (g1 )f (g2 )) = f −1 (f (g1 g2 )) = g1 g2 = f −1 (h1 )f −1 (h2 ). Thus f −1 : G2 7→ G1 is also a
group isomorphism.
(iii) Let f : G1 7→ G2 and h : G3 7→ G2 be group isomorphisms. By (ii), h−1 : G2 7→ G3
is an isomorphism. Thus by (i), g −1 f : G1 7→ G3 is a homomorphism. As the composition
of two bijections is also a bijection, g −1 f : G1 7→ G3 is an isomorphism.
(iv) Define, for groups G and H, that G ∼ H if G is isomorphic to H.
For any group G, the identity map i : G 7→ G is an isomorphism, and so G ∼ G. Thus
∼ is reflexive.
By (ii), if f : G 7→ H is an isomorphism, then f −1 : H 7→ G is also an isomorphism.
Thus ∼ is symmetric.
Suppose that f : G1 7→ G2 and h : G2 7→ G3 are isomorphisms. Then by (iii),
(hf ) : G1 7→ G3 is also an isomorphism, and so ∼ is transitive.
8
(2.42) This exercise gives some invariants of a group G. Let f : G 7→ H be an isomorphism.
(i) Prove that if a ∈ G has infinite order, then so does f (a), and if a has finite order n, then
so does f (a). Conclude that if G has an element of some order n and H does not, then
G∼
6 H.
=
(ii) Prove that if G ∼
= H, then for every divisor d of |G|, both G and H have the same
number of elements of order d.
Proof: Let a ∈ G be an element with order n and let f (a) ∈ H have order m. If n < ∞,
then (f (a))n = f (an ) = f (1G ) = 1H and so m|n. Thus m is also finite. Applying the same
argument to the isomorphism f −1 : H 7→ G, we conclude that if m < ∞, then n|m and so
n must also be finite.
Moreover, this also shows that if one of n or m is finite, then n|m and m|n, which imply
that n = m.
(ii) Let a1 , a2 , · · · , ak ∈ G be all the elements in G that are of order d. Then as
shown in (i), f (a1 , f (a2 ) · · · , f (ak ) ∈ H are k elements in H that are of order d.
If
y ∈ H −{f (a1 ), · · · , f (ak } has order d, then since f is an isomorphism, ∃x ∈ G−{a1 , · · · , ak }
such that f (x) = y and such that x also has order d, contrary to the assumption that
{a1 , a2 , · · · , ak } is the set of all order d elements in G. Thus both G and H have the same
number of elements of order d.
(2.43) Prove that A4 and D12 are nonisomorphic groups of order 12.
Proof: Since the elements of A4 has orders 1,2,or 3 but in D12 there are elements of order
6, By Exercise 2.42(ii), A4 and D12 are not isomorphic.
(2.45) Show that every group G with |G| < 6 is abelian.
Proof: We first prove the following claims for a finite group G.
Claim 1: Let H be a subgroup of G, then |H| divides |G|. This follows from the Lagrange
Theorem.
Claim 2: If |G| is prime, then G must be cyclic. For any g ∈ G with g 6= 1, H = hai ≤ G.
Since |G| is a prime and since |H > 1, by Claim 1, |H| = |G|, and so G = H is cyclic.
By Claim 2, if |G| ∈ {2, 3, 5}, then G is cyclic and so abelian. It remains to show the
case when |G| = 4.
If G has am element a of order 4, then G = hai is cyclic, and so abelian.
Hence by Claim 1, we conclude that every g ∈ G − {1} must have order 2. Thus it
9
suffices to show that if a group G in which every element a ∈ G, a2 = 1, then G is abelian.
Let G be such a group. ∀a, b ∈ G, we have (ab)2 = 1 = a2 b2 . This implies that
abab = aabb. Cancelling a from left and b from right, we have ba = ab, and so G is abelian.
(2.47) (i) If f : G 7→ H is a homomorphism and x ∈ G has order k, prove that f (x) ∈ H
has order m, where m|k.
