normality of subgroups of prime index

Proposition.

If $H$ is a subgroup of a finite group $G$ of index $p$, where $p$ is the smallest prime dividing the order of $G$, then $H$ is normal in $G$.

Proof.

Suppose $H\leq G$ with $|G|$ finite and $|G:H|=p$, where $p$ is the smallest prime divisor of $|G|$, let $G$ act on the set $L$ of left cosets of $H$ in $G$ by left , and let $\varphi:G\rightarrow S_{p}$ be the http://planetmath.org/node/3820homomorphism induced by this action. Now, if $g\in\ker\varphi$, then $gxH=xH$ for each $x\in G$, and in particular, $gH=H$, whence $g\in H$. Thus $K=\ker\varphi$ is a normal subgroup of $H$ (being contained in $H$ and normal in $G$). By the First Isomorphism Theorem, $G/K$ is isomorphic to a subgroup of $S_{p}$, and consequently $|G/K|=|G:K|$ must http://planetmath.org/node/923divide $p!$; moreover, any divisor of $|G:K|$ must also $|G|=|G:K||K|$, and because $p$ is the smallest divisor of $|G|$ different from $1$, the only possibilities are $|G:K|=p$ or $|G:K|=1$. But $|G:K|=|G:H||H:K|=p|H:K|\geq p$, which $|G:K|=p$, and consequently $|H:K|=1$, so that $H=K$, from which it follows that $H$ is normal in $G$. ∎

Title normality of subgroups of prime index NormalityOfSubgroupsOfPrimeIndex 2013-03-22 17:26:38 2013-03-22 17:26:38 azdbacks4234 (14155) azdbacks4234 (14155) 13 azdbacks4234 (14155) Theorem msc 20A05 Coset GroupAction ASubgroupOfIndex2IsNormal