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Let $G$ be a group which is generated by the set of its involutions, and assume that the product of every two involutions in $G$ has order a power of 2. Is it possible that $G$ has an element of odd order $\neq 1$?

Added on Sep 7, 2023: Dave Benson gave a nice answer for the case that $G$ is finite. However, the more interesting general case remains open so far.

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    $\begingroup$ The title looks different from the main question, I'd write "by its involutions" — reading the title I understood "by a set of involutions the product every 2 of which being of 2-power order", and this sounds much easier. $\endgroup$
    – YCor
    Commented Sep 5, 2023 at 19:03
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    $\begingroup$ @YCor Thank you! -- Fixed. $\endgroup$
    – Stefan Kohl ♦
    Commented Sep 5, 2023 at 19:06
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    $\begingroup$ @YCor The answer to the easier version of the question you suggested is yes. For example, the four involutions $(1,2)$, $(3,5)$, $(4,6)$ and $(1,3)(2,4)(5,6)$ generate ${\rm S}_6$, and the product of every two of them has order a power of 2. $\endgroup$
    – Stefan Kohl ♦
    Commented Sep 5, 2023 at 19:55
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    $\begingroup$ @MaxHorn No, it is not. For example, $(1,2)$ and $(2,3)$ are involutions in ${\rm S}_6$, but their product has order 3 (i.e. not a power of 2). $\endgroup$
    – Stefan Kohl ♦
    Commented Sep 5, 2023 at 20:32
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    $\begingroup$ About infinite groups: my belief it that it exists. Rough idea: I think it is known that the group $C_2\ast C_2\ast C_2$ admits a torsion quotient, and that it can arranged to be a 2-group. But the proof is quite flexible, and should imply that imposing that various elements are torsion can be made independently. So, one should be able to prescribe odd (or infinite) orders to suitable elements, beyond the products of two conjugates of generators. Details should be done carefully — still, I think it's accessible with early machinery of hyperbolic groups (and easier than bounded torsion). $\endgroup$
    – YCor
    Commented Sep 8, 2023 at 10:21

2 Answers 2

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The Baer-Suzuki theorem (Aschbacher, "Finite group theory", Theorem 39.6) says that if $X$ is a $p$-subgroup of a finite group $G$ then either $X\le O_p(G)$ or there exists $g\in G$ such that $\langle X,X^g\rangle$ is not a $p$-group. Taking $p=2$ shows that if $X=\{1,t\}$ has order two and its conjugates generates $G$, and the product of any two conjugates of $t$ generate a $2$-group then $G$ is a finite $2$-group. If there is more than one conjugacy class of involutions, this shows that each conjugacy class generates a subgroup of $O_2(G)$; they normalise each other, so together they generate a $2$-group.

This only answers the question if $G$ is finite, so any counterexample would have to be infinite.

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    $\begingroup$ Nice! However, the involutions are not necessarily all conjugate? A priori one could imagine that each single class of involutions generates a proper subgroup. $\endgroup$
    – Max Horn
    Commented Sep 5, 2023 at 21:46
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    $\begingroup$ Another detail that is true but you didn't write is that two involutions generate a 2-group if (and only if) their product has 2-power order. (This is pretty obvious too, as reduced words in $C_2 \ast C_2$ have the two letters occurring alternatingly. Odd length words are palindromic so have order $2$, and even length words are a power of $ab$ or $ba$.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Sep 6, 2023 at 0:37
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    $\begingroup$ @R.vanDobbendeBruyn: Or, two involutions generate a dihedral group of order $2n$, where $n$ is the order of their product. $\endgroup$
    – testaccount
    Commented Sep 6, 2023 at 4:34
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    $\begingroup$ @DaveBenson the problem, when you mod out (e.g., by a normal 2-subgroup), is that you can produce new elements of order 2, and then can break the assumption that the product of any two elements of order 2 is a 2-power. $\endgroup$
    – YCor
    Commented Sep 6, 2023 at 7:21
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    $\begingroup$ @YCor Yes, $O_p(G)$ is the $p$-core of $G$, the largest normal subgroup that is a $p$-group. Perhaps the easiest way to argue is that by the Baer-Suzuki Theorem either all involutions lie in $O_2(G)$, in which case $G$ is a $2$-group, or there exist two (conjugate) involutions that do not generate a $2$-group. $\endgroup$
    – Derek Holt
    Commented Sep 6, 2023 at 7:48
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This answer is written after taking into account Dave's answer https://mathoverflow.net/a/454042/14094 and the comments therein.

Here's the Baer-Suzuki theorem

If any two elements of a conjugacy class $C$ of a finite group generate a nilpotent subgroup, then all elements of the conjugacy class $C$ are contained in a nilpotent subgroup.

Also recall that in a nilpotent group, for each $p$ the set of elements of $p$-power order forms a subgroup (i.e., there's a unique $p$-Sylow).

Say that a group has Property $X(p)$ if it is generated by a conjugacy-invariant subset $S$ such that every element in $S$, and also the product of any two conjugate elements in $S$, has $p$-power order.

Note that Property $X(p)$ passes to quotients (unlike, a priori, the initial property of OP).

Claim: every finite group with Property $X(p)$ is a $p$-group.

Let $G$ be a counterexample of minimal cardinal, and $S$ as above. Let $T$ be a conjugacy class in $S$. By the Baer-Suzuki theorem, $T$ generates a nilpotent subgroup $N$, necessary a normal $p$-subgroup. So $G/N$ is also not a $p$-group, and inherits Property $X(p)$, contradiction.

Of course the property considered by OP (which requires that $S$ consists of those elements of order 2) implies $X(2)$. Hence, assuming finiteness, it forces the group to be a 2-group. However, the argument above doesn't directly work with OP's property, because a priori this property doesn't pass to quotients (due to the fact that passing to a quotient produces new elements of order 2).

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