# Torsion group with finitely many elements of order 2 but infinitely many elements of order 4

Does there exists a group $$G$$ satisfying all the following conditions?

1. $$G$$ is finitely generated,
2. $$G$$ is of bounded torsion (has finite exponent),
3. $$G$$ has finitely many elements of order $$2$$,
4. $$G$$ has infinitely many elements of order $$2^l$$ for some $$l$$ (for example $$l=2$$).

By 1. and 2. $$G$$ is a quotient of a free Burnside group.

Near-examples of such groups violating only Condition 2. are generalized dicyclic group $$Dic(A,y)$$, where $$A$$ is a finitely generated abelian group.

References on somehow related problems are welcome.

• Note that for an infinite group $G$, 4 is implied by 2 (so 4 is just "$G$ is infinite"). Also no infinite locally finite group satisfies (2 and 3) because it would have an infinite abelian subgroup with the same property and this can be discarded. I don't know either if $2$ is replaced by an odd prime, which may make the question easier or harder, I don't know.
– YCor
Commented Apr 28, 2020 at 16:55
• BTW: the assumption implies that the FC-center is non-trivial. Using the restricted Burnside, we can suppose that $G$ has no nontrivial finite quotient. Hence the FC-center is central. Also, because there's no locally finite example, the locally finite radical is finite, hence equals the FC-center, and hence equals the center. Then, modding out by a subgroup of index 2 in center yields an example in which the locally finite radical has order $2$. So we can reduce to this problem.
– YCor
Commented Apr 28, 2020 at 17:01
• @YCor In order for the implication (G infinite) 2=>4 to be true, the condition that the exponent of G is even is necessary. On the other hand, I have trouble to see why it is sufficient.
– PHL
Commented Apr 30, 2020 at 8:22
• Let $G$ have finite exponent, and finitely many elements of each 2-power order. So it has finitely many elements of 2-power order. Let $W$ be the subgroup generated by these elements. Since these elements of 2-power order have finite conjugacy classes, $W\subset FC(G)$, the FC-center of $G$, which is locally finite, and in particular $W$ is finite. This proves that (2 and not-4) implies (not-3). So (2 and 3) implies (4).
– YCor
Commented Apr 30, 2020 at 9:40

Here is an example with infinitely many $$4$$-torsion elements and only one $$2$$-torsion element. (The smallest number of generators I can do is $$4$$, and the exponent is pretty big and not a power of $$2$$.)

Example. Pick $$n$$ odd and $$d \geq 2$$ such that $$B(d,n)$$ is infinite. Let $$p$$ be an odd prime. Set $$A = \mathbf Z/2 \oplus \bigoplus_{g \in B(d,n)} (\mathbf Z/p)e_g,$$ with distinguished element $$y = (1,0,\ldots)$$ or order $$2$$. Then the generalised dicyclic group $$\operatorname{Dic}(A,y) = A \amalg Ax$$ satisfies all criteria except 1:

1. $$\operatorname{Dic}(A,y)$$ is not finitely generated because it has an index $$2$$ subgroup $$A$$ that is not finitely generated.
2. Every element of $$\operatorname{Dic}(A,y)$$ is killed by $$4p$$.
3. The only element of order $$2$$ is $$y \in A$$.
4. Every element in $$Ax$$ has order $$4$$, and $$Ax$$ is infinite by assumption.

Finally, the group $$B(d,n)$$ acts on $$A$$ fixing $$y$$ by $$(g,e_h) \mapsto e_{gh}$$. Thus this extends to an action on $$\operatorname{Dic}(A,y)$$, and we take $$G$$ to be the semidirect product $$G = \operatorname{Dic}(A,y) \rtimes B(d,n).$$ Then all criteria are satisfied:

1. If $$x_1,\ldots,x_d$$ are the standard generators of $$B(d,n)$$, then $$x, e_1, x_1,\ldots,x_d$$ generate $$G$$. Indeed, they generate the quotient group $$B(d,n)$$, hence we get all elements $$ge_1g^{-1} = e_g$$, hence we get everything.
2. Since $$B(d,n)$$ has exponent $$n$$ and $$\operatorname{Dic}(A,y)$$ has exponent $$4p$$, we conclude that $$G$$ has exponent dividing $$4pn$$.
3. and 4. Since $$B(d,n)$$ has odd exponent, all $$2$$-power torsion happens in $$\operatorname{Dic}(A,y)$$.

So $$G$$ is an example. $$\square$$

Remark. I have no idea if there are also examples of $$2$$-power exponent.

• Oh nice. I tried something similar and this failed because I stuck trying with 2-power exponent locally finite groups for which this is hopeless (whence my initial comment).
– YCor
Commented Apr 29, 2020 at 9:01

Here's an elaboration on R. van Dobben de Bruyn's answer.

