$\newcommand\Suz{\mathit{Suz}}$I recently noticed that the Suzuki group $\Suz$ has as subgroups classes of both $L_3(3)$ and $M_{12}$, both of which are also subgroups of the Mathieu groupoid $M_{13}$. Additionally, $\lvert M_{13}\rvert$ divides $\lvert\Suz\rvert$. Could it be that $M_{13}$ embeds into $\Suz$?
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$\begingroup$ Not familiar with groupoids, so I would be curious to see examples of a groupoid embedded into a finite group. (Other than subgroups of a finite group) $\endgroup$– spinAug 13, 2022 at 8:21
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2$\begingroup$ A necessary condition for a groupoid $(G,*)$ to embed into a group is that for all $a,b \in G$ then $a∗a^{−1}=b∗b^{−1}$. Is it true for $G=M_{13}$? If not, then $M_{13}$ cannot embed into a group, in particular $Suz$. Else, Thomas Connor and Dimitri Leemans may be able to help: leemans.dimitri.web.ulb.be/atlaslat $\endgroup$– Sebastien PalcouxAug 13, 2022 at 9:23
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$\begingroup$ @SebastienPalcoux $M_{13}$ can be constructed as a set of permutations on the 13 points of PG(2,3) and therefore embeds into the symmetric group $S_{13}$. $\endgroup$– Daniel SebaldAug 15, 2022 at 19:00
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4$\begingroup$ What does it even mean for a groupoid to embed in a group? $M_{13}$ may be defined in terms of permutation of 13 things, but that doesn't make it a subgroupoid (again, what even is that?) of $S_{13}$, because there are peculiar rules (en.wikipedia.org/wiki/Mathieu_groupoid#Construction). $\endgroup$– Sean EberhardAug 16, 2022 at 8:56
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1$\begingroup$ @Daniel what Sebastien's post means is the set of units, namely $a\ast a^{-1}$ as $a$ varies over all elements of the groupoid, consists in fact of a single element, because this is true in a group, and if the groupoid is embedded in a group, the multiplication and inverses are respected. I believe the groupoid embeds in the action groupoid of $S_{13}$ on the 13-element set used to build $M_{13}$, and not $S_{13}$ itself. $\endgroup$– David Roberts ♦Nov 21, 2022 at 5:04
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