Proving that if $(ab)^{p}=a^{p}\,b^{p}$, then the p-sylow subgroup is normal So, while studying Abstract Algebra, i ran into this problem (i.n. Herstein, second edition, chapter 2.12) and have been stuck since: 
Given a group of finite order, and a prime $p$ that divides $o(G)$, and suppose that $\forall\,a,b\in G$, $(ab)^{p}=a^{p}\,b^{p}$. Prove that the $p$-sylow subgroup is normal in $G$.
What I've tried: I defined a mapping $\varphi:G\to H=\lbrace x^{p}:\,x\in G\rbrace;\,\,\varphi(x)=x^{p}$, wich would be a surjective homomorphism. Then, I proved that $\ker(\varphi)\subseteq P$, where $P$ is a $p$-sylow subgroup. If I could prove either that $P\subseteq\ker(\varphi)$ or that $o(\ker(\varphi))=o(P)$, that would end it, because that would imply that $P=\ker(\varphi)$, and I know that $\ker(\varphi)\unlhd G$. 
One ideia to follow up on those would be to use the firs isomorphism theorem ($G/\ker(\varphi)\simeq Im(\varphi)$) to get $o(G)=o(\ker(\varphi))o(Im(\varphi))$ and from there work something out about the orders, but I cannot think of how to do that.
I've also tried proving that $G$ only has one $p$-sylow subgroup, using Sylow's third theorem, but I believe it's a dead end.
Any ideas?
 A: Let $\varphi:G\rightarrow G$ be defined by $\varphi(x)=x^p$.  $\varphi(x)$ is a homomorphism because $\varphi(xy)=(xy)^p=x^py^p$ by assumption.  Suppose that $p^q$ is the largest power of $p$ which divides $o(G)$.
Consider $\varphi^q$ so that $\varphi^q(x)=x^{p^q}$.  Since $\varphi^q$ is a composition of homomorphisms, it is a homomorphism.  Moreover, note that the kernel of $\varphi^q$ consists of all elements of $G$ whose order is $p^r$ for some $r$.
Since every element of order $p^r$ is in some Sylow-$p$ subgroup, this means that $\ker(\varphi^q)$ is exactly the union of all of the Sylow-$p$ subgroups.  Moreover, the order of $\ker(\varphi^q)$ is a power of $p$ (since otherwise, by Cauchy's theorem, there would be an element of order $p'$ where $p'$ is a prime not equal to $p$).  Hence, $\ker(\varphi^q)$ is a $p$-subgroup of $G$ containing a Sylow-$p$ subgroup of $G$, so $\ker(\varphi^q)$ must be a Sylow-$p$ subgroup.  Therefore, there is only one Sylow-$p$ subgroup and it is $\ker(\varphi^q)$ which is normal.
A: Let's try  mathematical induction with respect to the group order $n = |G|$.
The claim is obiviously true for $n=1$. Let it be true for $n=N$. To show it is true for $n=N+1$, let $|G| = N+1$ with $p^\alpha \big | |G|$ and $\alpha$ is the largest power of $p$ dividing $N+1$. And $(ab)^p = a^p b ^p$
Let $K = \ker(\phi)$
First notice that,  $p \big | |K|$. Consider the quotient group $G/K$, assuming $K$ is not the $p$ syllow s.g (otherwise we are done), 
By induction $G/K$ {has order $< N+1$, and satisfies $(ab)^p = a^p b ^p$ } contains a p-syllow s.g $\tilde{P}$ which is normal in $G/K$. Let $P = \{x\in G| xK \in \tilde{P}\}$, which is a normal subgroup in $G$. And we can notice that $p^\alpha \big | |P|$ and hence $P$ contains a p-syllow s.g of $G$ which is normal in $P$ (by induction and verify that $o(P)<N+1$). 
Since $P$ is normal in $G$, the p-syllow s.g contained in $P$ is also normal in $G$.
