Question

Let $\mathit{A}\mathbf{,}\mathit{B}\mathbf{,}\mathit{N}$be normal subgroups of a group$\mathit{G}$ such that $\mathit{N}\mathbf{\subseteq }\mathit{A}$,$\mathit{N}\mathbf{\subseteq }\mathit{B}$ . If $\mathit{G}\mathbf{=}\mathit{A}\mathit{B}$and $\mathit{A}\mathbf{\cap }\mathit{B}\mathbf{=}\mathit{N}$, prove that$\mathit{G}\mathbf{/}\mathit{N}\mathbf{\cong }\mathit{A}\mathbf{/}\mathit{N}\mathbf{\times }\mathit{B}\mathbf{/}\mathit{N}$ . ('The special case$\mathit{N}\mathbf{=}\mathbf{⟨}\mathbf{e}\mathbf{⟩}$ is Exercise 30 in Section 8 .2.)

Short answer

It is proved that, $G/N\cong A/N\times B/N$.

Step-by-step solution

Define the mapping

Consider$A,B,N$as normal subgroups of a group $G$,$G=AB$and $A\cap B=N$.

Define the mapping $f:A/N\times B/N\to G/N$as $f(\left[a\right],\left[b\right])=\left[ab\right]$.

Now, we have to show that $f$is an isomorphism.

First, show that $f$is well defined as:

Consider$(\left[a\right],\left[b\right]),(\left[c\right],\left[d\right])\in A/N\times B/N$such that $(\left[a\right],\left[b\right])=(\left[c\right],\left[d\right])$.

Then, we have $\left[a\right]=\left[c\right]$and $\left[b\right]=\left[d\right]$.

This implies that:

$\begin{array}{c}a{c}^{-1},b{d}^{-1}\in N\\ ab{d}^{-1}{c}^{-1}\in N\end{array}$

Thus,$b^{-1}a\in Z\left(G\right)$ this implies that$\left[a\right]=\left[b\right]$ in$G/Z\left(G\right)$ .

It is proved that, $f$is well-defined.

Prove that f is group isomorphism

First, show that f is group homomorphism as:

Assume that $(\left[a\right],\left[b\right]),(\left[c\right],\left[d\right])\in A/N\times B/N$.

Then, find $f(\left[a\right],\left[b\right])\cdot (\left[c\right],\left[d\right])$as:

$\begin{array}{c}f(\left[a\right],\left[b\right])\cdot (\left[c\right],\left[d\right])=f(\left[ac\right],\left[bd\right])\\ =\left[acbd\right]\\ =\left[a\right]\left[c\right]\left[b\right]\left[d\right]\\ =\left[a\right]\left[cb\right]\left[d\right]\end{array}$

Simplify the equation as:

$\begin{array}{c}f(\left[a\right],\left[b\right])\cdot (\left[c\right],\left[d\right])=\left[a\right]\left[bc\right]\left[d\right]\\ =\left[a\right]\left[b\right]\left[c\right]\left[d\right]\\ =\left[ab\right]\left[cd\right]\\ =f\left((\left[a\right],\left[b\right])\right)\cdot f\left((\left[c\right],\left[d\right])\right)\end{array}$

This implies that f is homomorphism.

Now, show that f is one-to-one and onto as:

Assume that $(\left[a\right],\left[b\right]),(\left[c\right],\left[d\right])\in A/N\times B/N$, such that

$\begin{array}{c}f\left((\left[a\right],\left[b\right])\right)=f\left((\left[c\right],\left[d\right])\right)\\ \left[ab\right]=\left[cd\right]\\ ab{d}^{-1}{c}^{-1}\in N=A\cap B\end{array}$

This implies that $b{d}^{-1}\in A\cap B$and $a{c}^{-1}\in A\cap B$.

That is,$\left[a\right]=\left[c\right]\text{\hspace{0.17em}and\hspace{0.17em}}\left[b\right]=\left[d\right]$

Hence, f is one-to-one.

Assume that $\left[g\right]\in G/N$.

As $G=AB$, for this $g\in G$there exists $a\in A$and $b\in B$such that $g=ab$.

Therefore, we have:

$\begin{array}{c}\left[g\right]=\left[ab\right]\\ =\left[a\right]\left[b\right]\\ f(\left[a\right],\left[b\right])\end{array}$

This implies that f is onto.

Hence, it is proved that f is group isomorphism.

Thus, it is proved that$G/N$ is isomorphic to $A/N\times B/N$.