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

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

• This is a more general version of the fact that a subgroup of index $2$ is always normal – So Lo Dec 11 '18 at 15:33
• 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 $HK = H$. – 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 $$K=H$$.