Homomorphism between $S_4$ and non zero real numbers under multiplication

Question

Let $S_4$ denote the symmetric group of 4 letters, and $\mathbb {R^*}$ be the multiplicative group of non-zero real numbers. If $f: S_4 \to \mathbb{R}^*$ is a homomorphism. Then the set $\{ x \in S_4 : f(x)=1\}$ has

1) atleast 12 elements

2) exactly 24 elements

3) atmost 12 elements

4) exactly 4 elements.

We know that $ ker(f)$ is a normal subgroup of $S_4$.

Answer 1

Since $|f(x)|$ must divide $24$, we have $f(x)=\pm1,\,\forall x$. But $f(e)=1$. Thus the order is at least $12$. Note, $f$ is $n$ to $1$, where $n=|\operatorname{ker}f|$, by the first isomorphism theorem.