Proving G cannot Equal HK When K Contains a Conjugate of H

[SiddharthM]

I've been trying to prove something that seems obvious but have had no success thus far:

say G is a finite group and H and K are proper subgroups, if K contains a conjugate of H, then it isn't possible to have G=HK.

Proof anybody? I'm happy if one can prove the special case below:

It's fairly easy to show one can't have the product of a proper subgroup and it's conjugate equal to a group i.e. it isn't possible that Hx^{-1}Hx=G unless H=G. Can you show it for an arbitrary product of conjugates? i.e. if $$\Pi_{x \in G} H^x=G$$ then H=G. Note that $$H^x=x^{-1}Hx$$

thanks for the help

 

[rasmhop]

Let c be an element such that $cHc^{-1} \subseteq K$.

Assume G=HK.

Express $c^{-1}$ as a product $c^{-1}=hk$ with h in H and k in K. Then
$$1 = chc^{-1}ck$$
$$(chc^{-1})^{-1}k^{-1} = c$$
both factors on the left hand side are in K so c is in K.

Since c is in K, $H \subseteq c^{-1}Kc = K$. We then have G=HK=K so K isn't a proper subgroup of G.

 

[SiddharthM]

thanks for the prompt reply, i appreciate the help.