question about normal subgroups If $N$ is a normal subgroup of $G$ and $M$ is a normal subgroup of $G$, and if $MN=\{mn|m\in M,n\in N\}$, prove that $MN$ is a subgroup of $G$ and that $MN$ is a normal subgroup of $G$.
The attempt:
I tried just starting by showing that $MN$ is a subgroup of $G$. I said let $a=m_1  n_1$ for some $m_1 \in M$ and $n_1 \in N $ and let $b=m_2 n_2$ for some $m_2 \in M$ and $n_2 \in N$, and we need to show $a*b^{-1}$ $\in MN$. 
So I get $a*b^{-1}$=$m_1n_1n_2^{-1}m_2^{-1}=m_1n_3m_2^{-1}$ but then I don't know how to show that this is in $MN$. Tips on this or the next part of the problem?

$$m_1n_3m_2^{-1}=m_1m_2^{-1}(m_2n_3m_2^{-1})\in MN$$
General Lemma: if $\,M,N\,$ are subgroups of $\,G\,$ , $\,MN\,$ is a subgroup iff $\,MN=NM\,$ .
In particular, if $\,M\triangleleft G\,$ or$\,N\triangleleft G\,$ ,then $\,MN=NM\,$