(ii) If f : G 7→ H is a homomorphism and if (|G|, |H|) = 1, prove that f (x) = 1 for all
x ∈ G.
Proof:
Since f : G 7→ H is a homomorphism, for x ∈ G of order k in G, (f (x))k =
f (xk ) = f (1G ) = 1H . If f (x) has order m, then m|k.
(ii) The problem assumes that both |G| and |H| are finite. By contradiction, suppose
that ∃x ∈ G with order k but f (x) 6= 1. Let the order of f (x) be m > 1. By (i), m|k. By
Lagrange, k divides |G|, and m divides |H|. It follows by m|k that m > 1 is a common
divisor of both |G| and |H|, contrary to the assumption that (|G|, |H|) = 1.
(2.50) (i) Show that if H is a subgraph with bH = Hb = {hb : h ∈ H} for every b ∈ G, then
H must be a normal subgroup.
(ii) Use part (i) to give a second proof of Proposition 2.62(ii): If H ≤ G has index 2, then
H G.
Proof:
(i) For any for every b ∈ G, bH = Hb = {hb : h ∈ H}. Then for any h ∈ H,
bh = h1 b for some h1 ∈ H and so bhd−1 = h1 ∈ H. Thus by definition of a normal subgroup,
H is a normal subgroup og G.
(ii) Since ∀b ∈ G − H, bH ∩ H = ∅ and [G : H] = 2, and since [G : H] counts the number
of left (right) cosets of H in G, we have G = H ∪ bH = H ∪ Hb. That means that ∀b ∈ G,
bH = Hb and so by (i), H G.
(2.53) Define W =< (12)(34) >, the cyclic subgroup of S4 generated by (12)(34). Show
that W is a normal subgroup of V , but that W is not a normal subgroup of S4 . Conclude
that normality is not transitive: W V and V G do not imply W G.
Proof: Since for every g ∈ S4 , x ∈ V , g −1 xg ∈ V , then V S4 .
Since V = {(1), (12)(34), (13)(24), (14)(23)}, W = h(12)(34)i = {(1), (12)(34)} is a subgroup of W of index 2, by Exercise 2.50, W V .
Since (13)−1 ((12)(34))(13) = (14)(23) 6∈ W , W 6 S4 .
(2.55) Give an example of a group G, a subgroup H ≤ G, and an element g ∈ G with
10
[G : H] = 3 and g 3 6∈ H.
Proof: Take G = S3 , H = h(12)i, g = (23). Then g ∈ S3 . By Lagrange, [G : H] = 3. But
g 3 = g 6∈ H.
(2.59) Recall that the group of quaternions Q consists of the 8 matrices in GL(2, C)
Q = {I, A, A2 , A3 , B, BA, BA2 , BA3 },
where
(i) Prove that −I is the only element in Q of order 2, and that all other elements M 6= I
satisfy M 2 = −I.
(ii) Prove that Q is non abelian group with operation matrix multiplication.
(iii) Prove that Q has a unique subgroup of order 2, and it is the center of Q.
Proof:
(i) Direct computation yields AB = BA3 , A2 = B 2 = −I, A4 = B 4 = I. Thus
(BA)2 = B(AB)A = B(BA3 )A = B 2 = −I, (BA2 )2 = BA(AB)A2 = BABA3 A2 =
BBA3 A5 = B 2 = −I, and (BA3 )2 = (AB)2 = BA3 (AB) = B 2 = −I. Thus −I is the only
order 2 element in Q, and if M ∈ Q − {I, −I}, then M 2 = −I.
(ii) As AB = BA3 6= BA, Q is not abelian.
(iii) Since every order 2 subgraph must be a cyclic subgroup generated by an order 2
element, and since (by (i)), −I is the only element of order 2, the subgroup h−Ii is the only
order 2 subgroup of Q.
