2
$\begingroup$

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?

  • 0
    @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.2012-06-18
  • 0
    Agreed. The question http://math.stackexchange.com/questions/68107/showing-a-normal-subgroup-contains-a-subgroup is similar, but not a duplicate.2012-06-18

3 Answers 3

1

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

$[G:H]$ 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,...

  • 0
    I did almost the same calculation but just didn't see it, thanks!2012-06-18
2

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

  • 0
    or equivalently $|KH:H|=|K:H \cap K|$2012-06-18
2

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 $[HK:\!H]$.

  • 0
    I could tell all of these things but not put them together... thank you!2012-06-18