Question

Question 23(b): Prove that part (a) is false for every non abelian group. [Hint: A counter example is insufficient here (Why?). So try Exercise 24 of Section 7.2.]

Short answer

It has been proved that$f\left(x\right)=x^2$is not homomorphism if G is non-abelian.

Step-by-step solution

Step-By-Step SolutionStep 1: Suppose that is abelian

Consider $f:G\to G$given by $f\left(x\right)=x^2$.

Suppose that G is non-abelian.

Prove that fab≠fafb

Now,

$$\begin{gathered} f\left(ab\right)={\left(ab\right)}^2 \\ =abab \end{gathered}$$

And,

$$\begin{gathered} f\left(a\right)f\left(b\right)=a^2{b}^2 \\ =aabb \end{gathered}$$

Now, if both of these are equal then, we get $ab=ba$

But this is a contradiction to our supposition that G is non-abelian.

Hence ,$f\left(ab\right)\ne f\left(a\right)f\left(b\right)$

Conclusion

Thus, fis a homomorphism only when G is abelian.

Here, fis not a homomorphism.