Residually nilpotent group with abelianization that is divisible by a prime need not be divisible by that prime

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Statement

It is possible to have a residually nilpotent group G and a prime number p such that the abelianization of G is p-divisible, but G itself is not p-divisible.

Proof

Further information: infinite dihedral group

Let G be the infinite dihedral group:

\langle a,x \mid xax^{-1} = a^{-1}, x = x^{-1} \rangle

and let p be any prime number other than 2. Then:

  • G is residually nilpotent: The c^{th} member of lower central series of G for c \ge 1, is \langle a^{2^c} \rangle. The intersection of these is trivial.
  • The abelianization of G is p-divisible: The abelianization of G is a Klein four-group, which is divisible for all odd primes.
  • G is not p-divisible: The element a \in G has no p^{th} roots.