Prove: $G \cong M \times N$ and $G$ is finite $\Rightarrow order(N)$ is not divisible by 5

Question

I have the following questions about things I have to prove for my math study:

Let $G$ be a group with the property that $(gh)^5 = g^5h^5$ for all $g, h \in G$.
Let $M = $ {$g \in G | g^5 = e$} and $N = $ {$g^5 | g\in G$}

Given that $f: M \hookrightarrow G \twoheadrightarrow G/N$ is an isomorphism,

Prove:
a) $G \cong M \times N$
b) $G$ is finite $\Rightarrow order(N)$ is not divisible by 5

I already proved that $M$ and $N$ are normal subgroups of $G$, and by the first isomorphism theorem that $G/M \cong N$. I also proved that $M \cap N =$ {$e$} (with $e$ of course the identity element) and $mn = nm$ for all $m \in M, n \in N$

Could you explain me how to solve a) and b)?

thanks in advance!

Answer 1

b) The map $\theta(n)=n^5$ is an automorphism of n. If would $5 \mid |N|$ then there would exist an element of $N$ that its order is 5, so $\theta$ wouldn't be a monomorphism. This contradiction shows that 5 must not divide the order of $N$.