If $G_1,G_2$ are groups, $\varphi:G_1 \rightarrow G_2$ is a homomorphism and $a \in(G_1)$, is an element of finite order, what can we say for the order $\varphi(a)$?
My attempt so far: Let $\operatorname{ord}(a)=r$; Then $(a^r)=e_1 $. But can I use this to claim that: ($\varphi(a)^r$)$=e_2$?
Yes, you can claim that because $\varphi$ is a homomorphism, hence $e_2=\varphi(a^r)=(\varphi(a))^r$ and then you can deduce that the order of $\varphi(a)$ divides $r$.