# Image of Group Homomorphism is Finite and Divides |Domain of Group| - Fraleigh p. 135 13.44

Let $\phi: G \rightarrow G'$ be a homomorphism. Show that if $|G|$ is finite, then $|\phi[G]|$ is finite and divides $|G|$.

Because $φ[G] = \{φ(g) \, | \, g ∈ G\}$, we see $|φ[G]| ≤ \quad |G|$ which the question presupposed is finite. By Theorem 13.15, there is a one-to-one correspondence between the elements of $φ[G]$ and the cosets of $Ker(φ)$ in $G$.

$\color{darkred}{ \text{ (1.) How do you envisage and envision to use Theorem 13.15? How does it relate to this question? } }$

Thus $|φ[G]| = |G|/|Ker(φ)|$, so $|φ[G]|$ divides $|G|$.

(2.) Where does $|φ[G]| = |G|/|Ker(φ)|$ crop up from? No isomorphism theorems are covered.

(3.) What's the intuition? Another proof

$\varphi$ partitions $G$ into $n$ equivalence classes, one for each preimage of an element of $\varphi(G)$. Do we know that each equivalence class is the same size, $m$? Then certainly the equivalence class that is the preimage of the only element we unquestionably know is in $G'$, the identity, (i.e., $\ker \varphi$) has size $m$. Then since $G$ is partitioned by these equivalence classes, $|G| = n \cdot m = |\varphi(G)| \ |\ker \varphi|$.