Let ϕ∈End(G) s.t. ∃n≥0,ker(ϕn)=G If kerϕ or [G:Inϕ] is finite, then G is finite.

Step 1
Assume that ${\displaystyle \varphi }$ has finite kernel. Consider the map
${\displaystyle \varphi :ker\left({\varphi }^{k+1}\right)\to ker\left({\varphi }^k\right)}$.
It has finite kernel. So if it has finite range, it has finite domain. By induction, if ${\displaystyle ker\left(\varphi \right)}$ is finite, so is every ${\displaystyle ker\left({\varphi }^n\right)}$ and in particular G.
If the image of ${\displaystyle \varphi }$ has finite index, then ${\displaystyle {\varphi }^k\left(\varphi \left(G\right)\right)}$ has finite index in ${\displaystyle {\varphi }^k\left(G\right)}$ for all k. It follows that ${\displaystyle 1={\varphi }^n\left(G\right)}$ has finite index in G so that G is finite.

Step 1
For the case where ${\displaystyle ker\varphi }$ is finite:
Note that if ${\displaystyle ker\left(\varphi \right)\le K\le H,}$ then
${\displaystyle [H:K]=[\varphi \left(H\right):\varphi \left(K\right)]}$,
by the isomorphism theorems.
Now, ${\displaystyle \varphi \left(ker{\varphi }^{i+1}\right)\le ker{\varphi }^i}$
Thus, in particular,
${\displaystyle [ker{\varphi }^2:ker\varphi ]=[\varphi \left(ker{\varphi }^2\right):\varphi \left(ker\varphi \right)]}$
${\displaystyle =[\varphi \left(ker{\varphi }^2\right):1]}$
${\displaystyle =\left|\varphi \left(ker{\varphi }^2\right)\right|}$
${\displaystyle \le \left|ker\varphi \right|}$
So if ${\displaystyle ker\varphi }$ is finite, then so is ${\displaystyle ker\left({\varphi }^2\right)}$. Now proceed by induction to show that ${\displaystyle ker{\varphi }^r}$ is finite.
If ${\displaystyle Im\left(\varphi \right)}$ has finite index, we can proceed similarly. Note that for arbitrary H and K, we know that
${\displaystyle \varphi \left(H\right)\stackrel{\sim }{=}\frac{H}{H\cap ker\varphi }\stackrel{\sim }{=}\frac{Hker\varphi }{ker\varphi }}$
So
${\displaystyle [G:Im\left(\varphi \right)]\ge [Gker\varphi :Im\left(\varphi \right)ker\varphi ]=[\varphi \left(Gker\varphi \right):\varphi \left(Im\left(\varphi \right)ker\varphi \right)]=[Im\left(\varphi \right):Im\left({\varphi }^2\right)]}$
So the ${\displaystyle I,\left({\varphi }^2\right)}$ also has finite index. Proceed by induction to conclude that G is finite.