Timeline for Can a group generated by its involutions, the product of every two of which has order a power of 2, have an element of odd order?
Current License: CC BY-SA 4.0
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| Sep 6, 2023 at 8:00 | comment | added | Dave Benson | Edited for clarity. | |
| Sep 6, 2023 at 7:59 | history | edited | Dave Benson | CC BY-SA 4.0 |
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| Sep 6, 2023 at 7:48 | comment | added | Derek Holt | @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. | |
| Sep 6, 2023 at 7:22 | comment | added | YCor | What does the notation $O_p(G)$ stand for? (my rough guess is: the largest normal $p$-subgroup) | |
| Sep 6, 2023 at 7:21 | comment | added | YCor | @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. | |
| Sep 6, 2023 at 4:34 | comment | added | testaccount | @R.vanDobbendeBruyn: Or, two involutions generate a dihedral group of order $2n$, where $n$ is the order of their product. | |
| Sep 6, 2023 at 0:37 | comment | added | R. van Dobben de Bruyn | 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 $ba$.) | |
| Sep 5, 2023 at 21:46 | comment | added | Max Horn | Nice! However, the involutions are not necessarily all conjugate? A priori one could imagine that each single class of involutions generates a proper subgroup. | |
| Sep 5, 2023 at 21:35 | comment | added | Dave Benson | en.wikipedia.org/wiki/Baer%E2%80%93Suzuki_theorem | |
| Sep 5, 2023 at 21:32 | history | answered | Dave Benson | CC BY-SA 4.0 |