Let Z = Z(Q) be the center of Q. Clearly I, −I ∈ Z. Since AB 6= BA, A, B 6∈ Z. It
follows by the fact that Z ≤ Q that the inverses A−1 = A3 and B −1 = BA2 are not in
Z. Since (BA)B = B(AB) = BBA3 = −A3 = A and B(BA) = B 2 A = −A, we observe
that (BA)B 6= B(BA), and so BA 6∈ Z. As Z ≤ G, (BA)−1 = BA3 6∈ Z. It follows that
Z = {I, −I} = h−Ii.
(2.60) Assume that there is a group of order 8 whose elements 1, i, j, k satisfy i2 = j 2 =
k 2 = −1, ij = k, jk = i, ki = j, and ij = −ji, ik = −ki, jk = −kj. Prove that G ∼
= Q and,
conversely, that Q is such a group.
Proof:
Define bijection σ by setting σ(1) = I, σ(i) = A, σ(j) = B, σ(k) = BA3 , σ(−1) =
A2 , σ(−i) = A3 , σ(−j) = BA2 , σ(−k) = BA. We can verify that this bijection keep the
operation and G ∼
= Q.
Section 2.6 Quotient Groups
(2.65) Prove U (Z9 ) ∼
= Z6 and U (Z15 ) ∼
= Z4 × Z2 .
11
Proof:
Note that U (Z9 ) = {[x] ∈ Z9 : (x, 9) = 1} = {[1], [2], [4], [5], [7], [8]}, and the
binary operation is multiplication. Direct computation yields that the order of [2] in the
multiplicative group U (Z9 ) is 6, and so U (Z9 ) = h[2]i ∼
= Z6 , as Z6 is also a cyclic group of
order 6.
Note that G = U (Z15 ) = {[x] ∈ Z15 : (x, 15) = 1} = {[1], [2], [4], [7], [8], [11], [13], [14]},
and the binary operation is multiplication.Direct computation reveals that the order of 4
elements are [2], [7], [8], [13] and order 2 elements are [4], [11], [14].
Let H = {[1], [2], [4], [8]}, and K = {[1], [11]}. Then G is abelian, H, K G, and
K ∩ H = {[1]}. Thus |HK| = |G| and so G = HK = {hk : h ∈ H, k ∈ K}. Moreover, if
−1 ∈ H ∩ K = {[1]}, and so
hk = h1 k1 for some h, h1 ∈ H and k, k1 ∈ K, then h−1
1 h = k1 k
h = h1 and k = k1 .
As [2] is a generator of H and [11] is a generator of K, and as every element in G
can be uniquely expressed as [2]i [11]j where 0 ≤ i ≤ 3 and 0 ≤ j ≤ 1, the following map
φ : U (Z15 ) 7→ Z4 × Z2 is an isomorphism.

φ=
[1]
[2]
[4]
[8]
[11]
[7]
[14]
[13]
(0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3)

.
(2.66) (i) Let H and K be groups. Without using the first isomorphism theorem, prove
that H ∗ = {(h, 1) : h ∈ H} and K ∗ = {(1, k) : k ∈ K} are normal subgroups of H × K
∼ H ∗ and K =
∼ K ∗ , and f : H 7→ (H × K)/K ∗ , defined by f (h) = (h, 1)K ∗ , is an
with H =
isomorphism.
(ii) Use the first isomorphism theorem to prove that K ∗ H ×K and that (H ×K)/K ∗ ∼
= H ∗.
Proof: (i) By the definition of H ∗ , we have H ∗ ⊆ H × K. Note that (1H , 1K ) ∈ H ∗ 6= ∅.
∈ H, and
∀(h1 , 1K ), (h2 , 1K ) ∈ H ∗ , we have h1 , h2 ∈ H. Since H is a group, h1 h−1
2
∗
∗
∗
so (h1 , 1K )(h2 , 1K )−1 = (h1 h−1
2 , 1K ) ∈ H , and so H ≤ H × K. ∀(h, 1) ∈ H and
−1
0
∗
∀(h1 , k1 ) ∈ H × K, (h1 , k1 )(h, 1)(h1 , k1 )−1 = (h1 hh−1
1 , k1 k1 ) = (h , 1K ) ∈ H , where
∗
∗
h0 = h1 hh−1
1 ∈ H. By definition, H H × K. Similarly, K H × K.
