Math 546
Homework 6
Due Wednesday, March 15.
Contents of this handout:
1. 546 Problems. All students (students enrolled in 546 and 701I) are required to attempt
these.
2. 701I Problems. Only students enrolled in 701I are required to attempt these. Students not
enrolled in 701I are welcome, of course, to attempt these as well.
3. Bonus Problems. All students (students in 546 and 701I) may turn these in for additional
credit.
4. Examples. I wrote solutions to certain non-assigned problems from the text. These may be
helpful for you when you are working on the problems.
1
546 Problems.
8.1. Carry out the indicated multiplications in S6 .
1 2 3 4 5 6
1 2 3 4 5 6
(a)
◦
.
3 6 1 4 2 5
5 4 3 2 1 6
1 2 3 4 5 6
1 2 3 4 5 6
(b)
◦
.
3 2 5 4 1 6
6 5 4 1 2 3
8.2. Write out each permutation as a product of disjoint cycles, and then as a product of
transpositions. Determine whether each permutation is even or odd.
1 2 3 4 5 6
(a)
.
3 6 1 4 2 5
1 2 3 4 5 6
(b)
.
2 4 6 1 3 5
1 2 3 4 5 6 7 8 9 10 11 12
8.3 c) Determine whether
∈ S12 is even or odd.
6 7 5 9 8 4 11 3 1 12 2 10
8.5. Write down all the elements of S4 , and indicate which ones are in A4 .
8.19 b) Let SZ = {f : Z → Z | f is a bijection}. Hence, SZ is the set of permutations of Z.
Let T be a finite subset of Z, and let K = {f ∈ SZ | for every t ∈ T , we have f (t) ∈ T }.
Is K a subgroup of SZ ? Why or why not?
8.20 Suppose that H ⊆ Sn is a subgroup. Prove that either all the elements of H are even or
else that |H| is even and exactly one-half of the elements of H are even.
Possible strategy: Mimic the proof of Theorem 8.5.
2
701I Problems.
8.21. Let f , g be distinct transpositions.
(a) Suppose that f and g are disjoint. Show that f g is expressible as the product of two
3-cycles.
(b) Suppose that f and g are not disjoint. Show that f g is expressible as a single 3-cycle.
8.22. Let n ≥ 3. Prove that every permutation in An is expressible as a product of 3-cycles.
3
Bonus Problems.
8.23. Find elements x, y ∈ SZ such that x and y each have finite order, yet xy infinite order.
4
Examples.
1 2 3 4 5 6
1 2 3 4 5 6
8.1 c) Carry out the indicated multiplication in S6 .
◦
.
2 4 6 1 3 5
5 6 3 4 1 2
Solution: We have
1 2 3 4 5 6
1
◦
2 4 6 1 3 5
5
1 2
= (1, 3, 6, 4)(2, 5) =
3 5
2 3 4 5 6
= [(1, 2, 4)(3, 6, 5)][(1, 5)(2, 6)]
6 3 4 1 2
3 4 5 6
.
6 1 2 4
1 2 3 4 5 6
8.2 c) Write the permutation
as a product of disjoint cycles, and then
3 6 1 4 2 5
as a product of transpositions. Determine whether the permutation is even or odd.
1 2 3 4 5 6
Solution: We have
= (1, 3)(2, 6, 5) = (1, 3)(2, 5)(2, 6). Therefore,
3 6 1 4 2 5
the permutation is odd.
8.16. Let X be a set, and let Y ⊆ X. Show that H = {f ∈ SX | f (y) = y for all y ∈ Y } is
a subgroup of SX .
Proof: We first show that H 6= ∅. We note, for all x ∈ X, that iX (x) = x. Since Y ⊆ X,
it follows, for all y ∈ Y , that iX (y) = y, which shows that iX ∈ H. To show that H is a
subgroup, we show that it meets the subgroup criteria.
• H is closed. To see this, let f , g ∈ H. Then for all y ∈ Y , we have f (y) = y and
g(y) = y. It follows that (f g)(y) = f (g(y)) = f (y) = y, which shows that f g ∈ H.
• H has inverses. To see this, let f ∈ H. Then for all y ∈ Y , we have f (y) = y. Since
f ∈ H ⊆ SX , we have f −1 ∈ SX . Since f (y) = y for all y ∈ Y , we must have
f −1 (y) = y for all y ∈ Y as well. It follows that f −1 ∈ H.
We conclude that H is a subgroup of SX .
8.19 a) Let H = {f ∈ SZ | for every n ∈ Z+ , we have f (n) ∈ Z+ }. Is H a subgroup of SZ ?
Why or why not?
Claim: H is not a subgroup of SZ .
Proof: To see this, we provide a counter-example. We define f : Z → Z for all n ∈ Z by
f (n) = n + 1. The function f is a bijection with inverse f −1 (n) = n − 1 for all n ∈ Z.
Hence, we see that f ∈ SZ . Furthermore, we observe that f ∈ H. However, since 1 ∈ Z+
but f −1 (1) = 1 − 1 = 0 6∈ Z+ , we see that f −1 6∈ H. Hence, H does not have inverses.
Therefore, it is not a subgroup.