# What examples are there of a group in which every element is nilpotent and has infinitely many square roots?

What examples are there of a group in which every element:

• is nilpotent
• has infinitely many square roots

(full disclosure: I recently asked a similar question but made a mistake of failing to specify nilpotency of the elements.)

By square roots, I mean $a$ is a square root of $b$ if $a\cdot a=b$ (apologies if that is obvious).

By nilpotent, I mean for every element $a$ there exists integer $n$ such that $a^n=e$ where $e$ is the group identity.

What have I tried? Well I'm a novice at group theory so this is tricky for me. But I think this much:

• Prufer p-group has all nilpotent elements, but every element in these has only $p$ many $p^{th}$ roots and I need infinitely many. So it would seem I might be looking for a direct product or sum of up to infinitely many Prufer p-groups.
• Since every element is a square root and is nilpotent, it follows every element $a$ must satisfy $a^{2^n}=e$
• This makes me think I need a product of infinitely many Prufer 2-groups
• @lulu Elements have only two square roots, not infinitely many Commented May 10, 2018 at 11:15
• Just a comment on notation. The condition that for every element $a$ of the group $G$ there exists an integer $n$ such that $a^n=e$ is usually referred to by saying that $G$ is a torsion group, that is, every element has finite order (period). Commented May 10, 2018 at 11:16
• @HagenvonEitzen Right. Will delete.
– lulu
Commented May 10, 2018 at 11:16
• @lulu but the direct sum of infinitely many copies of $\Bbb Q/\Bbb Z$ or copies of $\Bbb Z[\frac12]/\Bbb Z$ should work Commented May 10, 2018 at 11:17
• @AndreasCaranti The product of infinitely many torsion groups will no longer be a torsion group unless all factors have a common finite exponent (which is not the case if we take Prüfer groups). Commented May 10, 2018 at 11:26

So, you're looking for a torsion group in which every element has infinitely many square roots. If the group is also abelian, then this condition is equivalent to requiring that every element has at least one square root and that there are infinitely many elements of order $2$. So as Hagen von Eitzen says in the comments, the direct sum of infinitely many copies of the Prufer group $\mathbb{Z}[1/2]/\mathbb{Z}$ will work.