1. **Problem statement:** Given that $N$ and $M$ are normal subgroups of a group $G$, and $NM = \{ nm \mid n \in N, m \in M \}$ is a subgroup of $G$, prove that $NM$ is also a normal subgroup of $G$. 2. **Recall definitions:** - A subgroup $H$ of $G$ is normal if for every $g \in G$, $gHg^{-1} = H$. - Since $N$ and $M$ are normal, for all $g \in G$, $gNg^{-1} = N$ and $gMg^{-1} = M$. 3. **Goal:** Show that for all $g \in G$, $g(NM)g^{-1} = NM$. 4. **Start with conjugation:** $$ g(NM)g^{-1} = g\{ nm \mid n \in N, m \in M \}g^{-1} = \{ gnm g^{-1} \mid n \in N, m \in M \}. $$ 5. **Rewrite conjugation inside:** Since conjugation is a homomorphism, $$ gnm g^{-1} = (g n g^{-1})(g m g^{-1}). $$ 6. **Use normality of $N$ and $M$:** Because $N$ and $M$ are normal, $$ g n g^{-1} \in N, \quad g m g^{-1} \in M. $$ 7. **Therefore:** $$ g(NM)g^{-1} = \{ n' m' \mid n' \in N, m' \in M \} = NM. $$ 8. **Conclusion:** Since $g(NM)g^{-1} = NM$ for all $g \in G$, $NM$ is a normal subgroup of $G$. **Final answer:** $NM$ is normal in $G$.