# Abstract Nonsense

## A Clever Proof of a Common Fact

Point of Post: In this post we give a new proof that if $G$ is a finite group $H$ is a subgroup $G$ whose index is the smallest prime dividing $|G|$ then that subgroup is normal.

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Motivation

It is a commonly used theorem in finite group theory that if $G$ is a finite group and $H\leqslant G$ such that $\left[G:H\right]=p$ is the smallest prime dividing $|G|$ then $H\unlhd G$. We have already seen a proof of this fact by considering the homomorphism $G\to S_p$ which is the induced map from $G$ acting on $G/H$ by left multiplication, and proving that $\ker(G\to S_p)=H$. We now give an even shorter (and the just mentioned proof is already short) proof of this fact using double cosets.

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The Theorem

We recall that a double coset $HgK$ for subgroups $H,K\leqslant G$ is an orbit of the action of $H\times K$ on $G$ given by $(h,k)\cdot g=hgk^{-1}$. We also recall that $\#(HgK)=|H|\left[H:H\cap gKg^{-1}\right]$. With this we can easily prove the theorem

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Theorem: Let $G$ be a finite group and $p$ the smallest prime dividing $|G|$. Then, if $H\leqslant G$ is such that $\left[G:H\right]=p$ then $H\unlhd G$.

Proof: We know that $G$ decomposes as

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$G=H\sqcup Hg_1H\sqcup\cdots\sqcup Hg_mH$

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where $g_1,\cdots,g_m$. We see then that that

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$p=1+\left[H:H\cap g_1Hg_1^{-1}\right]+\cdots+\left[H:H\cap g_mHg_m^{-1}\right]$

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Now, we know then that $\left[H:H\cap g_k Hg_k^{-1}\right] and since it’s a divisor of $|G|$ we evidently mus have that $\left[H:H\cap g_k Hg_k^{-1}\right]=1$ and so $H=g_kHg_k^{-1}$. Since we can evidently choose $g_k$ arbritrarily in $G-H$  (since it is a representative for one of the double cosets) the conclusion follows. $\blacksquare$

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References:

References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2009