Prove that the order of homomorphic image divides the order of the group without using First Isomorphism Theorem. Suppose $G'$ is a homomorphic image of $G$.
Let $\phi:G\rightarrow G'$ be an epimorphism.
We want to show that $|G'|$ divides $|G|$.
It can be done easily by using the concept of cosets, quotient group and also First Isomorphism Theorem. But this exercise appears under the subtopic homomorphisms so those contents are not involved yet.  
From what I know, there is an identity that is
$$|G|=|\ker \phi||G'|$$
But I have no idea how to prove it.
Does this mean that the epimorphism will partition image of $G$ into $G'$ classes where each class have $|\ker \phi|$ elements?
I just write my answer in a proper way.
Since $\phi$ is surjective, the pre-images of every element in $G'$ will partition $G$. That is,
$$|G|=\sum_{y\in G'}|\phi^{-1}(y)|$$
Let $y\in G'$
Then $y=\phi(x)$ for some $x\in G$ due to the surjectivity.
Hence, $|\phi^{-1}(y)|=|\phi^{-1}(\phi(x))|=|x\cdot \ker \phi|=|\ker \phi|$.
This proves the identity.
 A: It means it indeed. Here is a self contained proof – which amounts more or less to prove the 1st isomorphism theorem:

Lemma: For any $g\in G$, we have $\phi^{-1}\bigl(\phi(g)\big)=g\cdot\ker\phi$.

This means  any $h$ such that $\phi(h)=\phi(g)$ has the form $h=gx$, for a (unique) $x\in\ker \phi$.
Indeed, $\phi(gx)=\phi(g)\phi(x)=\phi(g)e'=\phi(g)$ ($e'$ is the unit of $G'$). Conversely, if $\phi(h)=\phi(g)$, we have $\phi(g^{-1}h)=\phi(e)=e'$, so $x=g^{-1}h\in ker \phi$  and $h=g(g^{-1}h)$.
Since $\phi$ is surjective, this proves $G$ is partitioned into the  subsets  $\phi^{-1}\bigl(\phi(g)\big)$ which all have cardinality  $m=\lvert\ker g\rvert$. If $r$  is the number of elements in this partition, we have 
$$\lvert G\rvert=\underbrace{m+m+\dots+m}_{r \text{ terms}}=mr$$
by the shepherd's principle.
A: As $\phi$ is an epimorphism then it is a surjective homomorphism. As such $\operatorname{im}\phi=G'$ which is the subgroup of $G'$ that, since $\phi$ is an epimorphism, is the whole of $G'$, and $\ker\phi=K$, where $K$ is the subgroup of $G$ whose elements map to $1_{G'}$, the identity in $G'$.
Since we have an epimorphism: 
$$|G|=|\operatorname{im}\phi||\ker\phi|=|G'||\ker\phi|$$
and $G$ is thus partitioned into $\frac{|G|}{|K|}$subsets all of order $|\ker\phi|$ which map to $G'$ under $\phi$.
