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I am just learning to work with normal subgroups, and am trying to do the following proof. However, I have had some problems.

Let $N$ and $M$ be subgroups of a group $G$, where $M$ is a normal subgroup. Define $NM=\{nm : n \in N, m \in M\}$. We want to show that $NM$ is a subgroup of $G$.

In order to do this, we need to show three things: $NM$ is closed, it contains $G$'s identity element, and it has unique inverses.

Here is what I have:

Closure:

Let $n_1m_1,n_2m_2\in NM$. I attempted to conclude from the fact that $M$ is normal that $(n_1m_1)(n_2m_2)=(n_1n_2)(m_1m_2)\in NM$, but have since learned that this does not work: Are all normal subgroups Abelian?

However, I am having trouble coming up with an alternate method. The fact that there exists an $m$ which commutes with the given $n$ doesn't seem to help prove the desired result in general, and likewise with any statement about the cosets $nM$ and $Mn$.

Identity:

By the definition of a subgroup, $N$ and $M$ both contain $G$'s identity element, $e_G$. Thus, since $e_Ge_G=e_G$, we know that $e_G\in NM$.

Inverses:

I had a similar problem here as with closure. I wanted to say that $(nm)^{-1}=m^{-1}n^{-1}=n^{-1}m^{-1}\in NM$, the first step by the Socks-Shoes Property and the second step by the fact that $M$ is normal. This doesn't work for the same reason it doesn't work in proving closure, and I'm having trouble coming up with an alternate method for the same reasons.

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    $\begingroup$ You might also want to see The product of a subgroup and a normal subgroup is a subgroup. $\endgroup$ – Minus One-Twelfth Apr 22 at 0:50
  • $\begingroup$ Thank you very much for that link. Am I the only one who can't seem to find redundant questions on this site? I spent a good twenty minutes searching through results based on different relevant keywords before posting and didn't find that. In any case, thanks very much. $\endgroup$ – JustSomeGuy716 Apr 22 at 0:58
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In general $(n_1m_1)(n_2m_2)\ne(n_1n_2)(m_1m_2)$. However, $M$ being normal in $G$ means that $gM=Mg$ for all $g\in G$. Hence $n_2M=Mn_2$. So we know that $n_2m_2\in n_2M=Mn_2$ and hence there is some element $m_3\in M$ such that $n_2m_2=m_3n_2$. So now $(n_1m_1)(n_2m_2)=n_1(m_2m_3)n_2$. Now again, $m_2m_3n_2\in Mn_2=n_2M$ and hence there is some $m_4\in M$ such that $m_2m_3n_2=n_2m_4$. So now $(n_1m_1)(n_2m_2)=n_1(m_2m_3)n_2=(n_1n_2)m_4\in NM$. That proves closure. Now you can do the tricks to prove that $NM$ contains inverses.

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Since $M, N\le G$, we have $e=ee\in NM$. Hence $NM$ is nonempty.

Let $g, h\in NM$. Then $g=nm, h=n'm'$ for some $n, n'\in N$ and $m, m'\in M$. Consider

$$\begin{align} gh^{-1}&=(nm)(n'm')^{-1}\\ &=(nm)(m'^{-1}n'^{-1}) \\ &=n(mm'^{-1})n'^{-1}. \end{align}$$

We can write $n=\nu n'$ for some $\nu\in N$ since $N$ is a group and right multiplication by an element is a bijection on groups. Hence

$$gh^{-1}=\nu(n'(mm'^{-1})n'^{-1}),$$

but $n'(mm'^{-1})n'^{-1}\in M$ since $M\unlhd G$. Thus $gh^{-1}\in NM$.

Hence $NM\le G$ by the one-step subgroup lemma. $\square$

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If N is a subgroup and M is a normal subgroup, then $NM = MN.$

This doesn't mean that every elements of $M$ commute with elements of M. (If it were true, we would say $M$ is a subset of the centralizer of $N$)

But it does mean that for any pair of elements $n_1,m_1,$ in $N,M$ respectively, there exists a pair of elements $n_2,m_2$ such that $m_2n_2 = n_1m_1.$ Possibly, but not necessarily, $n_1 = n_2$ or $m_1 = m_2$

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