Prove that $NM$ is a normal subgroup of $G$ if $N$ and $M$ are normal. 
Prove that $NM$ is a normal subgroup of $G$ if $N$ and $M$ are normal.

Proof:
$N$ is normal $\implies$ $gNg^{−1}=N$ for all $g \in G$
$M$ is normal $\implies$ $gMg^{−1}=M$ for all $g \in G$
To prove that $NM$ is normal, we need to show that $gNMg^{−1}=NM$ for all $g \in G$
Take $gNg^{−1}=N$
Multiply both sides by $gMg^{−1}$
$(gNg^{−1})(gMg^{−1}) =NgMg^{−1}$
$gNMg^{−1}=NM$
Hence Proved.
Please correct me if this is wrong.
 A: It seems fine to me.
Alternatively, we have, for all $g\in G$,
$$\begin{align}
NM&=(gNg^{-1})(gMg^{-1})\\
&=gN(g^{-1}g)Mg^{-1}\\
&=gN(e)Mg^{-1}\\
&=g(NM)g^{-1}.
\end{align}$$
A: @Shaun answered your question. However, here I'm writing my complete solution with all details.
There are two facts to be proven: that $NM$ is a subgroup of $G$ at all, and that it is normal in $G$.
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To prove the first, we consider the product of two prototypical elements of $NM$, $n_{1}m_{1}$ and $n_{2}m_{2}$, where $n_{1}, n_{2} \in N$ and $m_{1}, m_{2}\in  M$. This product is $$(n_{1}m_{1})(n_{2}m_{2}) = n_{1}(m_{1}n_{2})m_{2}$$Note that,
because $n_{2} \in N$ and $m_{1} \in G$ and $N \unlhd G$, $m_{1}n_{2}m_{1}^{-1} \in N$; let us denote this expression by $ n_{2}^{'}$ for brevity, so that $m_{1}n_{2} = n_{2}^{'}m_{1}$. Then $n_{1}(m_{1}n_{2})m_{2}= n_{1}n_{2}^{'}m_{1}m_{2}$; since $n_{1}n_{2}^{'}\in N$ and $m_{1}m_{2}\in M$, this product is in $NM$ and we have shown that $NM$ is closed under multiplication.
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Similarly we can see that  $NM$ is closed under inversion. Indeed, note that  let us consider the inverse of a prototypical element of $NM$: $(nm)^{−1}= m^{−1}n
^{−1}$. As above, we may use normality to note that $n′= m^{−1}n^{−1}m \in N$, so $m^{−1}n^{−1} = n'm^{−1}$, and thus $(nm)^{−1}= n′m^{−1}\in N$, so $NM$ is closed under inversion.

Finally, we wish to show that $NM$ is itself normal in $G$, so that for any $g \in G$ and $nm \in NM$,
$gnmg^{−1} \in NM$. This is surprisingly easy: note that $gnmg^{−1}= (gng^{−1})(gmg^{−1})$, and that by
normality the first term of this product is in $N$, and the second in $M$.
