$\phi: Z_p \rightarrow Z_{q-1}$ and $p$ does not divide $q-1$, then the image must be trivial

Question

I have a trouble understanding the statement that goes:

If $\phi$ is a homomorphism such that $\phi: Z_p \rightarrow Z_{q-1}$ and $p$ does not divide $q-1$, then the image must be trivial (where $p, q$ are prime).

Any help would be appreciated!

Answer 1

Note that $f([1])\in \mathbb Z_{q-1}\implies o(f[1]) $ divides $q-1$.Also since $f$ is a homomorphism $f([1])$ must divide order of $[1]$ which is $p$.

Thus $o(f[1])$ is a common divisor of both $q-1,p$

Again since $p$ is prime so $o(f[1])=1$ or $o(f[1])=p$.

If $o(f[1])=1$ then $f[1]=[0]\implies f[n]=nf([1])=0\implies f$ is trivial.

If $o(f[1])=p$ then $p$ divides $q-1$ which is false.

Answer 2

Hint: Consider $\overbrace{\phi(1) + \cdots + \phi(1)}^{p} = p\,\phi(1)$.

Answer 3

Here's a general fact: let $G$, $H$ be groups and $\varphi: G \to H$ a homomorphism. $\varphi(G)$ is a subgroup of $H$ so $\# \varphi(G) | \# H $. By the first isomorphism theorem, $G/\ker \varphi \simeq \varphi(G)$, hence $\# G / \# \varphi(G) = \# \ker \varphi$, so $\# \varphi(G) | \# G$. Hence, $\# \varphi (G) | \gcd(\# G, \# H)$.

Now, under the assumptions do you think you can conclude?