A Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is Normal

Problem 470

Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$.
Then prove that $H$ is a normal subgroup of $G$.

(Michigan State University, Abstract Algebra Qualifying Exam)

Proof.

Let $G/H$ be the set of left cosets of $H$.
Then the group $G$ acts on $G/H$ by the left multiplication.
This action induces the permutation representation homomorphism
$\phi: G\to S_{G/H},$ where $S_{G/H}$ is the symmetric group on $G/H$.
For each $g\in G$, the map $\phi(g):G/H \to G/H$ is given by $x\mapsto gx$.

By the first isomorphism theorem, we have
\begin{align*}
G/\ker(\phi) \cong \im(\phi) < S_{G/H}.
\end{align*}
This implies the order $|G/\ker(\phi)|$ divides the order $|S_{G/H}|=p!$.

Since
$|G/\ker(\phi)|=\frac{|G|}{|\ker(\phi)|}=\frac{p^n}{|\ker(\phi)|}$ and $p!$ contains only one factor of $p$, we must have either $|\ker(\phi)|=p^n$ or $|\ker(\phi)|=p^{n-1}$.

Note that if $g\in \ker(\phi)$, then $\phi(g)=\id:G/H \to G/H$.
This yields that $gH=H$, and hence $g\in H$.
As a result, we have $\ker(\phi) \subset H$.

Since the index of $H$ is $p$, the order of $H$ is $p^{n-1}$.
Thus we conclude that $|\ker(\phi)|=p^{n-1}$ and
$\ker(\phi)=H.$

Since every kernel of a group homomorphism is a normal subgroup, the subgroup $H=\ker(\phi)$ is a normal subgroup of $G$.

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