Please can I have feedback on my answer to this question.
Let $\psi: G → H$ be a group homomorphism and $g \in G$. Prove that $\psi(g)^{|g|} = e$.
Answer:
By hypotesis $\psi: G → H$ is a group homomorphism and $g \in G$. We know that the order of $|g|$ divide $|G|$. Thus : $|G|=k|g|$ , where $k$ is an integer. Then $g^{|G|}= g^{k|g|} = (g^{|g|})^k$. Since $g^{|g|} = e$, then $(g^{|g|})^{k} = e^k=e$, by the definition of order.
By homomorphism property you have:
$$\psi(g^{|g|})=\psi(g)\psi(g^{|g|-1})$$
apply this property $|g|$ times, thus $$\psi(g)^{|g|}=\psi(g^{|g|})=\psi(e)=e$$