Product of normal subgroups is normal

Proposition (Product of normal subgroups). Let $G$ be a group and let $N,M \lhd G$ be normal subgroups . Define

Then $NM$ is a normal subgroup of $G$. Moreover, $NM=MN$.

Context. Products like $NM$ appear in building larger normal subgroups from smaller ones (e.g. in series and extensions). The equality $NM=MN$ is a typical “normality makes products commute setwise” phenomenon.

Proof sketch.

• Subgroup: If $n_1m_1,n_2m_2\in NM$, then $(n_1m_1)(n_2m_2)^{-1}=n_1m_1m_2^{-1}n_2^{-1}.$ Since $M\lhd G$, conjugation by $n_2^{-1}$ preserves $M$, so $m_1m_2^{-1}n_2^{-1}=n_2^{-1}m'$ for some $m'\in M$. Hence the product lies in $NM$.
• Normality: For $g\in G$ and $nm\in NM$, $g(nm)g^{-1}=(gng^{-1})(gmg^{-1})\in NM$ since $N$ and $M$ are normal.
• Setwise commutativity: For $n\in N,m\in M$, normality of $N$ gives $mnm^{-1}\in N$, so $nm=(mnm^{-1})m\in MN$, hence $NM\subseteq MN$; similarly $MN\subseteq NM$.