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

Carla Murphy

Carla Murphy

Answered

2022-01-12

Let
ϕEnd(G) s.t.
n0,ker(ϕn)=G
If kerϕ or [G:Inϕ] is finite, then G is finite.

Answer & Explanation

psor32

psor32

Expert

2022-01-13Added 33 answers

Step 1
Assume that ϕ has finite kernel. Consider the map
ϕ:ker(ϕk+1)ker(ϕk).
It has finite kernel. So if it has finite range, it has finite domain. By induction, if ker(ϕ) is finite, so is every ker(ϕn) and in particular G.
If the image of ϕ has finite index, then ϕk(ϕ(G)) has finite index in ϕk(G) for all k. It follows that 1=ϕn(G) has finite index in G so that G is finite.
aquariump9

aquariump9

Expert

2022-01-14Added 40 answers

Step 1
For the case where kerϕ is finite:
Note that if ker(ϕ)KH, then
[H:K]=[ϕ(H):ϕ(K)],
by the isomorphism theorems.
Now, ϕ(kerϕi+1)kerϕi
Thus, in particular,
[kerϕ2:kerϕ]=[ϕ(kerϕ2):ϕ(kerϕ)]
=[ϕ(kerϕ2):1]
=|ϕ(kerϕ2)|
|kerϕ|
So if kerϕ is finite, then so is ker(ϕ2). Now proceed by induction to show that kerϕr is finite.
If Im(ϕ) has finite index, we can proceed similarly. Note that for arbitrary H and K, we know that
ϕ(H)=HHkerϕ=Hkerϕkerϕ
So
[G:Im(ϕ)][Gkerϕ:Im(ϕ)kerϕ]=[ϕ(Gkerϕ):ϕ(Im(ϕ)kerϕ)]=[Im(ϕ):Im(ϕ2)]
So the I,(ϕ2) also has finite index. Proceed by induction to conclude that G is finite.

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