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MATH 451 Practice Midterm II (Fall 2013)
Instructor: Xiaowei Wang
Nov. 6th, 10:00am-11:15pm, 2013
Problem 1. Let σ ∈ Sn . Prove if σ is a cycle of odd length (, which is the number of
elements in its largest orbit), then σ 2 is a cycle.
Proof. By our assumption σ is a cycle, say it is of the form σ = (a1 , · · · , a2k+1 ) since it
is of odd length. To show σ 2 is a cycle, we need to show that
{a1 , a2 , · · · , a2k+1 } ⊂ {1, 2, · · · , n}
is exactly one orbit of σ 2 . That is, for each ai there is a l ∈ Z such that (σ 2 )l (a1 ) =
σ 2l (a1 ) = ai .
Since {a1 , a2 , · · · , a2k+1 } is an orbit of σ, by definition(cf. section 9) there is a m ∈ Z
such that σ m (a1 ) = ai . Thus if we can show the subgroup H 0 := hσ 2 i = G := hσi, then
we will have σ m = (σ 2 )l for some l ∈ Z. But this follows from Theorem 6.14, since
|H 0 | =
|G|
2k + 1
=
= 2k + 1 = |G| .
gcd(|G|, 2)
gcd(2k + 1, 2)
So the proof is completed.
Problem 2. Mark each of the following true or false (no explanation is required)
1. Every group is isomorphic to some group of permutations. (yes, by Cayley’s Theorem)
2. Every permutation is a one-to-one function. (yes, by definition of permutation)
3. Every group G is isomorphic to a subgroup of SG . (yes, by Cayley’s Theorem)
4. All generator of Z20 are prime numbers. (yes)
5. S7 is isomorphic to the subgroup of all those elements of S8 that leave the number
5 fixed. (yes)
6. A5 has 120 elements. (no, the order of A5 is 60.)
7. Sn is not cyclic for n ≥ 1. (no, S2 ∼
= Z2 is cyclic)
8. Every cycle is a permutation. (yes)
Problem 3. Find all subgroups of the given group and calculate the order of each subgroup.
1. Z27 .
2. Z24 .
Problem 4. Let Sn denote the symmetric group that are permutations of {1, 2, · · · , n}.
1
1. Express the permutation
1 2 3 4 5 6 7 8
ρ=
∈ S8
3 5 7 2 6 4 8 1
as a product of disjoint cycles, and then as a product of transpositions.
solution. (1,3,7,8)(2,5,6,4)=(1,8)(1,7)(1,3)(2,4)(2,6)(2,5).
2. Compute the product of cycles (1, 4, 5)(7, 8)(2, 5, 6) ∈ S8 .
solution.
1 2 3 4 5 6 7 8
4 1 3 5 6 2 8 7
Problem 5. Let A be a finite set, B ⊂ A be a subset, and let b ∈ B be one particular
element of B. Determine whether the given set is a subgroup of SA under induced
operation. Here σ[B] = {σ(x) | x ∈ B}
1. {σ ∈ SA | σ[B] = B}.
2. {σ ∈ SA | σ[B] ⊂ B}
Solution: We will only deal with the first part, since by our assumption A (,hence B
also) is a finite set then σ[B] ⊂ B is equivalent to σ[B] = B.
Let S := {σ ∈ SA | σ[B] = B}, which is a subset of the group SA . We claim that S
is a subgroup of SA .
To show that S is a subgroup, we only need to show that for any f, g ∈ S we will
havef ◦ g −1 ∈ S. To see this, one notice that f [B] = B, since f is a permutation. So we
have f −1 : B → B, this together with the fact that g : B → B imply the composition
g ◦ f −1 : B → B.
Problem 6. Section 9, Problem 13 and Section 8, Problem 21.
 
 
 
1
0
0
1
2
3
solution to Prob 21. Let e1 = 0,e2 = 1,e3 = 0, Then for σ =
we
a1 a2 a3
0
0
1
1 2 3
define a 3 × 3 matrix Aσ = [ea1 , ea2 , ea3 ]. E.g. if σ =
then
2 3 1


0 0 1
Aσ = [e2 , e3 , e1 ] = 1 0 0 .
0 1 0
Now prove the function
ϕ : S3 3 σ → Aσ ∈ M3 (R)( 3-by-3 square matrix )
defined as above is a group homomorphism, (Hint: prove ϕ(στ ) = Aσ Aτ for σ, τ ∈ S3 ,
where the multiplication inside ϕ is the multiplication in S3 and the multiplication on
the right is matrix multiplication)
solution to Prob 13. Done in the class last week.
2
Problem 7. Prove the following about Sn
1. Prove that (1, 2, 3, 4, 5) ∈ S5 can be written as a product of 4 transposition.
Proof. (1, 2, 3, 4, 5) = (1, 5)(1, 4)(1, 3)(1, 2)
2. Prove that every permutation in S5 can be written as a product of at most 4 transposition.
3. Prove that every permutation in Sn can be written as a product of at most n − 1
transposition.
Proof. see the solution to the 3rd part.
Proof. Let σ ∈ Sn , then by Theorem 9.8 σ = µ1 µ2 · · · µr with µi ’s being disjoint
cycles of length ni . Then we have n1 + n2 + · · · + nr = n. Since µi is a cycle,
say µi = (a1 , a2 , · · · , ani ) then by the argument after Definition 9.11 we have
(a1 , a2 , · · · , ani ) = (a1 , ani )(a1 , ani −1 ) · · · (a1 , a3 )(a1 , a2 ) this implies that each µi
can be written as ni − 1 transpositions. So σ can be written as n1 − 1 + n2 − 1 +
· · · + nr − 1 = n − r ≤ n − 1 transposition. So our proof is completed.
4. Prove that (1, 2, 3)(4, 5) ∈ S5 can be written as a product of at most 3 transposition.
Proof. (1, 2, 3)(4, 5) = (1, 3)(1, 2)(4, 5)
3