Finding all homomorphisms from $\mathbb Z_4$ to $\mathbb Q^*$
Let there be homomorphism $f: \mathbb Z_4 \rightarrow \mathbb Q^*$.
So there are several options for $|\ker f|: 1, 2, 4$
(1) $|\ker f|= 1 \rightarrow |\operatorname{im}f|=4$
So $f$ is monomorphism. So naturally $f(0)=1$. The order of $2 \in \mathbb Z_4$ is $2$, so it has to be mapped to $-1$ (because 1 was already mapped). The order of $1, 3 \in \mathbb Z_4$ is 4, but we already have mapped elements of order 1 and 2, and there aren't other elements in $\mathbb Q^*$ of such order or of order 4. So there is no such homomorphism.
Is there a more clever way of explaining that?
(2) $|\ker f|= 2 \rightarrow |\operatorname{im}f|=2$
$f(d)=1$ if $d=0\pmod2$, else $f(d)=-1$
(3) $|\ker f|=4$ trivial homomorphism.
Is this accurate? Thank you for any assistance!
Everything you've put is correct, but a clearer way to explain it:
Note that $f(1_4)$ must have order dividing $4$. But the only elements of order $1$, $2$ or $4$ in $\mathbb{Q}^*$ are $1$ and $-1$. Also, $1_4$ generates $\mathbb{Z}_4$, so the image of $1_4$ determines the homomorphism completely. So either $f(1_4) = 1$, in which case we have the trivial homomorphism, or $f(1_4) = -1$, in which case we have the homomorphism you describe in (2).