Let N and M be normal subgroups of a group G and assume that N and M have only one element in common. Prove that N is contained in $C_G(M)$.
First I concluded that |NM|=|N|*|M|.
Now I'm trying to show that nm=mn from every n in N and m in M. Since they are both normal I can write:
$mn=g^{-1}m_1gg^{-1}n_1g= g^{-1}n_1m_1g$
Now what? Any clues?
You have from normality that $n^{-1}mn\in M$, so by group closure, $m^{-1}n^{-1}mn\in M$. Also by normality, you have $m^{-1}n^{-1}m\in N$ (Using $m^{-1}$ as your conjugator and $n^{-1}$ as your element of $N$), so by closure and right multiplying by $n$ we have $m^{-1}n^{-1}mn\in N$. Hence $m^{-1}n^{-1}mn\in M\cap N=\{e\}$, so $m^{-1}n^{-1}mn=e$, and the result follows.