Let $N,M$ be normal subgroups of $G$ with $N\cap M=\{e\}$. Prove that $M\subset C_{G}(N)$ and $N\subset C_{G}(M)$. First consider the following definition:

Let $G$ be a group and $H$ a subgroup of $G$. The center:
$$ C_{G}(H)=\{g\in G\,:\,gh=hg,\,\forall h\in H\}$$

Now I'm trying to prove the following theorem:

Let $N,M$ be normal subgroups of $G$ so $N\cap M=\{e\}$.
Prove that $M\subset C_{G}(N)$ and $N\subset C_{G}(M)$.

What I tried:
We know that $N\triangleleft G$ so there is $n_{0}\in N$ so for every $n\in N$ and every $g\in G$, $n_{0}=g^{-1}ng$. We can convert it to be: $n=gn_0g^{-1}$.
We also know that $M\triangleleft G$ so there is $m_{0}\in M$ so $m=gm_0g^{-1}$.
We can see that:
$$ nm=(gn_0g^{-1})(gm_0g^{-1})=g(n_0m_0)g^{-1}\\
   mn=(gm_0g^{-1})(gn_0g^{-1})=g(m_0n_0)g^{-1}$$
But how I continue from here? As I understand I need to show that $nm=mn$.

It is useful to see $[M,N] := \langle mnm^{-1}n^{-1} |m\in M$ $ n\in N\rangle $. Shall show this is the trivial subgroup. $mnm^{-1} \in N$ and hence $mnm^{-1}n^{-1}\in N$ but then again $nm^{-1}n^{-1}\in M$ and hence $mnm^{-1} \in M$. So each generator is in $M\cap N=e$. Thus every element of $M$ commutes with every element if $N$. This solves your problem.
