# Math Help - Subgroup proof

1. ## Subgroup proof

Assume that H and K are subgroups of a group G and that K is normal in G. Then
HK is a subgroup of G, where

HK = {hk | h e H, k e K}

Thanks for any help

2. Originally Posted by jzellt
Assume that H and K are subgroups of a group G and that K is normal in G. Then
HK is a subgroup of G, where

HK = {hk | h e H, k e K}

Thanks for any help
First note that $e\in HK$ (easy to verify).

Let $h_1k_1, h_2k_2\in HK$, we seek to show that $(h_1k_1)(h_2k_2)^{-1}\in HK$

So $(h_1k_1)(h_2k_2)^{-1}=h_1k_1k_2^{-1}h_2^{-1}$.

Now, we see that $h_1k_1k_2^{-1}h_2^{-1}=(h_1h_2^{-1})(h_2 k_1k_2^{-1}h_2^{-1})$. Observe that $h_1h_2^{-1}\in H$ and since $K\lhd G$, $h_2(k_1k_2^{-1})h_2^{-1}\in K$.

Thus $(h_1k_1)(h_2k_2)^{-1}\in HK\implies HK\leq G$.

Does this make sense?