Originally Posted by

**topsquark** I can show that if H, K < G and HK = KH then HK < G. That's easy. I am having trouble with the other direction: If H, K < G and HK < G then HK = KH.

Here's what I have. (The language may not be quite as precise as it needs to be.)

We know that $\displaystyle hk \in HK$, where h is in H and k is in K, by definition. Since HK is a group we also know that $\displaystyle (hk)^{-1} \in HK$. But $\displaystyle (hk)^{-1} = k^{-1}h^{-1} \in KH$. Since h and k are arbitrary in H and K respectively we can state that HK = KH.

My difficulty with this line of reasoning is that I suspect myself of assuming that KH is also a subgroup of G. Does my argument have any flaws? Thanks.

-Dan