Here's my attempt:
I'll be using this for my proof:
If $\varphi : G \to G^\prime $ is a group homomorphism and $a \in G$
with $\text{ord} (a) = r$ then $\text{ord} (\varphi (a))$ divides $r$.
Let $G=\langle a \rangle$ where $\text{ord} (a) = p$ for some prime $p$ and $G^{\prime }$ be any other group. Also it is easy to notice if $\varphi$ is any homomorphism from $G$ to $G^{\prime }$ then $\varphi$ is determined by $\varphi (a)$. (In other words, to define the homomorphism we only need to define $\varphi (a)$.
Now if $\varphi (a) = e_{G^{\prime}}$ then for any $n \in \mathbb{Z}_p$, we have $\varphi (a^n)= (\varphi (a)) ^n = e_{G^{\prime}}$. Thus making $\varphi$ the trivial homomorphism. Now, suppose that $\varphi (a) \ne e_{G^{\prime}}$ then we need to show $\varphi$ is injective. Let $x,y \in G$. Then $x=a^m$ and $y=a^n$ for some $m,n \in \mathbb{Z}_p$. If $\varphi (x) = \varphi (y)$ then $\varphi (a^m ) = \varphi (a^n )$. It follows that $(\varphi (a))^{m-n} = e_{G^{\prime}}$. Now, by the corollary at the beginning of the proof, we must have that $\text{ord} (\varphi (a))$ divides $p$. Thus, it must be that $\text{ord} (\varphi (a)) = p$. And if $(\varphi (a))^{m-n} = e_{G^{\prime}}$ then $p$ divides $m-n$ but $m-n < p$ (since $m,n \in \mathbb{Z}_p$) thus it must be that $m=n$. And therefore, $x=y$ and thus $\varphi$ is injective.
Is my proof correct? Or am I missing out something?
Best Answer
Your proof is correct.
Let me offer another approach: If $\phi:G\to G'$ is a homomorphism, then $\ker(\phi)\lhd G$ is a subgroup. The only subgroups of $G$ are $G$ itself and $\langle e_G\rangle$. If $\ker(\phi)=G$, then $\phi$ is trivial. If $\ker(\phi)=\langle e_G\rangle$, then $\phi$ is injective.