View source for 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]] <math>G</math> and a [[prime number]] <math>p</math> such that the [[abelianization]] of <math>G</math> is <math>p</math>-[[divisible group for a set of primes|divisible]], but <math>G</math> itself is not <math>p</math>-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|[[particular example::infinite dihedral group]]}}

Let <math>G</math> be a infinite dihedral group:

<math>\langle a,x \mid xax^{-1} = a^{-1}, x = x^{-1} \rangle</math>

and let <math>p</math> be any [[prime number]] other than 2. Then:

* <math>G</math> is solvable: In fact, <math>G</math> is [[metacyclic group|metacyclic]]. It has a cyclic normal subgroup <math>\langle a \rangle</math> with a cyclic quotient group isomorphic to [[cyclic group:Z2]] (generated by the image of <math>x</math>).
* The abelianization of <math>G</math> is <math>p</math>-divisible: The abelianization of <math>G</math> is a [[Klein four-group]], which is divisible for all odd primes.
* <math>G</math> is not <math>p</math>-divisible: The element <math>a \in G</math> has no <math>p^{th}</math> roots.