$N$ and $M$ are two normal subgroups of $G$ with $NM=G$ and $N \cap M = \{e\}$
How would I show that the homomorphism $\varphi : M \times N \longrightarrow G/N \times G/M$ defined by $\varphi (m,n) = (mN, nM)$ is injective?
In other words, I can't figure out how to show that $n_1 \neq n_2 \Longrightarrow n_1M \neq n_2M$.
I'm guessing that $n_1M = n_2M$ would somehow imply that $n_1, n_2 \in M$ which contradicts the intersection $N \cap M$ being trivial, but I can't figure out how to show this.
If $n_1M = n_2M$ then $n_1^{-1}n_2 \in M$. Since this is also in $N$ we must have $n_1^{-1}n_2 = e$, meaning $n_1 = n_2$.
Remember: $aH = bH$ if and only if $H = a^{-1}bH$ if and only if $a^{-1}b \in H$. Or another way to look at it: $b = be \in bH$ so we must have $b = ah$ for $b$ to be in $aH$. But then $ah = b$ means $a^{-1}b = h \in H$.