# Thread: existence of a subgroup

1. ## existence of a subgroup

Let $G$ be a finite group in which $(ab)^{p}=a^{p}b^{p}$ for some prime $p$ dividing the order of $G$. Prove that if $P$ is the p-sylow subgroup of $G$, then there exists a normal subgroup $N$of $G$ with $P \cap N=(e)$ and $PN=G$.

All i can see is that the mapping $x \mapsto x^{p}$ is a homomorphism and the set $S=\{ x^{p} \ | \ x \in G\}$ and $K=\{ x^{p-1} \ | \ x \in G\}$ are normal and have trivial intersection.

2. Originally Posted by Chandru1
Let $G$ be a finite group in which $(ab)^{p}=a^{p}b^{p}$ for some prime $p$ dividing the order of $G$. Prove that if $P$ is the p-sylow subgroup of $G$, then there exists a normal subgroup $N$of $G$ with $P \cap N=(e)$ and $PN=G$.

All i can see is that the mapping $x \mapsto x^{p}$ is a homomorphism and the set $S=\{ x^{p} \ | \ x \in G\}$ and $K=\{ x^{p-1} \ | \ x \in G\}$ are normal and have trivial intersection.
let $|P|=p^m.$ define $f: G \longrightarrow G$ by $f(g)=g^{p^m}.$ since $(ab)^{p^m}=a^{p^m}b^{p^m},$ for all $a,b \in G,$ the map $f$ is a group homomorphism. the kernel of $f$ is a p-subgroup of $G$ which contains $P$ and thus

$\ker f=P.$ now let $N=\text{Im}(f).$ see that $N$ is a normal subgroup of $G$ and $P \cap N=\{1 \}.$ hence, since $G/P \cong N,$ we have $|PN|=|P||N|=|G|,$ and therefore $PN=G.$