Define a map φ : H 7→ H ∗ by φ(h) = (h, 1), ∀h ∈ H. Then ∀h1 , h2 ∈ H, φ(h1 h2 ) =
(h1 h2 , 1K ) = (h1 , 1K )(h2 , 1K ) = φ(h1 )φ(h2 ), and so φ is a homomorphism. If h ∈ ker(φ),
then (h, 1K ) = (1H , 1K ), and so h = 1H . Hence ker(φ) = {1H }, and so φ is an injection.
∀(h, 1K ) ∈ H ∗ , h ∈ H and so φ(h) = (h, 1K ), and so φ is surjective. Hence H ∼
= H ∗.
Similarly, K ∼
= K ∗.
To see that f is an isomorphism, we first check that f is a homomorphism. Let h1 , h2 ∈
H. Then
f (h1 h2 ) = (h1 h2 , 1K )K ∗ = (h1 , 1K )K ∗ (h2 , 1K )K ∗ = f (h1 )f (h2 ),
12
and so f is a homomorphism. ∀(h, k)K ∗ ∈ (H×K)/K ∗ , f (h) = (h, 1K )K ∗ = (h, 1K )(1H , k)K ∗ =
(h, k)K ∗ , and so f is surjective. If h1 , h2 ∈ H such that (h1 , 1K )K ∗ = f (h1 ) = f (h2 ) =
∗
(h2 , 1K )K ∗ , then (h2 , 1K )−1 (h1 , 1K ) ∈ K ∗ , and so (h−1
2 h1 , 1K ) ∈ K . It follows that
h−1
2 h1 = 1H , or equivalently, h1 = h2 . Thus f is injective, and so f is an isomorphism.
(ii) Define a map f : (H × K) 7→ H ∗ by f (h, k) = (h, 1K ). Then
f ((h1 , k1 )(h2 , k2 )) = f ((h1 h2 , k1 k2 ) = (h1 h2 , 1K ) = (h1 , 1K )(h2 , 1K ) = f (h1 , k1 )f (h2 , k2 ).
Thus f is a homomorphism. To determine the kernel of f , we note that
ker(f ) = {(h, k) : (h, 1K ) = (1H , 1K )} = {(1H , k) : ∀k ∈ K} = K ∗ .
Consider the image of f . ∀(h, 1K ) ∈ H ∗ , f (h, 1K ) = (h, 1K ) and so im(f ) = H ∗ . Therefore
by the 1st isomorphism theorem, (H × K)/K ∗ ∼
= H ∗.
(2.67) (i) Prove that Aut(V) ∼
= S3 and that Aut(S3 ) ∼
= S3 . Conclude that non isomorphic groups can have isomorphism groups.
(ii) Prove that Aut(Z) ∼
= Z2 . Conclude that an infinite group can have a finite automorphism group.
Proof: (i) Recall that
V = {(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}.
Let a1 = (1 2)(3 4), a2 = (1 3)(2 4) and a3 = (1 4)(2 3). Since any automorphism
f ∈ Aut(V) must fix the identity element, the map f is uniquely determined by its image
on f (a1 ), f (a2 ) and f (a3 ). Since ai aj = ak for {i, j, k} = {1, 2, 3}, f is uniquely determined
by its image on f (a1 ) and f (a2 ).
Direct computation shows that each ai has order 2. Let f ∈ Aut(V). Then as f (ai )
must have the same order of ai , we must have f (ai ) = af 0 (i) , ∀i ∈ {1, 2, 3}, where f 0 ∈ S3 .
Therefore, the map φ(f ) = f 0 is a map φ : Aut(V) 7→ S3 .
Let f1 , f2 ∈ Aut(V). Then ∀ai ∈ V,
(f1 f2 )(ai ) = f1 (f2 (ai )) = f1 af20 (i) = af10 (f20 (i)) .
