Let Z:G->H be a group Homomorphism and g an element of G. Prove that Z(g)^|g| = e?

Question

I know that A group homomorphism is a map between two groups such that the group operation is preserved: for all , where the product on the left-hand side is in and on the right-hand side in. But i dont know how to prove the above statement.

Answer 1

$$Z(g)^{|g|} = \underbrace{Z(g) \cdot Z(g) \cdots Z(g)}_{|g|} = Z(\underbrace{g \cdot g \cdots g}_{|g|})$$

Answer 2

Hint By induction on $n$ you can prove that $$Z(g)^n=Z(g^n)$$