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Let $G$ be group and $p$ be the least prime divisor of $|G|$.Then any subgroup of $G$ of index $p$ (if it exists) is normal in $G$.

My attempt $:$

Let $H \leq G$ be such that $[G:H] = p$.Let us consider the action of $G$ on the set $A$ of all left cosets of $H$ in $G$ (where the action is the left multiplication). Let $\pi_H$ denote the permutation representation of this action.Let $K = Ker\ \pi_H$. Then $K$ is a subgroup of $H$ (In fact $K$ is the largest normal subgroup of $G$ contained in $H$). Let $[H:K]=k$.Then $[G:K]=[G:H][H:K] = pk$. Since $\pi_H : G \longrightarrow Sym (A)$ is a homomorphism so by first isomorphism theorem we have $G/K \cong \pi_H (G)$. Since $|A| = [G:H] = p$. So $Sym (A) \cong S_p$. Hence $\pi_H (G)$ is a subgroup of $S_p$. This shows that $pk = |G/K| = |\pi_H (G)|$ divides $|S_p|=p!$ i.e. $k | (p-1)!$. Now since $k$ divides $|G|$ so if $k$ is prime or composite then by minimality of $p$, $k$ has to have a prime divisor $q \geq p$.Then $q | (p-1)!$ and since $q$ is prime so $q|(p-k)$ for some $1 \leq k \leq p-1$, a contradiction since $q \geq p$.Hence $k = 1$ and consequently $H = K$, completing the proof.

Is the above reasoning correct at all? Please verify it.

Thank you in advance.

Arnab Chattopadhyay.
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  • Your reasoning works fine. :) – pisco Aug 31 '17 at 07:24
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    Since you ask specifically about your own attempt, I won't mark this as a duplicate, but you might find the various answer to the question at https://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index/ interesting. Your answer here looks good though. – Tobias Kildetoft Aug 31 '17 at 07:29

1 Answers1

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Here is a totally other basic approach for finite groups.

Proposition Let $p$ be the smallest prime dividing the order of the finite group $G$, and assume that the subgroup $H$ has index $p$. Then $H \lhd G$.

Lemma Let $H \leq G$ with $G=HH^g$ for some $g \in G$. Then $G=H$.

Proof Apparently $g=hk^g$, for some $h,k \in H$. Hence $g=hg^{-1}kg$, from which one derives that $g=kh \in H$. Hence $H^g=H$, so $G=HH^g=HH=H$.

Let us proceed with a proof of the Proposition: assume that $H$ is not normal, so there exists a $g \in G$ with $H^g \ne H$. Since $H \ne G$, we have by the Lemma $|G| \gt |HH^g|$. But $|HH^g|=\frac{|H| \cdot |H^g|}{|H \cap H^g|}$. This shows that $|G:H|=p \gt |H^g:H \cap H^g|$. We conclude that $|H^g:H \cap H^g|=1$, as $p$ is the smallest prime dividing $|G|$. This means that $H^g=H \cap H^g$, so $H^g \subseteq H$, implying $H=H^g$, since $H^g$ and $H$ have the same order. We arrive at a contradiction.

Nicky Hekster
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