Thus φ(f1 f2 ) = f10 f20 = φ(f1 )φ(f2 ), and so φ is a group homomorphism. Let f ∈ ker(φ).
Then f (ai ) = ai for any i, which implies that f is the identity map in Aut(V), and so φ is
injective.
It remains to show that φ is surjective. For any s ∈ S3 , Define fs : V 7→ V by fs (1) = (1)
and fs (ai ) = as(i) . Then fs is a bijection. Moreover, as s is a permutation on {1, 2, 3}, and
13
as every ai has order 2, ∀ai , aj ∈ V, fs (ai aj ) = fs (ak ) = as(k) = as(i) as(j) = fs (ai )fs (aj ),
and so fs ∈ Aut(V) and φ(fs ) = s. Thus φ is onto. This proves that Aut(V) ∼
= S3 .
To show that Aut(S3 ) ∼
= S3 , we again first study how an f ∈ Aut(S3 ) would behave.
Let
a1 = (1 2), a2 = (1 3), a3 = (2 3), b1 = (1 2 3), b2 = (1 3 2).
Then as f (x) and x must have the same order, f (ai ) = aj where i, j ∈ {1, 2, 3}. Moreover, as a1 a2 = b2 and a1 a3 = b1 , when f is an automorphism, the image of f (b1 ) and
f (b2 ) are uniquely determined by f (a1 ), f (a2 ) and f (a3 ). As f (ai ) = as(i) for some s ∈ S3 ,
the isomorphism between Aut(S3 ) ∼
= S3 can be found similar to what we have done in
∼
Aut(V) = S3 .
(ii) Let f ∈ Aut(Z). Then consider the image of f (1). If f (1) = m 6∈ {1, −1}, then as
0 = f (0) = f (1 + (−1)) = f (1) + f (−1) = m + f (−1), we must have f (−1) = −m. It follows
by the assumption that f is an homomorphism, ∀n ∈ Z with n > 0, f (n) = f (1+1+· · ·+1) =
n · · · f (1) = nm ∈ mZ, and f (−n) = −mn. Thus f (Z) ⊆ mZ ⊆ Z − {−1, 1}, contrary to
the assumption that f is an automorphism. Therefore, either f (1) = 1 or f (1) = −1. Let
f0 , f1 ∈ Aut(Z) be such that f0 (1) = 1 and f1 (1) = −1. Then since f0 , f1 are homomorphisms, f0 (n) = n and f1 (n) = −n. The map composition yields that f1 f0 = f0 f1 = f1 ,
f1 f1 = f0 = f0 f0 . Therefore, a map φ : Aut(Z) 7→ Z2 given by φ(f0 ) = 0 and φ(f1 ) = 1 is
a group isomorphism.
(2.69) Prove that if G is a group for which G/Z(G) is cyclic, where Z(G) denotes the
center of G, then G is abelian.
Proof:
Let Z = Z(G). Since G/Z is cyclic, we may assume the for some a ∈ G,
G/Z = haZi. Let g, h ∈ G. (We want to show that gh = hg). Then gZ = ai Z and
hZ = aj Z, and so ∃z1 , z2 ∈ Z such that g = ai z1 and h = aj z2 . It follows by the definition
of Z(G) and by z1 , z2 ∈ Z that
gh = ai z1 aj z2 = ai aj z1 z2 = ai+j z2 z1 = aj ai z2 z1 = aj z2 ai z1 = hg.
(2.70) (i) Prove that Q/Z(Q) ∼
= V, where Q is the group of quaternions and V is the
four-group; conclude that the quotient of a group by its center can be abelian.
(ii) Prove that Q has no subgroup isomorphic to V . Conclude that the quotient Q/Z(Q) is
not isomorphic to a subgroup of Q.
14
Proof: Recall that
V = {(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}.
