Prove that if the element $a\in G$, where $G$ is a group, has order $n$ and its homomorphic image $f(a)$ has order $m$, then $m\mid n$

Question

Let $G$ be a group. If $a\in G$ has order $n$ and its homomorphic image $f(a)$ has order $m$, then $m\mid n$.

So I have a confusion with this question, what does it mean the order of $a$? wouldn't the order be of the group and the homomorphic group of $f(a)$?

besides from that i have tried using Lagrange's theorem but the theorem says that

The order of the subgroup of $G$ divides the group $G$ But the homomorphic image of $f(a)$ does not mean that it's a subgroup of $G$ or does it?

I appriciate all help! Thanks in advance!

Answer 1

An important distinction for you to know is the difference between the order of a group and the order of an element of a group. It seems like you already understand the former, so the latter:

Let $G$ be a group with identity element $e$, and let $a \in G$. The order of $a$ is the smallest positive integer $n$ such that $a^n = e$.

Does knowing this help?