48 Chapter 1. Groups I If h, h ∈ H, then hh ∈ H, because subgroups are closed under multiplication, and so HH ⊆ H. For the reverse inclusion, if h ∈ H, then h = h1 ∈ HH (because 1 ∈ H), and so H ⊆ HH. It is possible for two subsets X and Y in S(G) to commute even though their constituent elements do not commute. For example, if H is a nonabelian subgroup of G, then we have just seen that HH = H. Here is another example: let G = S3, let X be the cyclic subgroup generated by (1 2 3), and let Y be the one-point subset {(1 2) . Now (1 2) does not commute with (1 2 3) ∈ X, but (1 2)X = X(1 2). In fact, here is the converse of Exercise 1.57 on page 45. Lemma 1.71. A subgroup K of a group G is a normal subgroup if and only if gK = Kg for every g ∈ G. Thus, every right coset of a normal subgroup is also a left coset. Proof. Let gk ∈ gK. Since K is normal, gkg−1 ∈ K, say gkg−1 = k ∈ K, so that gk = (gkg−1)g = k g ∈ Kg, and so gK ⊆ Kg. For the reverse inclusion, let kg ∈ Kg. Since K is normal, (g−1)k(g−1)−1 = g−1kg ∈ K, say g−1kg = k ∈ K. Hence, kg = g(g−1kg) = gk ∈ gK and Kg ⊆ gK. Therefore, gK = Kg when K G. Conversely, if gK = Kg for every g ∈ G, then for each k ∈ K, there is k ∈ K with gk = k g that is, gkg−1 ∈ K for all g ∈ G, and so K G. • A natural question is whether HK is a subgroup when both H and K are subgroups. In general, HK need not be a subgroup. For example, let G = S3, let H = (1 2) , and let K = (1 3) . Then HK = {(1), (1 2), (1 3), (1 3 2)} is not a subgroup because it is not closed: (1 3)(1 2) = (1 2 3) / ∈ HK. Alternatively, HK cannot be a subgroup because |HK| = 4 is not a divisor of 6 = |S3|. Proposition 1.72. (i) If H and K are subgroups of a group G, at least one of which is normal, then HK is a subgroup of G moreover, HK = KH in this case. (ii) If both H and K are normal subgroups, then HK is a normal subgroup. Remark. Exercise 1.80 on page 59 shows that if H and K are subgroups of a group G, then HK is a subgroup if and only if HK = KH. Proof. (i) Assume first that K G. We claim that HK = KH. If hk ∈ HK, then k = hkh−1 ∈ K, because K G, and hk = hkh−1h = k h ∈ KH. Hence, HK ⊆ KH. For the reverse inclusion, write kh = hh−1kh = hk ∈ HK. (Note that the same argument shows that HK = KH if H G.)

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2010 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.