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

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

psor32 · Accepted Answer

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 · Accepted Answer

Step 1  For the case where kerϕ is finite:  Note that if ker(ϕ)≤K≤H, 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)=∼HH∩kerϕ=∼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.

Let \phi\in End(G) s.t. \exists n\geq0, \ker(\phi^{n})=G If \ker\phi or [G:In_{\phi}] is finite,

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