# Thread: existence of a subgroup

1. ## existence of a subgroup

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

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

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

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

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