Kernel of a homomorphism from $S_4$ to $\mathbb R^*$

Question

Let $S_4$ be the symmetric group on $4$ symbols. Let $f: S_4 \to \Bbb R^*$ be a homomorphism, where $\Bbb R^*$ denotes the non-zero real numbers. Then the number of elements in the set $\{ x\in S_4: f(x) =1\} $ is ....?

Answer 1

There is only one nonidentity element of finite order in $\mathbb R^*$, namely $-1$. Thus either the homomorphism is trivial or it has image of order $2$. Both are possible. Therefore there are two possible orders of the kernel, which I believe you can figure out.