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

Michigan State University Abstract Algebra Exam Problems and Solutions

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)

LoadingAdd to solve later


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
G/\ker(\phi) \cong \im(\phi) < S_{G/H}.
This implies the order $|G/\ker(\phi)|$ divides the order $|S_{G/H}|=p!$.

\[|G/\ker(\phi)|=\frac{|G|}{|\ker(\phi)|}=\frac{p^n}{|\ker(\phi)|}\] 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

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

LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory
Purdue University Abstract Algebra Exam Problems and Solutions
If Squares of Elements in a Group Lie in a Subgroup, then It is a Normal Subgroup

Let $H$ be a subgroup of a group $G$. Suppose that for each element $x\in G$, we have $x^2\in H$....