How to find all group homomorphisms $\left(\mathbb{Z}/3\mathbb{Z} \right) ^\times\to \left(\mathbb{Z}/5\mathbb{Z}\right )^\times$

Question

Are there other group homomorphisms $f: \left(\mathbb{Z}/3\mathbb{Z} \right) ^\times\to \left(\mathbb{Z}/5\mathbb{Z}\right )^\times$ other than

$$f: \left(\mathbb{Z}/3\mathbb{Z} \right) ^\times\to \left(\mathbb{Z}/5\mathbb{Z}\right )^\times, ~x \mapsto e$$

with $e$ being the identity element - and if yes, how can I find them?

Answer 1

$1$ must be mapped to $1$.

$2^2=1$ in $(\mathbb Z/3\mathbb Z)^\times$,

so $2$ must be mapped to $1$ (which is the solution you mentioned) or $4$ in $(\mathbb Z/5\mathbb Z)^\times$.

Answer 2

Hint. If all you care about is the multiplicative structure these are cyclic groups of order $2$ and $4$. The fact that they come from fields is irrelevant.

Answer 3

Since $(\mathbb Z/p\mathbb Z)^\times\cong \mathbb Z/(p-1)\mathbb Z$ for all primes $p,$ a homomorphism $(\mathbb Z/3\mathbb Z)^\times\rightarrow (\mathbb Z/5\mathbb Z)^\times$ is the same as a group homomorphism $\mathbb Z/2\mathbb Z\rightarrow \mathbb Z/4\mathbb Z.$ Now, $0$ has to go to $0,$ but $1$ can go to either $0$ or $2,$ (since $f(1)+f(1)=f(1+1)=f(0)=0$) so there are two such homomorphisms.

Answer 4

Note that, $\overline 2\in \Bbb Z_3^×$ is a generator of order $2$. So, $\overline 2\mapsto \overline 4\in \Bbb Z_5^×$ is the only possible non-trivial homomorphism.