= Product of a subgroup with a normal subgroup
{title2=$H\leq G,\ K\trianglelefteq G\ \Longrightarrow\ HK\leq G$}
If $H$ is a <subgroup> and $K$ a <normal subgroup> of a <group> $G$, their product set $HK=\{hk:h\in H,k\in K\}$ is a <subgroup>. Normality gives $(h_1k_1)(h_2k_2)=h_1h_2(h_2^{-1}k_1h_2)k_2\in HK$ and $(hk)^{-1}=h^{-1}(hk^{-1}h^{-1})\in HK$. Together with the <identity element> this proves the <subgroup test>. For finite <subgroups>, $|HK|=|H||K|/|H\cap K|$: the pairs representing one product differ exactly by multiplying the first factor by an element of $H\cap K$ and its second factor by its inverse.
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