Let $G$ be a finite group and $f: G \to \Bbb Z$ be a homomorphism of groups. Prove that for all $x \in G$, $f(x)=0$

Question

I took elements in $G$ as {$a_1,a_2,...,a_n$} considering order of $G$ to be as $n$. I am not able to define a function to prove $f(x)=0$ but I am not able to prove it. Please help.

Answer 1

Note that $f(G)$ is a finite subgroup of $\mathbb{Z}$.
But since the only finite subgroup of $\mathbb{Z}$ is the trivial group $\{0\}$, we must have $$f(G)=\{0\}.$$ That is, $f(x)=0$ for all $x\in G$.

Answer 2

Hint: Two relevant facts for you to prove: $f(x^n) = n f(x)$ and $f(1) = 0$

Answer 3

$e$ being the identity of $G$, we have

$f(e) = 0, \tag 1$

since group homomorphisms map identities to identities; if

$\exists x \in G, \; f(x) \ne 0, \tag 2$

then

$x \ne e \tag 3$

and

$\exists \Bbb Z \ni m \ge 2 , \; x^m = e; \tag 4$

thus we have

$0 \ne mf(x) = f(x^m) = f(e) = 0, \tag 5$

and this contradiction implies there is no $x$ as in (2); it follows that

$f(x) = 0, \; \forall x \in G. \tag 6$