MHB Is Any Subgroup of a Finite Group with Smallest Prime Order Divisor Normal?

  • Thread starter Thread starter Euge
  • Start date Start date
Euge
Gold Member
MHB
POTW Director
Messages
2,072
Reaction score
245
Here is this week's POTW:

-----
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.

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
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 $H\lhd G$.
 
Back
Top