Question

Question: Let$f:G\to H$ be a homomorphism of groups and suppose that $a\in G$has finite orderk .

(a) Prove that $f{\left(a\right)}^k=e$. [Hint: Exercise 15]

Short answer

It has been proved that$f{\left(a\right)}^k=e$.

Step-by-step solution

Step-By-Step SolutionStep 1: Show that fak=e

Given that $f:G\to H$is a homomorphism of groups.

Let $a\in G$has order, it means $a^k=e$.

Show that fak=e

Now, $f{\left(a\right)}^k=f\left(a^k\right)$

therefore,

$$\begin{gathered} f{\left(a\right)}^k=f\left(a^k\right) \\ =f\left(e\right) \\ =e \end{gathered}$$

Conclusion

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