Let $h:S_4 \rightarrow \mathbb{Z}$ be a group homomorphism. Prove that $h(x) =0$ for all $x \in S_4$

Question

This a 3 part question. I struggle working with $S_4$ but seem to understand the questions if $S_4$ was not the codomain.

1. Let $f:\mathbb{Z} \rightarrow S_4$ be a group homomorphism such that $f(1)=(1243)$. Precisely describe im($f$) and ker($f$).

2. Let G be a group, and let $x \in G$ such that ord$(x)$ is finite. Let $K$ be another group and let $g:G \rightarrow K$ be a group homomorphism. Explain why ord(g($x$)) in $K$ is a divisor of ord($x$).

3. Now let $h:S_4 \rightarrow \mathbb{Z}$ be a group homomorphism. Prove that $h(x) =0$ for all $x \in S_4$.

Answer 1

$S_4$ is the group of permutations of 4 elements. Let $*$ represent composition of two permutations.

Part 1: Since identity must map to identity, $$f(0)=()$$ $$f(1)=(1243)$$ $$f(2)=f(1+1) = f(1)*f(1) = (14)(23)$$ $$f(3)=f(2+1) = f(2)*f(1) = (1342)$$ $$f(4)=f(3+1) = f(3)*f(1) = ()$$ Let $n=4q+r$, where $0\le r<4$ and $n,q\in \Bbb Z$. $$f(n) = f(4q+r) = f(4q)*f(r)=f(4)^q*f(r)=()^q*f(r)=f(r)$$

So $$Im(f)=\{(),(1243),(14)(23),(1342)\}$$ $$Ker(f)=\{4n|n\in \Bbb Z\}$$

Part 2: Let $m=ord(x)$ and $n=ord(g(x))$ where $x \in G$. So, $x^m=1_G$:-

Since identity must map to identity, $$g(1_G)=g(x^m)=(g(x))^m = 1_K$$ $$(g(x))^m = 1_K \implies m=nk $$ Thus n divides m.

Part 3: Let $x \in S_4$ such that $|h(x)| = M$ is maximum. Then $h(x*x)=h(x)+h(x)=2h(x)$. By assumption, $|h(x)|\ge|2h(x)|$. Therefore $|h(x)|=0$. Since maximum is zero, and $|h(x)|\ge0$, $h(t)=0$ for all $t\in S_4$.