Let $φ : G → G_0$ be a group homomorphism. Show that if |G| is finite, then $|φ[G]|$ is finite and is a divisor of |G|. Q: Let $φ : G → G_0$ be a group homomorphism. Show that if |G| is nite, then $|φ[G]|$ is nite and is a divisor of |G|. φ[G] is the image of a subset under its domain. A: φ[G] = {φ(g): g $\in$ G} so |φ[G]| $\leq$ |G| (Would these be equal if φ[G] was the image of its domain instead of a subset?). ... I'm not sure how to answer the second part of the question. I saw the solution & it uses a kernel but I'm having trouble understanding it.

By the first isomorphism theorem, you have $G/\ker(\varphi)\simeq\varphi[G]$. Since your group is finite, this just means $$ |\varphi[G]|=[G:\ker(\varphi)]=|G|/|\ker(\varphi)|. $$ Rearranging, since every cardinality is just an integer, $$ |\ker(\varphi)|=\frac{|G|}{|\varphi[G]|}. $$ So $|\varphi[G]|$ divides $|G|$.