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?
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.