Let $$A$$ be an abelian group. Define the group $$\mathrm{Di}(A)$$ as follows: (a) consider the direct product $$A'=A\times C_2$$, denoting $$y$$ the nontrivial element of $$C_2$$. (b) perform the semidirect product $$A''=C_4\ltimes A$$, with $$\pm$$-action. Denote by $$z$$ the element of order 2 of the acting $$C_4$$: it remains central in $$A''$$, and so does $$y$$. (c) Obtain $$\mathrm{Di}(A)$$ (generalized dicyclic group on $$A$$) by modding out $$A''$$ by the central subgroup of order 2 $$\langle z^{-1}y\rangle$$. Still denote by $$y$$ the image of $$y$$ in $$\mathrm{Di}(A)$$, it it central of order $$2$$. Note that $$\mathrm{Di}(A)/\langle y\rangle$$ is the dihedral product $$C_2\ltimes_\pm A$$.

In $$A''$$, denoting by $$t$$ a generator of $$C_4=\{1,t,z,t^{-1}\}$$ and writing $$A$$ additively, we have $$(t^{\pm},a)^2=(z,0)$$, $$(z,a)^2=(1,a)^2=(1,2a)$$ for $$a\in A'$$. In particular, $$\eta=(z,y)$$ (which is killed in $$\mathrm{Di}(A)$$ is not a square. Hence the elements of order $$\le 2$$ in $$\mathrm{Di}(A)$$ are the images of elements of order $$2$$ in $$A''$$, which are the elements of the form $$(1,a),(z,2a)$$ with $$2a=0$$. In particular, if $$A$$ has no element of order $$2$$, these elements are $$(1,0),(1,y),(z,0),(z,y)$$, which in $$\mathrm{Di}(A)$$ is reduced to $$\{1,y\}$$. That is, if $$A$$ has no element of order $$2$$ then the only element of order $$2$$ in $$\mathrm{Di}(A)$$ is $$y$$.

Also this shows that all elements $$(t^\pm,a)$$ map to elements of order $$4$$ in $$\mathrm{Di}(A)$$, which therefore has infinitely many elements of order $$4$$ if $$A$$ is an arbitrary infinite abelian group.

Now the construction $$A\mapsto\mathrm{Di}(A)$$ is clearly functorial under group isomorphisms. Hence, if $$\Gamma$$ acts on $$A$$ by automorphisms, then it naturally acts on $$\mathrm{Di}(A)$$ by automorphisms: in steps: (a) extend in the trivial way the action to $$A'=A\times C_2$$, then (b) extend in the trivial way to $$C_4\ltimes A'$$ (acting trivially on $$C_4$$): this works because the $$C_4$$-action, by $$\pm$$, commutes with the $$\Gamma$$-action on $$A'$$; finally this action fixes $$\eta=(z,y)$$ and hence passes to the quotient to an action on $$\mathrm{Di}(A)$$, defining a semidirect product $$\mathrm{Di}\rtimes\Gamma$$.

From what's preceding, we immediately get: if $$\Gamma$$ has no element of order $$2$$, then the only element of order $$2$$ in $$\mathrm{Di}(A)\rtimes\Gamma$$ is $$y$$; if $$\Gamma$$ is infinite then it contains $$\mathrm{Di}(A)$$ hence has infinitely many elements of order $$4$$.

Finally we can choose $$A=C_n^{(\Gamma)}=\bigoplus_{\gamma\in\Gamma}C_n$$ for some odd $$n>1$$, and choose $$\Gamma$$ of finite odd exponent $$q>1$$. Here $$n$$ and $$q$$ are unrelated, can be chosen equal or not (they could be chosen even for the construction but then this will produce infinitely elements of order $$2$$). Then the resulting group $$G=\mathrm{Di}(C_n^{(\Gamma)})\rtimes \Gamma$$ works: it has a single element of order $$2$$, infinitely of order $$4$$, and has exponent dividing $$nq$$. Actually modding out by $$\langle y\rangle$$, the resulting group admits, as subgroup of order $$2$$, the standard wreath product $$C_n\wr \Gamma$$ (in particular, all elements of order $$4$$ in $$G$$ lie in the nontrivial coset of the unique subgroup of index $$2$$).

(In general— arbitrary abelian group $$A$$, arbitrary $$\Gamma$$-action on $$A$$, we always get this subquotient killing the center $$\langle y\rangle$$ and passing to a subgroup of index $$2$$, which yields the semidirect product $$A\rtimes\Gamma$$. For instance, if we want $$G$$ to have Kazhdan's Property T, the permutational action is hopeless, but possibly some choice of $$\Gamma$$-module works, namely we need $$A\rtimes\Gamma$$ to have Kazhdan's Property T.)