Originally Posted by
Drexel28 So, for the sake of notational convenience let's call our homomorphism $\displaystyle \theta:G\to G'$.
Also, I'll assume these are finite groups.
So, let $\displaystyle G=\langle g\rangle$. We claim that $\displaystyle G'=\langle \theta(g)\rangle$. To see this let $\displaystyle g'\in G$ be arbitrary. By assumption $\displaystyle g'=\theta(h)$ for some $\displaystyle h\in G$. But, that means that $\displaystyle h=g^n$ for some $\displaystyle n$ and so $\displaystyle g'=\theta(h)=\theta(g^n)=\theta(g)^n$.
Work with that.
How much group theory do you know? You should know by the FIT that $\displaystyle G/\ker\theta\cong G'$ and so $\displaystyle |G'|=\left|G/\ker\theta\right|=\frac{|G|}{|\ker\theta|}\implies \frac{|G|}{|G'|}=|\ker\theta|$