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subgroups of a nilpotent group are nilpotent

oblixps

Member
May 20, 2012
38
i know that this can be proved easily using the lower central series, but i am having a hard time trying to prove this using the upper central series definition.

I saw a proof which said that if H is a subgroup of G, [tex] Z_{r}(H) \geq Z_{r}(G) \cap H [/tex] for all r. So if G is nilpotent which means that [tex] Z_{n}(G) = G [/tex] for some G, then this implies that [tex] Z_{n}(H) = H [/tex] for some n and therefore H is nilpotent.

i am having trouble understanding why [tex] Z_{r}(H) \geq Z_{r}(G) \cap H [/tex] for all r. I can see that the result is true for r = 1 since we are dealing only with the centers of G and H, but i am having trouble proving the induction step in order to prove this for all r. Can someone give me some hints on this?