Carla Murphy

Answered

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 $[G:I{n}_{\varphi}]$ is finite, then G is finite.

If

Answer & Explanation

psor32

Expert

2022-01-13Added 33 answers

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.

Assume that

It has finite kernel. So if it has finite range, it has finite domain. By induction, if

If the image of

aquariump9

Expert

2022-01-14Added 40 answers

Step 1

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

Note that if$ker\left(\varphi \right)\le K\le H,$ then

$[H:K]=[\varphi \left(H\right):\varphi \left(K\right)]$ ,

by the isomorphism theorems.

Now,$\varphi \left(ker{\varphi}^{i+1}\right)\le ker{\varphi}^{i}$

Thus, in particular,

$[ker{\varphi}^{2}:ker\varphi ]=[\varphi \left(ker{\varphi}^{2}\right):\varphi \left(ker\varphi \right)]$

$=[\varphi \left(ker{\varphi}^{2}\right):1]$

$=\left|\varphi \left(ker{\varphi}^{2}\right)\right|$

$\le \left|ker\varphi \right|$

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

$[G:Im\left(\varphi \right)]\ge [Gker\varphi :Im\left(\varphi \right)ker\varphi ]=[\varphi \left(Gker\varphi \right):\varphi \left(Im\left(\varphi \right)ker\varphi \right)]=[Im\left(\varphi \right):Im\left({\varphi}^{2}\right)]$

So the$I,\left({\varphi}^{2}\right)$ also has finite index. Proceed by induction to conclude that G is finite.

For the case where

Note that if

by the isomorphism theorems.

Now,

Thus, in particular,

So if

If

So

So the

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