Let $G$ be a finite group, and let $H$ be a subgroup of $G$ such that the index $(G\colon H)$ of $H$ in $G$ is the smallest prime that divides the order of $G$. Can we say anything about whether or not $H$ is normal in $G$?

The answer is of course yes if $G$ is abelian.

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    $\begingroup$ Yes $H$ is normal in $G$. Consider the permutation representation of $G$ on the cosets of $H$, and show that the image must have order $|G:H|$, so $H$ must be the kernel. $\endgroup$ – Derek Holt Apr 18 '14 at 7:53
  • $\begingroup$ Derek Holt, what is the permutation representation of $G$ on the cosets of $H$? $\endgroup$ – Saaqib Mahmood Apr 18 '14 at 8:05

Denote set of left cosets by $G/H$ and let $S\left(G/H\right)$ denote the symmetric group of bijections on it.

Then $f:G\rightarrow S\left(G/H\right)$ defined by $g\mapsto\lambda_{g}$, where $\lambda_{g}\in S\left(G/H\right)$ on its turn is defined by $xH\mapsto gxH$, is a grouphomomophism with kernel $N:=\cap_{x\in G}xHx^{-1}$.

Here $S\left(G/H\right)$ and $S_{p}$ are isomorphic and $G/N\simeq f\left(G\right)\leq S\left(G/H\right)$ so $\left[G:N\right]$ divides $p!$.

We have $N\subset H\subset G$ with $\left[G:N\right]=p\left[H:N\right]$ so that $\left[H:N\right]$ divides $\left(p-1\right)!$ and $\left|G\right|$.

But $\left|G\right|$ and $\left(p-1\right)!$ are coprime leading to $\left[H:N\right]=1$ or equivalently $H=N$.

As kernel of $f$ subgroup $H$ is normal.

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    $\begingroup$ I suppose you wanted to say $\lambda_g(xH) = gxH$. $\endgroup$ – Christoph Apr 18 '14 at 9:20
  • $\begingroup$ @Christoph Indeed. Thank you. Repaired. $\endgroup$ – drhab Apr 18 '14 at 9:27

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