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.
Best Answer
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.