If $\phi: G \to G^{'} $ is a group homomorphism and $g \in G$ is an element of finite order, then the order of $\phi(g)$ divides the order of $g$ I've checked there are other posts with the same question but I have some more to ask, so I would really appreciate it if you could read through my post. Thank you ahead, my saviours.
The problem I'm working on:
If $\phi: G \to G^{'} $ is a group homomorphism and $g \in G$ is an element of finite order, then the order of $\phi(g)$ divides the order of $g$
I wrote my questions at the bottom.
My thought:
Let $\mid g \mid = k $. Then $[\phi(g)]^k = \phi(g^k) = e'$ where $e'$ is the identity in $G'$.
So $\mid \phi(g) \mid$ divides $\mid g \mid$.
My solution is incomplete because the result seems to say  $\mid \phi(g) \mid = \mid g \mid = k$. But this is not true because $\mid \phi(g) \mid$ can be smaller than $k$.
Q1. Am I right about the comment I made on my solution?
Another solution I found is:
Let $n = \mid g \mid, v = \mid \phi(g) \mid$. Since $g^n = e$ and $\phi(g^n) = \phi(g)^n,$ we have $v \leq n.$
Assume $v \nmid n$. Let $n = kv + r,$ with $r < v$.
Then $e = \phi(g)^n = (\phi(g)^v)^k \phi(g)^r = \phi(g)^r$, which contradicts the definition of $v$. Hence $v \mid n.$
Q2. Why can we say $v \leq n$? Is it because $v$ is the smallest integer that satisfy $\phi(g)^v = e'$, and if $v > n$, it doesn't make sense since $\phi(g)^n $is already $e'$
Q3. When the solution says that "which contradicts the definition of $v$" at the very last part of the proof, what does that mean?
 A: Remember that the order of an element $x$ is the smallest positive integer $n$ such that $x^n$ is the identity. So $\phi(g)^k=e'$ does not by itself imply the order of $\phi(g)$ is $k$.
For question 2, your thoughts are correct. $v\leq n$ for the reasons you point out. $v$ is the smallest positive integer satisfying $\phi(g)^v=e$. This also answers your third question. The contradiction is that you have $\phi(g)^r=e$ and $r<v$, contradicting the definition of $v$ as the smallest integer with this property. (It would be better if the proof emphasized the fact that $r$ is nonzero.)
Edit: To help clarify the situation, there are really two separate facts in play here.

*

*If $\phi\colon G\to H$ is a group homomorphism, and $g\in G$ is such that $g^n=e_G$, then $\phi(g)^n=e_H$.


*Suppose $G$ is a group and $g\in G$ is such that $g^n=e$ for some $n>0$. Then the order of $g$ divides $n$.
These two facts together answer your question, and it appears that the central issue is how to prove the second fact. The argument is essentially what you wrote before your question 2. But I will write it a little differently. (In particular, the approach by contradiction and the first part about observing $v\leq n$ aren't necessary.)
Proof of (2). Let $v$ be the order of $g$ and suppose $g^n=e$. We want to show $v$ divides $n$. Write $n=qv+r$ where $q$ and $r$ are integers with $0\leq r<v$. Then
$$
e=g^n=g^{qv+r}=(g^v)^q g^r=e^qg^r=g^r.
$$
Since $r<v$ and $v$ is the order of $g$, we must have $r=0$. So $n=qv$ and we see that $v$ divides $n$.
