# If $N$ is a subgroup of $G$ and $M$ is a normal subgroup of $G$, show that $NM$ is a subgroup of $G$.

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.

• You might also want to see The product of a subgroup and a normal subgroup is a subgroup. – Minus One-Twelfth Apr 22 at 0:50
• 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. – JustSomeGuy716 Apr 22 at 0:58

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.

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. $$\square$$

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 $$m_1 = m_2$$