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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.

Please help :)

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

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  • $\begingroup$ Ah, of course! Thank you :) $\endgroup$ – Shaun Apr 18 at 14:15
  • $\begingroup$ But does the highlighted bit make sense or have I sweeped stuff under the carpet? $\endgroup$ – Shaun Apr 18 at 14:16
  • $\begingroup$ 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 $N$. $\endgroup$ – Shaun Apr 18 at 14:21
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    $\begingroup$ No, that's perfect. Thank you :) $\endgroup$ – Shaun Apr 18 at 14:26
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    $\begingroup$ No problem! Glad to help :) $\endgroup$ – EpsilonDelta Apr 18 at 14:27

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