Let $\varphi\in{\rm End}(G)$ s.t. $\exists n\ge 0$, $\ker(\varphi^n)=G$. If $\ker\varphi$ or $[G:{\rm Im}\varphi]$ is finite, then $G$ is finite

Question

Let $G$ be a group and let $\varphi \in \operatorname{End}(G)$ be such that there exists $n\geq 0$ such that $\ker (\varphi ^n)=G$.

Prove that if $\ker \varphi$ is finite or $[G:\operatorname{Im}\varphi ]$ is finite then $G$ is finite.

My attempt:

WLOG we can let $n$ be the smallest non-negative integer such that $\ker \varphi =G$. Then it is clear that $\ker \varphi ^k\lhd \ker \varphi ^{k+1}$ for $0\leq k<n$, hence $1,\ker \varphi ,\ker \varphi ^2,\ldots ,\ker \varphi ^{n-1},G$ is a normal series of $G$. I also proved that $\operatorname{Im}(\varphi ^{n-1})<\ker \varphi$, hence $\varphi ^{n-1}(G)$ is finite.

But I don't know how to use the hypothesis of $\ker \varphi$ being finite to prove that $G$ is finite (maybe we can show that if $\operatorname{Im}(\varphi ^k)$ is finite then $\operatorname{Im}(\varphi ^{k-1})$ is finite).

I also believe that I should prove that $[G:\operatorname{Im}\varphi ]$ is finite, then $\ker \varphi$ is finite and finish using the result above.

Answer 1

Assume that $\varphi$ has finite kernel. Consider the map $\varphi:\ker(\varphi^{k+1}) \rightarrow \ker(\varphi^k)$. It has finite kernel. So if it has finite range, it has finite domain. By induction, if $\ker(\varphi)$ is finite, so is every $\ker(\varphi^n)$ and in particular $G$.

If the image of $\varphi$ has finite index, then $\varphi^k(\varphi(G))$ has finite index in $\varphi^{k}(G)$ for all $k$. It follows that $1=\varphi^n(G)$ has finite index in $G$ so that $G$ is finite.

Answer 2

For the case where $\ker\varphi$ is finite:

Note that if $\ker(\varphi)\leq K\leq H$, then $[H:K]=[\varphi(H):\varphi(K)]$, by the isomorphism theorems.

Now, $\varphi(\ker\varphi^{i+1})\leq\ker\varphi^i$. Thus, in particular, $$\begin{align*} [\ker\varphi^2:\ker\varphi]&= [\varphi(\ker\varphi^2):\varphi(\ker\varphi)]\\ &=[\varphi(\ker\varphi^2):1]\\ &= |\varphi(\ker\varphi^2)|\\ &\leq |\ker\varphi|. \end{align*}$$ So if $\ker\varphi$ is finite, then so is $\ker(\varphi^2)$. Now proceed by induction to show that $\ker\varphi^r$ is finite.

If $\mathrm{Im}(\varphi)$ has finite index, we can proceed similarly. Note that for arbitrary $H$ and $K$, we know that $\varphi(H) \cong \frac{H}{H\cap\ker\varphi} \cong \frac{H\ker\varphi}{\ker\varphi}$. So $$[G:\mathrm{Im}(\varphi)] \geq [G\ker\varphi:\mathrm{Im}(\varphi)\ker\varphi] = [\varphi(G\ker\varphi):\varphi(\mathrm{Im}(\varphi)\ker\varphi)] = [\mathrm{Im}(\varphi):\mathrm{Im}(\varphi^2)].$$ So the $\mathrm{Im}(\varphi^2)$ also has finite index. Proceed by induction to conclude that $G$ is finite.