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Let $G$ be a finite group and $H$ a subgroup of $G$ such that $|G:H|=2$. Suppose $K$ a subgroup of $G$ of odd order. Show $K$ is contained in $H$.

I'm stuck. Need a hint.

Matt Samuel
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Tuo
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2 Answers2

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Hint: Show that $H$ is a normal subgroup (this is because the left cosets are the same as the right cosets), then consider the image of $K$ in the quotient by $H$.

Matt Samuel
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  • Alright thanks I'll work it out. And theres nothing to show, all subgroups of index 2 are normal. – Tuo Jun 04 '15 at 22:23
  • @Andrew this is true, but simply stating something true doesn't prove that it's true. If you're allowed to use it for this then skip that step. – Matt Samuel Jun 04 '15 at 22:24
  • So the image of $K$ in $G/H$ is $\pi(K) = {kH : k \in K}$ and $\pi(K)$ is a subgroup of $G/H$ which is isomorphic to $Z_{2}$. So $\pi(K)$ is either the trivial subgroup (in which case $K$ is a subgroup of $H$) or $\pi(K) \cong Z_{2}$ in which case $\pi(K)$ contains an element of order $2$. But the order order of $\pi(k)$ for any $k\in K$ divides the order of $k$, which is not divisible by $2$ because $K$ has odd order, so we can't have the latter case. I think this works. – Tuo Jun 05 '15 at 00:20
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Since $H$ is normal, going modulo $H$ we can assume that $|G|=2$. Now in group of order $2$, what could be the subgroup of odd order?

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