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Let $G$ be a finite group and $K$ be a normal subgroup of $G$. If $\gcd([G:K],|K|)=1$, then $K=\{x^i | x \in G\}$, where $i=[G:K]$.

I know that if we consider $G/K$, since the index is $i$, $g^i \in K$ for all $g\in G$. Hence $\{x^i | x \in G\}\subseteq K.$ But I don't know how to prove the other direction using that they're relatively prime.

It seems like if $K$ is abelian, then by consider the homomorphism $\phi(k)=k^i$, we can have that $\phi$ is one-to-one since $Ker\phi=\{g\in K|g^i=e\}$ can only be trivial. (Otherwise $K$ would have a subgroup with order divides $i$, which is a contradiction.) So every element of $K$ is of the form of $k^i$. This leads to the other direction. However here we don't have abelian. How to show it when $K$ is just normal?

Thanks for any help.

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  • $\begingroup$ Such a subgroup is called a "Hall subgroup", by the way. $\endgroup$ – pg1989 Apr 2 '16 at 20:23
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Let $g\in K$ have order $m$ which is relatively prime to $i$, since $m$ divides $|K|$. Then there exist integers $x, y$ such that $mx+iy=1$. Then $g=g^1=g^{mx+iy}=g^{iy}$, since $g$ has order $m$. Hence, $g=(g^y)^i$, so every element of $K$ is an $i$th power.

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