Question

Question: Let $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{H}$be a homomorphism of groups. Prove that for each $a\in G$and each integern ,$\mathit{f}\left(a^n\right)\mathbf{=}\mathit{f}{\left(a\right)}^{\mathbf{n}}$

Short answer

It is proved that .$f\left(a^n\right)=f{\left(a\right)}^n$

Step-by-step solution

Step-By-Step Solution Step 1: Show that  f(an)=f(a)nfor positive integers

Since $f$is a homomorphism

then , $f\left(a.a\right)=f\left(a\right)f\left(a\right)$

that is $f\left(a^2\right)=f{\left(a\right)}^2$

On generalization, it is clear that$f\left(a^n\right)=f{\left(a\right)}^n$for all positive integers.

Show that f(an)=f(a)n for negative integers

Recall Theorem 7.20, which states that, “If $f:G\to H$is a homomorphism, then $f\left(a^{-1}\right)=f{\left(a\right)}^{-1}$for every $a\in G$”.

Then, using the result for positive integers to get,

$$\begin{gathered} f\left(a^{=n}\right)=f\left({\left(a^{-1}\right)}^n\right) \\ =f{\left(a^{-1}\right)}^n \\ ={\left(f{\left(a\right)}^{-1}\right)}^n \\ =f{\left(a\right)}^{-n} \end{gathered}$$

Hence ,$f\left(a^n\right)=f{\left(a\right)}^n$for all negative integers.

Conclusion

Hence ,$f\left(a^n\right)=f{\left(a\right)}^n$