# A step in a proof that $M, N\unlhd G$ implies $MN\unlhd G$.

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 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.

• $\mu = m(m')^{-1}$ works – ZeroXLR Apr 18 at 14:24

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 $$\mu m' =\phi_{m'}(\mu) = m$$.

• Ah, of course! Thank you :) – Shaun Apr 18 at 14:15
• But does the highlighted bit make sense or have I sweeped stuff under the carpet? – Shaun Apr 18 at 14:16
• I don't follow. The $m'$ is needed on the left hand side of $nn'^{-1}m'^{-1}$ in order for it to be a conjugate, so in $N$. – Shaun Apr 18 at 14:21
• No, that's perfect. Thank you :) – Shaun Apr 18 at 14:26
• No problem! Glad to help :) – EpsilonDelta Apr 18 at 14:27