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Problem Of The Week # 329 - Oct 23, 2018

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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,929
Here is this week's POTW:

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If $p$ is the smallest prime divisor of the order of a finite group $G$, prove that any subgroup of $G$ of index $p$ is normal.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,929
This week's problem was solved correctly by Olinguito and castor28 . You can read castor28 's solution below.


Let $G$ be a finite group, and $H$ a subgroup of $G$ with $(G:H)=p$, where $p$ is the smallest prime divisor of $|G|$.

The action of $G$ by left multiplication on the cosets of $H$ defines a homomorphism $\varphi:G\to S_p$ where $S_p$ is the symmetric group on $p$ points. Let $K=\ker\varphi$.

If $g\in K$, then $gH=H$; this shows that $K\subset H$. The image of $\varphi$ has order:
$$
(G:K) = (G:H)(H:K) = p(H:K)
$$
This image is a subgroup of $S_p$, which has order $p!$. This shows that $p(H:K)$ divides $p!$, and $(H:K)$ divides $(p-1)!$.

Now, $(H:K) \mid (G:K) \mid |G|$. As the smallest prime divisor of $|G|$ is $p$ and all the prime divisors of $(p-1)!$ are smaller than $p$, we conclude that $(H:K)=1$ and $K=H$. As $K$ is normal in $G$ as the kernel of a homomorphism, this shows that $H\lhd G$.
 
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