Question

Question: If $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{H}$ is a surjective homomorphism of groups and G is abelian, prove that H is abelian.

Short answer

It has been proved that H is abelian.

Step-by-step solution

Step-By-Step SolutionStep 1: Suppose two elements in H

let , $a,b\in H$

Since $f$is surjective therefore $a=f\left(x\right)$, and $b=f\left(y\right)$for some$x,y\in G$.

Since $f$is a homomorphism,

Therefore,$f\left(xy\right)=f\left(x\right)f\left(y\right)$

Show that ab = ba

Consider,

$$\begin{gathered} ab=f\left(x\right)f\left(y\right) \\ =f\left(xy\right) \end{gathered}$$

Since G is abelian, $xy=yx$

So, $f\left(xy\right)=f\left(yx\right)$

Thus,

$$\begin{gathered} ab=f\left(yx\right) \\ =f\left(y\right)f\left(x\right) \\ =ba \end{gathered}$$

hence ,$ab=ba$

Conclusion

Since $ab=ba$for any $a,b\in H$. Therefore, H is abelian.