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Let $G$ be a finite group. $ H,K \leq G $ and $ K \lhd G $.

$G:H$ and $|K|$ are coprime. Show that $K \leq H $

I started like this:

$G:H = (G:KH)(KH:H)$

Therefore, both $(G:KH)$ and $(KH:H)$ are coprime to $|K|$, but have no idea how to continue. Any clues?

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@JackSchmidt: Not quite; in the one you link to, we have $N\triangleleft G$, $[G:N]$ coprime to $H$ (index of the normal subgroup. order of the other subgroup). Here we the index of the other subgroup and the order of the normal subgroup. – Arturo Magidin Jun 18 '12 at 16:38
Agreed. The question… is similar, but not a duplicate. – Jack Schmidt Jun 18 '12 at 16:59
up vote 1 down vote accepted

$\eqalign{ & [G:KH]=\frac{\vert G\vert}{\vert KH \vert} \cr & =\frac{\vert G \vert}{\vert H \vert}\frac{\vert H\cap K\vert}{\vert K \vert} \cr & =[G:H]\frac{\vert H\cap K\vert}{\vert K \vert} }$

$[G:H]$ and $\vert K\vert$ are coprime,hence $\vert K \vert$ divides $\vert H\cap K\vert$. But $\vert H\cap K\vert \leq \vert K \vert$,so,...

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I did almost the same calculation but just didn't see it, thanks! – catch22 Jun 18 '12 at 12:03

The key is to show that $|HK|=|H||K|/|H\cap K|$

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or equivalently $|KH:H|=|K:H \cap K|$ – Ben Jun 18 '12 at 9:30

From the second isomorphism theorem you know that $HK/K\cong H/(H\cap K)$. Thus, $$[HK:\!K]=[H:\!H\cap K]\;,$$ and therefore

$$[HK:\!K]|K|=|HK|=[HK:\!H][H:\!H\cap K]|H\cap K|=[HK:\!H][HK:\!K]|H\cap K|\;.$$

Cancelling $[HK:\!K]$, we get $$|K|=[HK:\!H]|H\cap K|\;.$$

Now just use your observation about $[HK:\!H]$.

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I could tell all of these things but not put them together... thank you! – catch22 Jun 18 '12 at 12:58

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