I am searching a nonabelian group in which every element has finite order and every natural number is an order of some element. Obviously the group cannot be finite. It has to have infinite order. There are groups in which every natural number is an order for example, $GL_n(F)$ for suitable fields, but they do not fit here, because there are elements of infinite order. Some subgroup of these general linear group might be a potential example. There are some examples in abelian case : $\mathbb Q / \mathbb Z,$ so I suspect that there are some examples in nonabelian case aswell.
Group with every finite order elements
examples-counterexamplesgroup-theory
Best Answer
What about the semi-direct product $\mathbb{Q}/\mathbb{Z}\rtimes \mathbb{Z}_2$ with the homomorphism $\mathbb{Z}_2\rightarrow \mathrm{Aut}(\mathbb{Q}/\mathbb{Z})$ given by (generator) $\mapsto (-1)$? In it every element has finite order and every natural number appears as an order. Denoting by $\mathbb{Z}_2=\{\pm 1\}$, the multiplication is given explicitly by $$\begin{split}(\mathbb{Q}/\mathbb{Z}\times \mathbb{Z}_2)\times (\mathbb{Q}/\mathbb{Z}\times \mathbb{Z}_2)&\rightarrow (\mathbb{Q}/\mathbb{Z}\times \mathbb{Z}_2),\\ \left( ([p],a),([q],b)\right) &\mapsto ([p+aq], ab)\end{split} $$