1. ## prime index

Prove that if G is a finite group and $H\leq G$ whose index is the least prime dividing the order of G, then $H\triangleleft G$.

Proof:

This is all I could come up with-
$|G|=|G:H|\times |H|$, $|G:H|=p$, and let $|H|=m$.

Now, $|G|=pm$ and $p| \ |G|$.

Now what do I do?

2. ## Re: prime index

Originally Posted by dwsmith
Prove that if G is a finite group and $H\leq G$ whose index is the least prime dividing the order of G, then $H\triangleleft G$.

Proof:

This is all I could come up with-
$|G|=|G:H|\times |H|$, $|G:H|=p$, and let $|H|=m$.

Now, $|G|=pm$ and $p| \ |G|$.

Now what do I do?
Here is the classic way to do this problem. Let $G$ act on $G/H$ (the coset space, not the quotient group) in the usual way. This gives a homomorphism $\phi:G\to S_{[G:H]}$. It's trivial that $\ker\phi\leqslant H$. That said, note that since $[G:\ker\phi]$ divides $|G|$ it cannot be divisible by any smaller prime than $[G:H]$ yet since $[G:\ker\phi]$ divides $|S_{[G:H]}|=[G:H]!$ (by FIT) it clearly then follows that $[G:\ker\phi]=[G:H]$ and so $|\ker\phi|=|H|$. Since $\ker\phi\leqslant H$ this implies that $\ker\phi=H$ and so $H\unlhd G$.

A more interesting proof, if you know double cosets, can be found here on my blog.