proof: every $g \in G$ can be written as $g =nm$ Let $G$ be a group, and $N$ and $M$ normal subgroups of $G$. Also $N \cap M =$ {$e$}
Why is it true that if G is generated by $N \cup M \Rightarrow$ every $g \in G$ can be written as $g = nm$ with $n \in N, m\in M$?
Someone stated this in my group theory book, but I can't understand why it is true.
Can you explain it to me?
 A: Suppose you know that $G = \langle M \cup N\rangle$ and $M \cap N = \{e\}$.
Fact 1:  $MN \subseteq \langle M \cup N\rangle$, because $MN = \{mn: m\in M,n\in N\}$ is a subset of $G = \langle M \cup N\rangle$.
Fact 2: $\langle M \cup N\rangle \subseteq MN$ for any two subgroups $M,N$ of any group $G$, whenever $MN$ is a group.
Why? Because if $m \in M$, then $m = me$, with $m \in M, e \in N$. Similarly, if $n \in N$ then $n = en \in MN$.
Thus $M \subseteq MN$, and $N \subseteq MN$, so that $M \cup N \subseteq MN$.
Since $\langle M \cup N\rangle$ is the smallest subgroup of $G$ containing $M \cup N$, we have $G = MN$.
(Note: the condition $MN = NM$ is precisely the condition required for $MN$ to be a group. For this to happen, only ONE of $M,N$ need to be normal in $G$. So the condition that both subgroups be normal, is much stronger than we need to show that $G = MN$).
Finally, since $G = MN$, we know that for $g \in G,\ g = mn$ for some $m \in M,n \in N$. Suppose that $g = mn = m'n'$. Then $m'^{-1}m = n'n^{-1} \in M \cap N$, so $m = m', n = n'$, so not only can $g$ be so factored, but this factorization is unique.
