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

Question

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.

Answer 1

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.

Answer 2

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.)