Let G be a group, G′=[G,G] and G′′=[G′,G′] the first and second derived subgroups and assume G′′ is cyclic. Prove that G′′⊂Z(G′).

manpreet Best Answer 2 years ago

I'm trying to prove the following, but I'm stuck and I don't see how to continue. Any help is much appreciated!

Let GG be a group, G′=[G,G]G′=[G,G] and G′′=[G′,G′]G″=[G′,G′] the first and second derived subgroups and assume G′′G″ is cyclic. Prove that G′′⊂Z(G′)G″⊂Z(G′).

My work this far: Since G′′G″ is cyclic, it is abelian. Since it is the commutator subgroup of G′G′, it is also known that G′′G″ is normal. Thus (using a theorem from my syllabus), there exists a homomorphism g:G′/G′′→Aut(G′′)g:G′/G″→Aut(G″) such that