This is an easy question for me to verify, I suppose, but, for whatever reason, I'm having a crisis of confidence.
This is a proof-verification question.
The bit I'm unsure of is highlighted in bold text.
The Question:
Suppose $G$ is a group with $M,N\unlhd G$. Show that $MN\unlhd G$.
My Proof:
Since $M,N\unlhd G$, we have $e\in M\cap N$. Hence $e=ee\in MN$, so $MN\neq \emptyset$.
Let $g,h\in MN$. Then $g=mn$ and $h=m'n'$ for some $m,m'\in M$, $n,n'\in N$. Consider
$$\begin{align} gh^{-1}&=(mn)(m'n')^{-1}\\ &=(mn)(n'^{-1}m'^{-1}) \\ &=m(nn'^{-1})m'^{-1}. \end{align}$$
We may write $m=\mu m'$ for some $\mu\in M$ since $M$ is a group and right multiplication of an element of a group is a bijection on that group. Hence
$$gh^{-1}=\mu(m'(nn'^{-1})m'^{-1}),$$
but $m'(nn'^{-1})m'^{-1}\in N$ since $N\unlhd G$. Thus $gh^{-1}\in MN$.
Hence $MN\le G$ by the one-step subgroup test.
Now let $mn\in MN$ and $g\in G$. Then
$$g^{-1}mng=(g^{-1}mg)(g^{-1}Ng),$$
where $g^{-1}mg\in M$ as $M\unlhd G$ and $g^{-1}ng\in N$ as $N\unlhd G$. Hence $g^{-1}mng\in MN$.
Thus $MN\unlhd G$. $\square$
Why am I unsure?
I don't know. What says that $\mu\in M$? I mean: I know it's because of the reason stated above but I feel like I've waved my hands a little bit there.
Please help :)
Here is how you can formally justify the step you have a doubt on:
Consider the map
$$\phi_{m'}:M \to M: g\to gm'$$
This map is a bijection (an inverse is easy to construct, while constructing this you will see what $\mu$ actually is - also shown in the comments)
Thus, in particular, the map is surjective and thus for $m \in M$, there exists $\mu \in M$ such that $\mu m' =\phi_{m'}(\mu) = m$.