# Proving a subgroup is normal

Problem

Let $G$ be a group with $|G|=pm$, $p$ prime and $p \geq m$. Suppose there is $H$ subgroup of $G$ with $[G:H]=p$. Show that $H$ is normal.

This problem was given to me in class just after learning the following proposition:

Proposition

Let $G$ be a group and $H \leq G$ with $[G:H]=n$. Then there exists a morphism $\phi: G \to S_n$ such that $K=ker(\phi) \leq H$, $[G:K] |n!$

I've tried to solve the problem using this proposition but I couldn't arrive to the solution:

If $[H:K]=1$ then $H=K$ and by the proposition we are done.

If $[H:K]>1$, (1) let $q | [H:K]$. We have (2) $[G:K]=\dfrac{[G]}{[H]}\dfrac{[H]}{[K]} | p!$ again by the proposition.

By (1) and (2) it follows $q | (p-1)!$, so there is $1<s \leq p-1$ such that $q$ divides $s$. By the problem hypothesis, we have $q$ divides $m$.

Here I got stuck, maybe this isn't an appropiate approach, if my attempt at a solution can be easily completed, I would appreciate hints or suggestions to do so, if my approach has a dead end, then I would appreciate any suggestions to an alternative solution.

• it is true when $p$ is the smallest prime dividing $|G|$, Be sre about your question. – mesel Nov 26 '14 at 21:43
• With your hypotheses, the Sylow p subgroup is normal. Maybe this was supposed to be the question? – johng Nov 28 '14 at 5:36

$G = D_{10} := \left < \{r. s| r^5 = 1, s^2 = 1, rs = sr^{-1} \} \right>$ be the dihedral group of order $10$ with the standard generators $r, s.$ Let $H := \left< s \right>.$ Then $[G : H ] = 5.$ But $H$ is not normal in $G$, since $rsr^{-1} = r^2s = sr^{-2} \neq s, e.$ [Here $p = 5, m = 2.$]