S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 3, Solutions
Name:
1. Let G be a group and let H be a subgroup of G. Let a ∈ G. Prove that the set
aHa−1 = {aha−1 | where h ∈ H} is a subgroup of G.
Proof:
• Claim 1: aHa−1 ⊂ G, i.e. is a subset of G.
Proof of Claim 1: If x ∈ aHa−1 , then x = aha−1 for some h ∈ H.
Since H ⊂ G, then h ∈ G. Since G is a group it is closed under inverses and multiplications. Therefore aha−1 ∈ G.Therefore x ∈ G. Therefore aHa−1 ⊂ G.
• Claim 2: aHa−1 6= ∅, i.e. is a nonempty set.
Proof of Claim 2: Since H is a subgroup of G, the identity of G is in H, i.e. e ∈ H.
Therefore aea−1 ∈ aHa−1 . So e = aea−1 using inverse and identity properties in a group.
Therefore e ∈ aHa−1 . Therefore aHa−1 6= ∅.
• Claim 3: If x, y ∈ aHa−1 , then xy ∈ aHa−1 .
Proof of Claim 3: Let x, y ∈ aHa−1 .
Then, there exist h1 , h2 ∈ H such that x = ah1 a−1 and y = ah2 a−1 , by defn of aHa−1 .
Therefore xy = (ah1 a−1 )(ah2 a−1 ) = ah1 eh2 a−1 = ah1 h2 a−1 = aha−1 where h = h1 h2 .
Then h ∈ H (H is closed under operation since it is a subgroup).
Therefore xy ∈ aHa−1 .
• Claim 4: If x ∈ aHa−1 , then x−1 ∈ aHa−1 .
Proof of Claim 4: Let x ∈ aHa−1 .
Then there exist an h ∈ H such that x = aha−1 .
x−1 = (aha−1 )−1 = (a−1 )−1 h−1 a−1 = ah−1 a−1 .
Since H is a subgroup, it is closed under inverses. Therefore h−1 ∈ H. So x−1 ∈ aHa−1 .
• Conclusion: aHa−1 is a subgroup in G.
This follows by the Theorem on subgroups: A nonempty subset of a group is a subgroup
if it is closed under group operation and inverses.
2. (a) Find the conjugate of (1234)(56) by a = (25) in S7 .
Definition: A conjugate of σ by a is φa (σ) = aσa−1 .
Remark: If order of an element a in a group is |a| = n, then a−1 = an−1 .
If a = (25) then |a| = 2. Therefore a−1 = a.
φa (σ) = φ(25) (1234)(56) = (25)(1234)(56)(25)−1 = (25)(1234)(56)(25) = (1534)(26).
(b) Find the conjugate of (1234)(56) by a = (27) in S7 .
The same arguments as in (a).
φa (σ) = φ(27) (1234)(56) = (27)(1234)(56)(27)−1 = (27)(1234)(56)(27) = (1734)(56).
(c) Find the conjugate of (1234)(56) by a = (37) in S7 .
The same arguments as in (a).
φa (σ) = φ(37) (1234)(56) = (37)(1234)(56)(37)−1 = (37)(1234)(56)(37) = (1274)(56).
1
S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 3, Solutions
Name:
1 2 3 4 5 6 7 8
3. Consider the permutation α =
∈ S8 .
2 3 7 1 8 6 4 5
(a) Write α as a product of disjoint cycles.
α = (12374)(58)(6) = (12374)(58)
(b) Write α as a product of transpositions.
α = (12374)(58) = (14)(17)(13)(12)(58)
(c) Write α as a product of transpositions in two different ways.
α = (12374)(58) = (14)(17)(13)(12)(58) = (58)(14)(17)(13)(12)
(disjoint cycles commute)
α = (12374)(58) = (23741)(58) = (21)(24)(27)(23)(58)
α = (12374)(58) = (74)(34)(24)(14)(58)
α = (12374)(58) = (14)(17)(13)(12)(58)(45)(45)
(multiplication by identity)
α = (12374)(58) = (25)(14)(25)(17)(13)(12)(58)
(multiplication by identity and using the fact that disjoint cycles commute)
(d) Write α in three other ways, which are different from any of the above expressions.
α = (12374)(58) = (14)(17)(13)(12)(58) = (174)(13)(12)(58) = (174)(123)(58) = (741)(123)(58) =
(741)(123)(45)(45)(58) = (741)(123)(45)(458) (etc,etc....)
(e) Is α even or odd permutation? (explain)
α = (12374)(58) = (14)(17)(13)(12)(58)
(α is ODD permutation since it is written as a product of odd number of transpositions
and we stated the Theorem which says that the parity of the number of transpositions will
not change when the same permutation is written in a different way.)
(f) What is the order of α? (explain)
|k − cycle| = k
If β and γ are disjoint cycles then |β γ| = lcm(|β||γ|)
|α| = |(12374)(58)| = lcm(|(12374)|, |(58)|) = lcm(5, 2) = 10
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 3, Solutions
Name:
4. How many permutations of order 6 are there in S10 . Make sure that you explain.
Disjoint cycle partitions:
Number of permutations:
(6,3,1)
10 6!
6 6
·
4 3!
3 3
= 201600
(6,2,2)
10 6!
6 6
·
4 2!
2 2
·
(6,2,1,1)
10 6!
6 6
·
4 2!
2 2
= 151200
(6,1,1,1,1)
10 6!
6 6
= 25200
(3,3,2,2)
10 3!
3 3
·
7 3!
3 3
·
1
2!
·
4 2!
2 2
·
(3,3,2,1,1)
10 3!
3 3
·
7 3!
3 3
·
1
2!
·
4 2!
2 2
= 50400
(3,2,2,2,1)
10 3!
3 3
·
7 2!
2 2
·
5 2!
2 2
·
(3,2,2,1,1,1)
10 3!
3 3
·
7 2!
2 2
·
5 2!
2 2
·
(3,2,1,1,1,1,1)
10 3!
3 3
·
7 2!
2 2
= 5040
2 2!
2 2
·
1
2!
= 75600
3 2!
2 2
1
2!
2 2!
2 2
·
1
3!
·
= 25200
= 25200
= 25200
The number of permutations of order 6 in S10 is 584640.
3
1
2!
S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 3, Solutions
Name:
5. Find all possible disjoint cycle decompositions of elements of order 5 in S14 .
Answer: (5,5,1,1,1,1), (5,1,1,1,1,1,1,1,1,1). (lcm of the lengths must be 5)
6. Find all possible disjoint cycle decompositions of elements of order 10 in S14 .
7. How many elements of order 15 are there in S8 ?
Answer: If σ = αβ where α and β are disjoint cycles, then |σ| = lcm(|α|, |β|).
Also order of a k-cycle is k.
Therefore the only possible disjoint cycle decompositions for a permutation σ ∈ S8 with
|σ| = 15 is (5, 3).
The number of such permutations is 85 5!5 33 3!3 = 2688.
8. How many elements of order 15 are there in S7 ?
9. How many elements of order 15 are there in S9 ?
10. How many elements of disjoint cycle decomposition (2, 2), i.e. of the form (a b)(c d) are there
in S4 ? Write down all of them.
11. How many elements of disjoint cycle decomposition (2, 2, 2), i.e. of the form (a b)(c d)(e f )
are there in S6 ?
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