S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions)
3 PERMUTATIONS
1
Some Definitions
For your convenience, we recall some of the definitions:
• A group G is called simple if it has no proper normal subgroups, i.e. the only normal subgroups
are {e} and G.
• Let x ∈ G be an element of a group G. A conjugate of x by g ∈ G is the element gxg −1 .
• Let x ∈ G be an element of a group G. A conjugacy class of x is C(x) = {gxg −1 | g ∈ G}.
• Let H < G be a subgroup of a group G. A conjugate of H by g ∈ G is the subgroup gHg −1 .
• Let H < G be a subgroup of a group G. A conjugacy class of H is C(H) = {gHg −1 | g ∈ G}.
2
Basic properties
1. Let x ∈ G be an element of a group G. Prove that |x| = |gxg −1 | for any g ∈ G.
2. Let H < G be a subgroup of a finite group G. Prove that |H| = |gHg −1 | for any g ∈ G.
Proof: Consider the map f : H → gHg −1 given by f (x) = gxg −1 .
- f is a map H → gHg −1 since f (x) ∈ gHg −1 for every x ∈ H.
- f is one to one: Suppose f (x1 ) = f (x2 ). Then gx1 g −1 = gx2 g −1 .
- Multiply by g −1 on left and g on the right, use associative law, inverse and identity, and get
x1 = x2 . Therefore f is one-to-one.
- f is onto: Any element gxg −1 ∈ gHg −1 is f (x) for some x ∈ H.
- Therefore f is a bijection between finite sets. (No need to check group homomorphism for
this, since the question was only about the number of elements.)
- So |H| = |f (H)| = |gHg −1 |.
3
Permutations
1. Let α = (24367) ∈ S8 , be a permutation in the permutation group on 8 letters: {1, 2, 3, 4, 5, 6, 7, 8}.
(a) Find (28)(24367)
(b) Find (24367)(28)
(c) Find (28)(24367)(28)−1
(d) Find (28)(24367)(28)
(e) Find (281)(24367)
(f) Find (281)−1
(g) Find (281)(24367)(281)−1
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions)
4 CYCLIC GROUPS
2. Prove that a 5-cycle in any permutation group has order 5.
3. Let S9 be the permutation group on 9 letters: {1, 2, 3, 4, 5, 6, 7, 8, 9}.
(a) How many cycles of order 5 are there in S9 ?
Answer: There are 95 5!5 5-cycles.
(b) How many permutations of order 5 are there in S9 ?
Answer: There are 95 5!5 5-cycles.
- Order of 5-cycle is 5.
- Order of a product of disjoint cycles is least common multiple of the orders of the cycles.
- Therefore the only possibility for order 5 in S9 is 5 = lcm(5, 1, 1, 1, 1), hence 5-cycles.
(c) How many permutations of order 10 are there in S9 ?
Answer:
- There are 95 5!5 · 42 2!2 · 22 2!2 · 2!1 permutations with cycle decomposition (5, 2, 2)
- There are 95 5!5 · 42 2!2 permutations with cycle decomposition (5, 2, 1, 1)
- The only possibilities to get permutations of order 10, are the cycle decompositions:
(5, 2, 2) and (5, 2, 1, 1) since lcm(5, 2, 2) = 10 and lcm(5, 2, 1, 1) = 10.
(d) How many permutations of order 11 are there in S9 ?
Answer: None. Reason:
- Since 11 is prime, need to get 11-cycle in order to get permutation of order 11.
- There are no partitions of 9, which have 11 elements
(e) How many permutations of order 12 are there in S9 ?
4. Let α = (abc) and let β = (ad). Prove that the conjugate of α by β is (dbc).
5. Let α = (abcdef g) and let β = (bmf ). Prove that the conjugate of α by β is (amcdebg).
6. Let α = (abcdef g) and let β = (arbsct). What is the conjugate of α by β? Prove it.
7. Let α = (236). Find β such that βαβ −1 = (143).
8. Let α = (1236). Find β such that βαβ −1 = (1435).
9. Let α = (1236). Prove that there is no β such that βαβ −1 = (14)(36).
10. Let α = (1236). Prove that there is no β such that βαβ −1 = (14).
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Cyclic Groups
1. Write the definition of a cyclic group.
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions)
4 CYCLIC GROUPS
2. Prove that a subgroup of a finite cyclic group is cyclic group.
Proof: Let G be a finite cyclic group. Then there exists an element a ∈ G such that G = hai.
