4
$\begingroup$

Suppose that $H$ is a subgroup of a finite group $G$ and that $|H|$ and $\big([G:H]-1\big)!$ are relatively prime. Prove that $H$ is normal in G

Let $[G:H]=m$

Let $G$ act on set $A$ of left cosets of $H$ in $G$ under left multiplication, then we have

$$\phi : G \to S_A\simeq S_m$$

where $$K= \ker\phi \subseteq H$$ $$\frac{G}{K} \simeq \phi(G)\leq S_m$$

So $$[G:K] \mid m!$$

$$\Rightarrow [G:H][H:K]\mid m!$$

$$\Rightarrow [H:K]|(m-1)!$$

But $\gcd\big(|H|,(m-1)!\big) = 1 \Rightarrow |K|>1 $

How do I proceed to prove $K=H \lhd G$?

Or is this approach wrong

$\endgroup$
  • $\begingroup$ This is a more general version of the fact that a subgroup of index $2$ is always normal $\endgroup$ – So Lo Dec 11 '18 at 15:33
  • $\begingroup$ Instead you can note that the condition implies that any prime divisor in $|H|$ is greater than the index of $H$. Then if $K$ is any subgroup of the same order as $H$ you can see from the possible order that $HK = H$. $\endgroup$ – Tobias Kildetoft Dec 11 '18 at 15:39
4
$\begingroup$

Note that $[H:K]$ divides both $|H|$ and $\big([G:H]-1\big)!=(m-1)!$. Therefore, it divides the greatest common divisor of $|H|$ and $\big([G:H]-1\big)!$, which is $1$. Hence, $[H:K]=1$, making $K=H$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.