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

Statement

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

Related facts

• Residually nilpotent group with abelianization that is divisible by a prime need not be divisible by that prime
• Equivalence of definitions of nilpotent group that is divisible for a set of primes

Proof

Further information: infinite dihedral group

Let $G$ be a 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 solvable: In fact, $G$ is metacyclic. It has a cyclic normal subgroup $\langle a \rangle$ with a cyclic quotient group isomorphic to cyclic group:Z2 (generated by the image of $x$).
• 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.