- Let n be the smallest positive integer such that an = e, where e is the identity of G.
- Then G = {e, a, a2 , a3 , . . . , an−1 }.
- Let H < G be a subgroup of G.
- Let i be the smallest positive integer such that ai ∈ H.
- Claim: H = hai i, i.e. every element of H is of the form (ai )j for some j.
- Proof of the Claim: Let x ∈ H. Since x ∈ H ⊂ G we have x ∈ G.
- Then x = am for some non-negative integer m, since G = hai.
- Let d = gcd(i, m). Then there exist integers α, β ∈ Z such that d = αi + βm.
- So ad = aαi+βm = (ai )α (am )β .
- Since ai ∈ H it follows that (ai )α ∈ H by closure of subgroup under group operation.
- Since am ∈ H it follows that (am )β ∈ H by closure of subgroup under group operation.
- Since (ai )α ∈ H and (am )β ∈ H it follows that aαi+βm ∈ H again by closure of subgroup
under group operation.
- Therefore ad ∈ H.
- Since i is the smallest positive such integer, it follows that d = i.
- Therefore i | m. So m = ij for some integer j.
- Hence x = am = aij = (ai )j for some j, i.e. every element x ∈ H can be expressed in terms
of ai .
3. Prove that a quotient of a finite cyclic group is cyclic group.
4. Prove that a cyclic group of order 60 is isomorphic to Z60 .
5. Consider a cyclic group of order 16, i.e. G = hai where |a| = 16.
(a) Write all generators of G.
Solution: If G = hai and |a| = n, then generators of G are all ai such that gcd(i, n) = 1.
- Generators of G = hai where |a| = 16 are: {a = a1 , a3 , a5 , a7 , a9 , a11 , a13 , a15 }
(b) Write the formula for the number of generators of a cyclic group of order n and check
that you got the correct number of generators according to the formula.
Solution: The number of positive integers smaller then n and relatively prime to n is
given by Euler phi function φ(n).
- We have φ(pk ) = pk−1 (p − 1)
- Therefore φ(16) = φ(24 ) = 23 (2 − 1) = 8 (agrees with the above 8 generators).
(c) Write all elements of order 8 in G.
Solution: If G = hai, |a| = n and k|n, then
|an/k | = k and
|ai | = |aj | if and only if gcd(i, n) = gcd(j, n).
- Therefore |a16/8 | = 8, i.e. |a2 | = 8.
- To find the other elements of order 8, need to find other integers j < 16, so that
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions)
4 CYCLIC GROUPS
gcd(j, 16) = gcd(2, 16) = 2. So j = 2, 6, 10, 14
- Therefore the elements of order 8 are: a2 , a6 , a10 , a14 .
(d) Write the formula for the number of elements of order 8 in a cyclic group of order 16 and
check that you got the correct number of elements according to the formula.
Solution: If G = hai, |a| = n, k|n, then the number of elements of order k in G is φ(k).
- So the number of elements of order k = 8 is φ(8) = φ(23 ) = 22 (2 − 1) = 4, which is
exactly how many we got in the previous part.
(e) Write all elements of order 4 in G.
Solution: |a16/4 | = 4, i.e. |a4 | = 4.
- To find the other elements of order 4, need to find other integers j < 16, so that
gcd(j, 16) = gcd(4, 16) = 4. So j = 4, 12
- Therefore the elements of order 4 are: a4 , a12 .
(f) Write the formula for the number of elements of order 4 in a cyclic group of order 16 and
check that you got the correct number of elements according to the formula.
Solution: The number of elements of order k = 4 is φ(4) = φ(22 ) = 21 (2 − 1) = 2, which
is exactly how many we got in the previous part.
(g) Write all elements of order 2 in G.
Solution: |a16/2 | = 2, i.e. |a8 | = 2.
- To find the other elements of order 2, need to find other integers j < 16, so that
gcd(j, 16) = gcd(8, 16) = 8. So j = 8
- Therefore the elements of order 2 are: a8 .
(h) Write the formula for the number of elements of order 2 in a cyclic group of order 16 and
check that you got the correct number of elements according to the formula.
Solution: The number of elements of order k = 2 is φ(2) = φ(21 ) = 20 (2 − 1) = 1, which
is exactly how many we got in the previous part.
6. Consider the group Z16 . (Remember, operation is addition modulo 16).
(a) Write all generators of Z16 .
