Question 933859
Suppose HK is a subgroup. Let *[tex \large h \in H] and *[tex \large k \in K]. It follows that *[tex \large h, k \in HK] since we can write h = h*1, k = k*1. However, HK is closed under products, so *[tex \large kh \in HK]. Therefore *[tex \large KH \subseteq HK].

Also note that *[tex \large (hk)^{-1} \in HK]. Then we can write *[tex \large (hk)^{-1} = xy] where *[tex \large x \in H], *[tex \large y \in K]. This implies that *[tex \large ((hk)^{-1})^{-1} = (xy)^{-1} \Rightarrow hk = (xy)^{-1} = y^{-1}x^{-1} \in KH]. Therefore *[tex \large HK \subseteq KH], so *[tex \large HK = KH].


Now suppose that HK = KH. Let *[tex \large x, y \in HK], so that *[tex \large x = h_1k_1], *[tex \large y = h_2k_2]. Then *[tex \large k_1h_2 \in KH = HK] (by assumption), so suppose *[tex \large k_1 h_2 = hk] for *[tex \large h \in H], *[tex \large k \in K]. Consider the product *[tex \large xy]:


*[tex \large xy = h_1k_1h_2k_2 = h_1hkk_2]


Since H and K are subgroups, *[tex \large h_1h \in H] and *[tex \large kk_2 \in K] by closure. Therefore *[tex \large xy \in HK].


Also note that for any *[tex \large x = h_1k_1 \in HK], we have


*[tex \large x^{-1} = (h_1k_1)^{-1} = k_1^{-1}h_1^{-1} \in KH = HK]


Therefore elements in HK are closed under products and inverses, so HK is a subgroup.