Note that φ : V 7→ Z2 × Z2 given below

φ=
(1)
(0, 0)
(1 2)(3 4) (1 3)(2 4) (1 4)(2 3)
(0, 1)
(1, 0)
(1, 1)


is an isomorphism. Therefore we can have the following proofs. The following proofs can
also be given directly without going through Z2 × Z2 by replacing (x, y) ∈ Z2 × Z2 by
φ−1 (x, y).
(i) Let Z = Z(Q). Note that Q/Z = {Z, AZ, BZ, (BA)Z}, as Z = A2 Z, AZ = A3 Z,
BZ = (BA2 )Z and (BA)Z = BA3 )Z.
Define a map f : Q/Z 7→ V by f (Z) = (0, 0), f (AZ) = (1, 0), f (BZ) = (0, 1) and
f ((BA)Z) = (1, 1). Since (AZ)(BZ) = (AB)Z = (BA3 )Z = (BA)Z = (BZ)(AZ),
(BAZ)(AZ) = (−B)Z = BZ = (BA3 A)Z = ((AB)A)Z) = (AZ)((BA)Z), and (BZ)((BA)Z) =
(−A)Z = (BBA3 )Z = ((BA)Z)(BZ), and since in V , (1, 0) +(0, 1) = (1, 1), (1, 1) +(1, 0) =
(0, 1) and (0, 1)+(1, 1) = (1, 0), f is a bijective homomorphism, and so f is an isomorphism.
(ii) Since V has 3 elements of order 2, and Q has only one element of order 2 (Exercise
(2.59)(iii)), no subgroup of Q can be isomorphic to V .
(2.72) If H and K are subgroups of a group G, prove that HK is a subgroup of G if
and only if HK = KH.
Proof:
Suppose first that HK ≤ G. Note that ∀h ∈ H an d∀k ∈ K, as h = h · 1 ∈ HK
and k = 1 · k ∈ HK, kh ∈ HK, and so KH ⊆ HK. On the other hand, since HK ≤ G,
HK = {x−1 : x ∈ HK}. Therefore, ∀x ∈ HK, we can write x = (h)−1 for some h ∈ H and
k ∈ K. However, (hk)−1 = k −1 h−1 ∈ KH, and so HK ⊆ KH.
Conversely, we assume that HK = KH. Then 1 ∈ HK 6= ∅. For x, y ∈ HK, we can
write x = h1 k1 and y = h2 k2 for some h1 , h2 ∈ H and k1 , k2 ∈ K. Then xy −1 = h1 k1 k2−1 h−1
2 .
∈ KH. As KH = HK, we can write
Note that since K ≤ G and H ≤ G k1 k2−1 h−1
2
−1 = h k k −1 h−1 = (h h)k ∈ HK,
k1 k2−1 h−1
1 1 2
1
2 = hk for some h ∈ H and k ∈ K. Thus xy
2
and so HK ≤ G.
(2.76) If H and K are normal subgroups of a group G with HK = G, prove that G/(H ∩
K) ∼
= (G/H) × (G/K).
Proof:
Since H and K are normal in G, G/H and G/K are groups. Define a map
15
f : G 7→ (G/H) × (G/K) by f (x) = (xH, xK). Then ∀x, y ∈ G,
f (xy) = ((xy)H, (xy)K) = (xH, xK)(yH, yK) = f (x)f (y),
and so f is a homomorphism. Let L = ker(f ). Then
L = {x ∈ G : f (x) = (H, K)} = {x ∈ G : xH = H&xK = K}
= {x ∈ G : x ∈ H&x ∈ K} = H ∩ K.
Since HK = G, by Exercise 2.72, G = KH. For any (aH, bK) ∈ (G/H) × (G/K), we
can write a = aK aH with aH ∈ H and aK ∈ K, and b = bH bK with bH ∈ H and
bK ∈ K. Let x = aK bH =∈ G. Then f (x) = ((aK bH )H, (aK bH )K) = (aK H, bH K) =
((aK aH )H, (bH bK )K) = (aH, bK). Therefore, f is onto. By the first isomorphism Theorem,
G/(H ∩ K) ∼
= (G/H) × (G/K).
16