(b) Write the formula for the number of generators of the group Z16 and check that you got
the correct number of generators according to the formula.
(c) Write all elements of order 8 in Z16 .
(d) Write the formula for the number of elements of order 8 in the group Z16 and check that
you got the correct number of elements according to the formula.
(e) Write all elements of order 4 in Z16 .
(f) Write the formula for the number of elements of order 4 in the group Z16 and check that
you got the correct number of elements according to the formula.
(g) Write all elements of order 2 in Z16 .
(h) Write the formula for the number of elements of order 2 in the group Z16 and check that
you got the correct number of elements according to the formula.
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice
5 EXTERNAL
Some Solutions)
PRODUCTS OF GROUP
7. Consider the group Z49 . (Remember, operation is addition modulo 49).
(a) What are the possible orders of elements?
Solution: Since Z49 is cyclic group of order 49, the only possible orders of elements are:
1, 7, 49.
(b) List all the elements and their orders. Which formula are you using? Does it agree with
your computations?
Solution: Since Zn is cyclic group of order n, we can use everything we know about cyclic
groups, but use additive notation!!!
- Use additive version of the two statements:
|an/k | = k, which is |(n/k)a| = k and
|ai | = |aj | if and only if gcd(i, n) = gcd(j, n), which is
|ia| = |ja| if and only if gcd(i, n) = gcd(j, n)
- Notice that Zn is generated by 1 (additively!!!)
- Elements of order 49 : (49/49)1 = 1 · 1 = 1
- Other elements of order 49 are j < 49 such that gcd(j, 49) = gcd(1, 49) = 1
- Elements of order 49: 1,2,3,4,5,6,8,9,10,11,12,13,15,16,17,18,19,20,22,23,24,25,26,27,29,30,31,32,33,34,3
- Notice φ(49) = φ(72 ) = 72−1 (7 − 1) = 7 · 6 = 42
- Elements of order 7: 7,14,21,28,35,42
- Notice φ(7) = 7 − 1 = 6
- Elements of order 1: 0
5
External Products of Group
1. Consider G = Z4 ⊕ Z3 .
(a) How many elements does G have?
Solution: 4 × 3 = 12
(b) What is the order of the element (0, 1) ∈ Z4 ⊕ Z3 ?
Solution: 3
(c) What is the order of the element (0, 2) ∈ Z4 ⊕ Z3 ?
Solution: 3
(d) What is the order of the element (1, 1) ∈ Z4 ⊕ Z3 ?
Solution: |(1, 1)| = lcm(|1|, |1|) = lcm(4, 3) = 12
(e) What is the order of the element (3, 2) ∈ Z4 ⊕ Z3 ?
(f) What are all the possible orders of elements (a, b) ∈ Z4 ⊕ Z3 ?
|(a, b)| = lcm(|a|, |b|), |a| = 1, 2, 4 and |b| = 1, 3
(g) Describe all elements (a, b) ∈ Z4 ⊕ Z3 of order 3?
(h) Describe all elements (a, b) ∈ Z4 ⊕ Z3 of order 4?
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice 7Some
GROUP
Solutions)
HOMOMORPHISMS
(i) Describe all elements (a, b) ∈ Z4 ⊕ Z3 of order 12?
2. Prove that Z4 ⊕ Z3 ∼
= Z12 .
6
Abelian Groups
1. Describe all Abelian groups (up to isomorphism) of order 120.
2. Describe all Abelian groups (up to isomorphism) of order 100 which are not cyclic.
Answer: 100 = 22 52
- Abelian groups of order 22 , up to isomorphism are:
Z22 (which corresponds to the partition (2) of 2
Z2 ⊕ Z2 (which corresponds to the partition (1,1) of 2.
- Abelian groups of order 52 , up to isomorphism are:
Z52 (which corresponds to the partition (2) of 2
Z5 ⊕ Z5 (which corresponds to the partition (1,1) of 2.
- Abelian groups of order 100 = 22 52 , up to isomorphism are the 4 possible combinations:
Z22 ⊕ Z52
Z22 ⊕ Z5 ⊕ Z5
Z2 ⊕ Z2 ⊕ Z52
Z2 ⊕ Z2 ⊕ Z5 ⊕ Z5
- Use that Zm ⊕ Zn ∼
= Zmn when gcd(m, n) = 1. So
∼
Z22 ⊕ Z52 = Z100 which is cyclic !!!
