Subgroup proof

**jzellt** · November 19th 2010, 12:49 PM

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

**Chris L T521** · November 19th 2010, 01:06 PM

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?

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