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

  • $\begingroup$ 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

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 $K=H$.


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