Z22 ⊕ Z5 ⊕ Z5 ∼
= Z20 ⊕ Z5 , which is NOT cyclic
Z2 ⊕ Z2 ⊕ Z52 ∼
= Z2 ⊕ Z50 , which is NOT cyclic
Z2 ⊕ Z2 ⊕ Z5 ⊕ Z5 ∼
= Z10 ⊕ Z10 again NOT cyclic.
7
Group Homomorphisms
1. Give an example of a group homomorphism which is onto but not one-to-one.
2. Give an example of a group homomorphism which is one-to-one but not onto.
3. Give an example of a group homomorphism which is neither one-to-one not onto.
4. Write all isomorphisms Z12 → Z4 ⊕ Z3 .
5. Consider group homomorphism f : Z12 → Z4 ⊕ Z3 given by f (a) = (a, 0).
(a) Describe Ker(f ). Write down all the elements.
(b) Describe Im(f ). Write down all the elements.
6. Consider group homomorphism g : Z12 → Z24 given by g(a) = 2a.
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions)
9 SYLOW THEORY
(a) Describe Ker(g). Write down all the elements of Ker(g).
(b) Describe Im(g). Write down all the elements of Im(g).
7. Consider group homomorphism ϕ : Z12 → Z24 given by ϕ(a) = 4a.
(a) Describe Ker(ϕ). Write down all the elements of Ker(ϕ).
(b) Describe Im(ϕ). Write down all the elements of Im(ϕ).
8. Consider group homomorphism φ : Z12 → Z24 given by φ(a) = 6a.
(a) Describe Ker(φ). Write down all the elements of Ker(φ).
(b) Describe Im(φ). Write down all the elements of Im(φ).
9. Prove that there is no group homomorphism ω : Z12 → Z24 such that ω(a) = a.
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Group Actions
1. Let G be group of order 25, i.e. |G| = 25. Prove that the center of G, Z(G) 6= {e}.
Proof: Let G act on itself (i.e. X = G) by conjugation.
- The action ϕP
: G × G → G isP
given by ϕ(g, x) = gxg −1 .P
- |X| = |G| = |Ox |=1 (|Ox |) + |Ox |>1 (|Ox |) = |Z(G)| + |Ox |>1 (|Ox |).
- |Ox | |kG| = 25. So if |Ox | =
6 1 then 5 | |Ox |.
- Also 5 | |G|.
- Therefore 5 | |Z(G)|.
- Since e ∈ Z(G), we know that Z(G) 6= 0, so Z(G) is a non-zero multiple of 5.
- Therefore Z(G) has at least 5 elements.
2. Let G be group of order 81, i.e. |G| = 81. Prove that G has a normal subgroup.
3. Let G be group of order pk , p a prime and k ≥ 1, k ∈ Z. Prove that G has a normal subgroup.
9
Sylow Theory
We’ve defined and proved the following in class, so you may use this:
• Definition 1 A group G is called a p-group if every element of G has order power of p.
• Definition 2 A subgroup H of a group G is called a p-subgroup if every element of H has order
power of p.
• Definition 3 A subgroup P of G is called a p-Sylow subgroup of G if it is a maximal p-subgroup
of G, (i.e. there is no p-subgroup Q of G such that P Q G.)
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions)
9 SYLOW THEORY
• Theorem 1 Let G be a group such that |G| = pk m where p is prime and p does not divide m.
Then there exists a subgroup P of order pk .
• Proposition Let G be a group such that |G| = pk m where p is prime and p does not divide m.
Let P be a subgroup of G of order pk . Let C = {gP g −1 | g ∈ G} = {P = P1 , P2 , . . . , Pn } be
all conjugates of P . Then any p-subgroup of G is contained in one of the Pi ∈ C.
DO THE FOLLOWING PROBLEMS
1. Let X be a p-subgroup of a group G and a ∈ G. Prove that aXa−1 is a p-subgroup of G.
Proof: Every element x ∈ X has order power of p, by Defn 2 of a p-subgroup.
- Every element of aXa−1 is of the form axa−1 for some x ∈ X.
- Then the order of axa−1 is equal to the order of x by Problem1 (in 2 Basic Properties).
- Therefore the order of any element in aXa−1 is a power of p.
- Therefore aXa−1 is a p-subgroup by definition of p-subgroup.
