Question

Let $\mathit{M}$ and $\mathit{N}$ be normal subgroups of a group $\mathit{G}$ such that $\mathit{M}\mathbf{\cap }\mathit{N}\mathbf{=}\mathbf{⟨}\mathbf{e}\mathbf{⟩}$. Prove that $\mathit{G}$ is isomorphic to a subgroup of $\mathit{G}\mathbf{/}\mathit{M}\mathbf{\times }\mathit{G}\mathbf{/}\mathit{N}$.

Short answer

It is proved that $G$ is isomorphic to a subgroup of $G/M\times G/N$.

Step-by-step solution

Prove that it is a group homomorphism 

Consider that $M$ and $N$ to be two normal subgroups of a group $G$with the condition $M\cap N=⟨e⟩$.

Define the map $f:G\to G/M\times G/N$ as $f\left(g\right)=(gM,gN)$.

Assume two arbitrary elements $g_1,g_2\in G$; this implies that:

$\begin{array}{c}f\left(g_1{g_2}^{-1}\right)=(g_1{g_2}^{-1}M,g_1{g_2}^{-1}N)\\ =(g_{1}M{g_2}^{-1}M,g_{1}N{g_2}^{-1}N)\\ =(g_{1}M,g_{1}N){(g_{1}M,g_{2}N)}^{-1}\\ =f\left(g_1\right)f{\left(g_2\right)}^{-1}\end{array}$

Therefore $f$ is homomorphism.

Prove the result

Prove that $f$ is one-to-one:

Assume two arbitrary elements $g_1,g_2\in G$; this implies that:

$$\begin{gathered} f\left(g_1\right)=f\left(g_2\right) \\ {g}_{1}M=g_{2}M\text{\hspace{0.17em}and\hspace{0.17em}}{g}_{1}N=g_{2}N \end{gathered}$$

This is equivalent to:

${g_2}^{-1}{g}_1\in M\cap N$

This implies that:

$\begin{array}{c}{g}_1{g_2}^{-1}=e\\ g_1=g_2\end{array}$

Therefore, $f$ is one-to-one.

This means that $f$ is isomorphism onto its image $f\left(G\right)$; this implies that $G$ is isomorphic to a subgroup of $G/M\times G/N$.

Hence, it is proved that $G$ is isomorphic to a subgroup of $G/M\times G/N$.

$\begin{array}{c}{g}_1{g_2}^{-1}=e\\ g_1=g_2\end{array}$