# Thread: prime index

1. ## prime index

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

Proof:

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

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

Now what do I do?

2. ## Re: prime index

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

Proof:

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

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

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

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