2. Let G be a group such that |G| = pk m where p is prime and p does not divide m. Let P
be a subgroup of G of order pk . Let C = {gP g −1 | g ∈ G} = {P = P1 , P2 , . . . , Pn } be all
conjugates of P . Then any p-subgroup of G is contained in one of the Pi ∈ C.
Prove that any maximal p-subgroup of G is equal to one of the Pi ’s in C.
Proof: The existence of a subgroup P such that |P | = pk is guaranteed by the above Thm 1.
-
Let Q be any maximal p-subgroup of G. Then Q is a p-subgroup.
By the above Proposition Q is contained in one of the subgroups Pi ∈ C.
Let Q ⊂ Pi0 ∈ C. But Pi0 is also a p-subgroup since it is a conjugate to a p-subgroup.
Then Q = Pi0 since Q is a maximal p-subgroup.
3. Let G be a group such that |G| = pk m where p is prime and p does not divide m. Let P
be a subgroup of G of order pk . Let C = {gP g −1 | g ∈ G} = {P = P1 , P2 , . . . , Pn } be all
conjugates of P .
Prove that any p-Sylow subgroup of G is equal to one of the Pi ’s in C.
This is only one line proof with the previous problem and the definitions.
Proof: Let Y be a p-Sylow subgroup of G.
- Then Y is a maximal p-subgroup of G by definition and
- Y is equal to one of the Pi ∈ C be the previous problem.
4. Theorem All p-Sylow subgroups of a finite group G are conjugate to each other.
Prove this theorem (i.e. prove that there is only one conjugacy class).
Proof: Let |G| = pk m with p 6 |m.
- Let P be a subgroup of order |P | = pk .
- Then any p-Sylow subgroup is equal to one of the Pi ∈ C. (By the previous problem.)
- Therefore any p-Sylow subgroup is conjugate to P , since C consist of ALL conjugates of P .
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions)
9 SYLOW THEORY
-
Let Y and Z be two p-Sylow subgroups of G.
So both Y and Z are conjugate to P .
Therefore Y = aP a−1 and Z = bP b−1 for some a, b ∈ G.
So Y = aP a−1 = a(b−1 Zb)a−1 = (ab−1 )Z(ab−1 )−1
Hence Y is conjugate to Z.
5. Theorem Let G be a finite group. Then #{p-Sylow subgroups of G} divides |G|.
Proof: Let G act on the set X = {p-Sylow subgroups of G} by conjugation,i.e.
ϕ : G × {p-Sylow subgroups of G} → P
{p-Sylow subgroups of G} is given by ϕ(g, P ) := gP g −1
|X| = |{p-Sylow subgroups of G}| = disjoint |orbits|
- By the Theorem in Problem 4., we know that all p-Sylow subgroups of G are conjugate to
each other, i.e. there is only one conjugacy class.
- Therefore |X| = |{p-Sylow subgroups of G}| = |orbit of P | = |only one orbit|
- |Orbit| | |G| for any action of a group.
- Therefore |{p-Sylow subgroups of G}| | |G|.
6. Theorem Let G be a finite group. Then #{p-Sylow subgroups of G} ≡ 1(mod p).
Prove this.
7. Let G be a group of order |G| = 35.
(a) What are the possible #{5-Sylow subgroups of G} using Problem 5?
Answer: 1,5,7,35
(b) What are the possible #{5-Sylow subgroups of G} using Problem 6?
Answer: 1,6,11,16, 21, 26,31
(c) How many 5-Sylow subgroups does G have?
Answer: 1
(d) Let P5 be a 5-Sylow subgroup of G. How many elements does P5 have?
Answer: |P5 | = 5
(e) What are the possible #{7-Sylow subgroups of G} using Problem 5?
Answer: 1,5,7,35
(f) What are the possible #{7-Sylow subgroups of G} using Problem 6?
Answer: 1,8,15,22,29
(g) How many 7-Sylow subgroups does G have?
Answer: 1
(h) Let P7 be a 7-Sylow subgroup of G. Find |P7 |.
Answer: |P7 | = 7
8. Find ALL Sylow subgroups of Z12 . Prove that these are all.
(Hint, you may want to use Sylow theorems for each prime p such that p||Z12 in order to
determine the possible numbers of the Sylow p-subgroups, for each of the primes.)
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S10MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions)
10 X
9. Determine the Sylow subgroups of the alternating group A4 (the even permutations of {1, 2, 3, 4}.
10
x
1. Prove that the conjugacy class of (2314) in S4 has exactly 6 elements.
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