# A Subgroup of the Smallest Prime Divisor Index of a Group is Normal

## Problem 105

Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$.

Then prove that any subgroup of index $p$ is a normal subgroup of $G$.

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Contents

## Hint.

Consider the action of the group $G$ on the left cosets $G/H$ by left multiplication.

## Proof.

Let $H$ be a subgroup of index $p$.
Then the group $G$ acts on the left cosets $G/H$ by left multiplication.

It induces the permutation representation $\rho: G \to S_p$.

Let $K=\ker \rho$ be the kernel of $\rho$.
Since $kH=H$ for $k\in K$, we have $K\subset H$.
Let $[H:K]=m$.

By the first isomorphism theorem, the quotient group $G/K$ is isomorphic to the subgroup of $S_p$, thus $[G:K]$ divides $|S_p|=p!$ by Lagrange’s theorem.
Since $[G:K]=[G:H][H:K]=pm$, we have $pm|p!$ and hence $m|(p-1)!$.

If $m$ has a prime factor $q$, then $q\geq p$ since the minimality of $p$ but the factors of $(p-1)!$ are only prime numbers less than $p$.
Thus $m|(p-1)!$ implies that $m=1$, hence $H=K$. Therefore $H$ is normal since a kernel is always normal.

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