$G$ is a finite group s.t. $(ab)^p=a^pb^p$; $P$ is its $p$-Sylow subgroup. Then there exists $N \unlhd G$ such that $N \cap P = (e)$ and $NP=G$.

Question

This is problem 2.12.15 in Herstein's Topics in Algebra, second edition. The full problem is

Let $G$ be a finite group in which $(ab)^p=a^pb^p$ for every $a,b \in G$, where $p$ is a prime dividing $o(G)$. Prove

(a) The $p$-Sylow subgroup of $G$ is normal in $G$.

$^*$(b) If $P$ is the $p$-Sylow subgroup of $G$, then there exists a normal subgroup $N$ of $G$ with $P \cap N = (e)$ and $PN=G$.

(c) $G$ has a nontrivial center.

I have done part (a) but am stuck at part (b). I have tried different approaches, but they all seem ineffective. I should also say that, at this point in the book, Herstein hasn't even formally introduced direct products, only the Sylow theorems. How am I supposed to solve this?

The next problem, problem 16, basically asks to prove a result called Burnside's p-complement theorem or possibly another called the Schur-Zassenhaus theorem (according to other threads on this website), but that's a double-starred problem, so I was thinking that there must be a simpler solution to this one-star problem.

Answer 1

The map $\phi:a\mapsto a^p$ is a homomorphism from $G$ to $G$. If $|P|=p^r$, then $\phi^s$ has kernel $P$ wherever $s\ge r$. Let $|G|=p^rm$. Choose $s\ge r$ with $p^s\equiv 1\pmod m$. Let $N$ be the image of $\phi^s$. It has order $m$, by the first isomorphism theorem. Its elements all satisfy $a^m=e$. But if $a^m=e$ then $\phi^s(a)=a$, so $N=\{a\in G:a^m=e\}$, and so is a characteristic subgroup of $G$. But obviously, $P\cap N$ is trivial, and so $PN$ must have order $p^rm=|G|$.

Answer 2

Let $N$ be set of elements of $G$ whose order is not divisible by $p$. Let $a,b\in N$ and set $n=lcm(|a|,|b|)$. Since $gcd(p,n)=1$, there exists positive $k$ such that $p^k\equiv 1 \ mod \ n$ (one may take $k=\varphi(n$)).

Now, $$(ab)^{p^k}=a^{p^k}b^{p^{k}}=ab.$$

This shows that $ab\in \langle (ab)^p \rangle$, and so $|ab|=|(ab)^p|$. Thus, the order of $ab$ is not divisible by $p$. It follows that $N$ is a subgroup of $G$. Since $a$ and $g^{-1}ag$ are of same order, $N$ is a normal subgroup of $G$. Clearly, $N\cap P=1$ and $G/N$ is a $p$-group. Thus, $G=NP$ as desired.