Math 4281 – Selected solutions to Homework 4
5.22. Prove that if G is a group, a ∈ G, and ab = b for some b ∈ G, then a is the identity
element of G.
Proof. Suppose that ab = b. Multiplying both sides of this equation by b−1 on the right gives
(ab)b−1 = bb−1
⇒ a(bb−1 ) = e (associativity)
⇒ ae = e
⇒ a = e.
7.13. Prove that if H and K are subgroups of a group G, then H ∩ K is a subgroup of G.
Proof. We apply the subgroup test (Theorem 7.1) to H ∩ K.
• Because H and K are both subgroups, they both contain the identity element e, and
therefore e ∈ H ∩ K. In particular, H ∩ K is non-empty.
• Suppose x, y ∈ H ∩ K. This means x, y ∈ H and x, y ∈ H. Since H is a subgroup of
G, we can conclude xy ∈ H. Likewise, K is a subgroup so we conclude xy ∈ K. Thus
xy ∈ H ∩ K; that is H ∩ K is closed under products.
• Suppose x ∈ H ∩ K, so that x ∈ H and x ∈ K. Then x−1 ∈ H and x−1 ∈ K since H
and K are subgroups. Therefore x−1 ∈ H ∩ K; that is, H ∩ K is closed under inverses.
#2. Find the set I of all permutations in S4 which are their own inverses. Is I a subgroup of
S4 ?
Solution. The set I is
{1234, 1243, 1324, 1432, 2134, 2143, 3214, 3412, 4231, 4321}.
It is not a subgroup: although it is certainly closed under inverses and contains the identity,
it isn’t closed under products: 2143 · 3214 = 4123, but 4123 ∈
/ I (since 4123−1 = 2341).
In the case of S4 it’s feasible to find I just by checking all 24 elements of S4 and their
inverses, but there is a more clever way to do this. Remember that if σ ∈ Sn is a permutation,
σ −1 is defined by the property that σ(i) = j if and only if σ −1 (j) = i. Thus, saying that
σ = σ −1 is equivalent to saying σ(i) = j if and only if σ(j) = i for every i, j. That is, σ
switches the numbers 1, 2, . . . , n in pairs somehow (or leaves some of them alone).
For example, σ = 463152 is its own inverse, and sends 1 to 4 and 4 to 1; it sends 2 to 6
and 6 to 2; it sends 3 to 3; and it sends 5 to 5. On the other hand, σ = 436152 is not its own
inverse, which you can tell because it sends 2 to 3 but doesn’t send 3 to 2.
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So, what you can do to compute I is to find all the possible ways of pairing off the numbers
1, 2, 3, 4, where some numbers may remain unpaired. Here they are, along with the corresponding self-inverse permutations (sometimes called “involutions”):
1, 2, 3, 4
1234
12, 3, 4
2134
13, 2, 4
3214
14, 2, 3
4231
23, 1, 4
1324
24, 1, 3
1432
34, 1, 2
1243
12, 34
2143
13, 24
3412
14, 23
4321
My notation here is that, for example, “13, 2, 4” means to pair 1 with 3 and leave 2, 4 unpaired.
Thus 13, 2, 4 corresponds with the self-inverse permutation 3214, since that switches 1 and 3
and maps 2 and 4 to themselves.
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