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.
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2$\begingroup$ For each $n\in \mathbb N$ consider a non-abelian group $G_n$ of finite order which contains an element of order $n$ and take the product group $\Pi G_n$ (consisting of sequences which are eventually identity). This might work. $\endgroup$– NewBMay 28 at 5:59
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5$\begingroup$ If you have some abelian example A, consider its direct product with S_3, the symmetric group of order 3. $\endgroup$– Mariano Suárez-ÁlvarezMay 28 at 6:08
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1$\begingroup$ I wonder if you can find such a group which is finitely generated... $\endgroup$– Matthew TowersMay 28 at 8:24
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$\begingroup$ @MatthewTowers This is a old problem in group theory, you may see here : en.wikipedia.org/wiki/Burnside_problem $\endgroup$– 3f183201Jun 2 at 17:15
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3 Answers
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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 $$\begin{split}(\mathbb{Q}/\mathbb{Z}\times \mathbb{Z}_2)\times (\mathbb{Q}/\mathbb{Z}\times \mathbb{Z}_2)&\rightarrow (\mathbb{Q}/\mathbb{Z}\times \mathbb{Z}_2),\\ \left( ([p],a),([q],b)\right) &\mapsto ([p+aq], ab)\end{split} $$
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Yes, Take $GL_n(R)$ as mother group. Construct $G_n$ be cyclic subgroup of $GL_n(R)$generated by an element order $n$.
Define, $G=S_3×G_2×G_3×.......$ This is non abelian with every elements of finite order having one element of order n
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The ascending union of $G_p=\bigcup_n GL_n(\mathbb F_p)$ (imbedded by blocks in the upper left corners) has elements of every order not divisible by prime $p$ (and, yes, some with orders divisible by $p$, but none by $p^2$). Thus, for distinct primes $p,q$, the product $G_p\times G_q$ has elements of every (finite) order, none of infinite order.
answered May 28 at 17:15