 Carla Murphy

2022-01-12

Let
$\varphi \in End\left(G\right)$ s.t.
$\mathrm{\exists }n\ge 0,ker\left({\varphi }^{n}\right)=G$
If $ker\varphi$ or $\left[G:I{n}_{\varphi }\right]$ is finite, then G is finite. psor32

Expert

Step 1
Assume that $\varphi$ has finite kernel. Consider the map
$\varphi :ker\left({\varphi }^{k+1}\right)\to ker\left({\varphi }^{k}\right)$.
It has finite kernel. So if it has finite range, it has finite domain. By induction, if $ker\left(\varphi \right)$ is finite, so is every $ker\left({\varphi }^{n}\right)$ and in particular G.
If the image of $\varphi$ has finite index, then ${\varphi }^{k}\left(\varphi \left(G\right)\right)$ has finite index in ${\varphi }^{k}\left(G\right)$ for all k. It follows that $1={\varphi }^{n}\left(G\right)$ has finite index in G so that G is finite. aquariump9

Expert

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

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