Question

Let G be an abelian group.

Prove that $\mathit{G}\mathbf{/}\mathit{K}\mathbf{\cong }\mathit{H}$. Define a surjective homomorphism from G to H with kernel K

Short answer

It is proved that,$G/K\cong H$ .

Step-by-step solution

First IsomorphismTheorem

Theorem 8.20

Let $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{H}$be a surjective homomorphism of a group with kernel K. Then quotient group$\mathit{G}\mathbf{/}\mathit{K}$ is isomorphic to H.

Proving thatG/K≅  H

Referring to part(b) of this exercise, we know$H=\text{\hspace{0.17em}}\left\{x^2|x\in G\right\}$ .

As given in the hint, we define ,$f:G\to H$ by $f\left(x\right)=x^2$.

For , $x,y\in G$we have:

$\begin{array}{c}f\left(xy\right)={\left(xy\right)}^2\\ =x^2{y}^2\\ =f\left(x\right)f\left(y\right)\end{array}$

Therefore, our assumption is true and f is a homomorphism.

Since$H=\text{\hspace{0.17em}}\{x^2:x\in G\}$ , therefore we can define$\ker f$ as,$\ker f=\{x\in g,x^2=e\}$ .

From the above expression of$\ker f$ it is seen that f is surjective.

Therefore, f is a surjective homomorphism.

Hence, from the above result and byFirst Isomorphism Theorem, we can conclude that $G/K\cong H$.