$f:G\rightarrow G'$ is a non trivial homomorphism, show that if $\text{ord}(G)$ is prime then $f$ is one-to-one

Question

Let $f:G\rightarrow G'$ be a non trivial homomorphism,

Prove that if $|G|$ is prime, then $f$ is one-to-one.

My Attempt:

As $f$ is homomorphism,

$f(gg')=f(g)f(g')$ for all $g,g'\in G$.

If $|G|$ is prime then what?

Answer 1

By Lagrange's theorem, $G$ has no non-trivial subgroups. Since $\ker f$ is a proper subgroup of $G$, it must be $ \ker f = \{ 1_G \}$. So $f$ is injective.

However, you cannot prove that $f$ is onto without further hypothesis. An example is any embedding $G \longrightarrow G \times H$ where $H$ a non-trivial group.

Answer 2

The first isomorphism theorem gives

$$G/\ker (f) \cong \text{im}(f).$$ Taking the order of both sides, and using Lagrange's theorem gives

$$|G|=|\ker(f)||\text{im}(f)| $$

As $\ker (f) \leq G $ is a proper subgroup, and $|G|$ is prime, Lagrange's theorem gives $\ker(f)=1$. Thus $\text{im}(f)\leq G'$ is a subgroup of the same order as $G$.