Let G be a finite group in which $(ab)^p=a^pb^p$ for every a,b in G where p is prime dividing order of G Then if P is p- Sylow subgroup of G then there exists a normal subgroup N of G with $P\cap N=e $ and PN=G.
I PROVED that P is normal. But unable to proceed further .any hint will be useful
On
An observation in a different direction: let $\phi: G \rightarrow G$ be defined by $\phi(g)=g^p$. Then $\phi$ is a homomorphism. Let $P \in Syl_p(G)$ and put $p^e=exp(P)$, the smallest power of $p$ such that $x^{p^e}=1$ for all $x \in Syl_p(G)$. Observe that $\phi^e$ is a homomorphism, since it is an iteration of mappings ($\phi^e(g)=\phi(\phi(\phi \cdots (g) \cdots))$, $e$ times). Note that $P \subseteq ker(\phi^e)$. On the other hand, $ker(\phi^e)$ is a $p$-subgroup. We conclude that $P=ker(\phi^e)$ and hence $P$ is normal. Now we can use the theorem of Schur-Zassenhaus to provide a complement subgroup $H$ to $P$: $G=HP$ and $H \cap P=1$.
Look at the set of $p$-th powers in $G$, they form a subgroup because $a^pb^p=(ab)^p$, and is clearly characteristic. Moreover, it is a proper subgroup of $G$ by cardinality, so let $N_1$ be this characteristic subgroup. We claim $N_1P=G$ (easy).
Unfortunately $N_1\cap P$ need not be trivial, so look at the characteristic subgroup $N_2$ of $p$-th powers in $N_1$ as needed, repeat. Since the exponent of $P$ is at most $\lvert P\rvert$, this process stops with an $N$ that works.
In fact, it is easy to see that we have $$N=\bigcap_{n=1}^\infty\phi^n(G)\text{ and }P=\bigcup_{n=1}^\infty\ker(\phi^n),$$ where $\phi\colon G\to G$ is the homomorphism $g\mapsto g^p$.