In the lower central series of a group GG, how to show each of the subgroups is fully invariant?

If we have that G=γ1(G)≥γ2(G)≥⋯,G = \gamma_1(G) \ge \gamma_2(G) \ge \cdots, is a LSC and γi+1(G)=[γi(G),G]\gamma_{i+1}(G)=[\gamma_{i}(G),G].

I can see that for every endomorphism of GG, it is true that f([x,g])∈[G,G]f([x,g])\in [G,G], where [x,g]∈[γi(G),G][x,g] \in [\gamma_i(G),G], but I can’t see how to show f([x,g])∈[γi(G),G]f([x,g]) \in [\gamma_i(G),G].



1 Answer



(1) For any i\;i\; , γi(G)⊲G\;\gamma_i(G)\lhd G\;, and a subgroup \;N\le G\;\;N\le G\; is normal iff \;[N,G]\le N\;\;[N,G]\le N\; .

(2) Induction, with \;\gamma_i(G)=[G,G]\;\;\gamma_i(G)=[G,G]\; fully invariant being (almost…or without almost) trivial.



Thank you. Now using induction, assuming the result true for i=ki=k, If [x,g] \in \gamma_{k+1}(G)[x,g] \in \gamma_{k+1}(G), then because f(\gamma_{k+1}(G)) \le f(\gamma_k(G))f(\gamma_{k+1}(G)) \le f(\gamma_k(G)), it follows that f([x,g]) = [f(x),f(g)] \in [f(\gamma_k(G),G] \le [\gamma_k(G),G] = \gamma_{k+1}(G)f([x,g]) = [f(x),f(g)] \in [f(\gamma_k(G),G] \le [\gamma_k(G),G] = \gamma_{k+1}(G). Is this correct?
– J. Doe
2 days ago



@J.Doe That looks correct.
– DonAntonio
2 days ago