Abstract Algebra – Normal Subgroup of Prime Index

abstract-algebrafinite-groupsgroup-theorynormal-subgroups

Generalizing the case $p=2$ we would like to know if the statement below is true.

Let $p$ the smallest prime dividing the order of $G$. If $H$ is a subgroup of $G$ with index $p$ then $H$ is normal.

Best Answer

This is a standard exercise, and the answer is that the statement is true, but the proof is rather different from the elementary way in which the $p=2$ case can be proven.

Let $H$ be a subgroup of index $p$ where $p$ is the smallest prime that divides $|G|$. Then $G$ acts on the set of left cosets of $H$, $\{gH\mid g\in G\}$ by left multiplication, $x\cdot(gH) = xgH$.

This action induces a homomorphism from $G\to S_p$, whose kernel is contained in $H$. Let $K$ be the kernel. Then $G/K$ is isomorphic to a subgroup of $S_p$, and so has order dividing $p!$. But it must also have order dividing $|G|$, and since $p$ is the smallest prime that divides $|G|$, it follows that $|G/K|=p$. Since $|G/K| = [G:K]=[G:H][H:K] = p[H:K]$, it follows that $[H:K]=1$, so $K=H$. Since $K$ is normal, $H$ was in fact normal.

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