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Given a finite group $G$, is $G$ determined by its category of maps from abelian groups? Specifically, we can form the category $G_A$ of "abelian points" of $G$ with objects pairs $(A,\phi)$, with $\phi:A\rightarrow G$, for $A$ abelian, with morphisms $\xi:(A,\phi)\rightarrow (B,\psi)$ given by morphisms of groups $\xi:B\rightarrow A$ such that $\phi \circ \xi = \psi$.

Very similar questions have been asked before, but they focused on the maps from cyclic groups, and the (small) counterexamples given for the analagous question do not have isomorphic categories of abelian points.

The following are the similar questions I refer to: Is a finite group uniquely determined by the orders of its elements? If I know the order of every element in a group, do I know the group?

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    $\begingroup$ In terms of the category $G_A$, are you asking where an equivalence (or isomorphism?) between $G_A$ and $H_A$ implies $G\cong H$? $\endgroup$ – Ittay Weiss May 27 '18 at 13:09
  • $\begingroup$ Yea, that's a good way of phrasing it. $\endgroup$ – user277182 May 27 '18 at 13:37
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    $\begingroup$ Although the question is about finite groups, it seems worth mentioning that if we allow infinite groups, any two Tarski $p$-monsters for the same prime $p$ provide a counterexample. $\endgroup$ – Ravi Fernando May 29 '18 at 0:20
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No. Consider two semidirect products $(\mathbf{Z}/7\mathbf{Z})^2\rtimes(\mathbf{Z}/3\mathbf{Z})$; where the canonical generator of $\mathbf{Z}/3\mathbf{Z}$ acts in the first case by the matrix $\begin{pmatrix}2 & 0\\0 & 2\end{pmatrix}$, and in the second case by $\begin{pmatrix}2 & 0\\0 & 4\end{pmatrix}$. They're clearly not isomorphic, but they have the same "combinatorics" of abelian subgroups: the unique $7$-Sylow subgroup (elementary abelian of order $7^2$) and its subgroups, and $7^2$ subgroups of order 3, and all these have pairwise trivial intersection.

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  • $\begingroup$ PS they're not isomorphic because for the second one the determinant of the action on the 7-Sylow is one, and not for the first one. $\endgroup$ – YCor Jun 1 '18 at 17:06
  • $\begingroup$ Or because all subgroups of $(\mathbb Z/7\mathbb Z)^2$ are normal in the first, but not in the second. $\endgroup$ – Ravi Fernando Jun 1 '18 at 17:09
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    $\begingroup$ Yes indeed. Actually both arguments are ad-hoc. If we consider the group $G_{m,n}$ defined as $(Z/7Z)^{m+n}\rtimes (Z/3Z)$ with action by the diagonal matrix with $m$ diagonal entries equal to $2$ and $n$ equal to $4$, then the only nontrivial isomorphisms are between $G_{m,n}$ and $G_{n,m}$. $\endgroup$ – YCor Jun 1 '18 at 17:22
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Not a full answer but hopefully a good starting point. Let $G_1,G_2$ be two groups and $\varphi\colon G_1 \to G_2$ a homomorphism. We'll say $\varphi$ is good if in the semi-direct product $G_1 \rtimes_\varphi G_2$ elements $x,y$ commute iff either $x,y\in G_1$ or $x,y\in G_2$. In that case, any homomorphism $A\to G_1\rtimes_\varphi G_2$ from an abelian group lands in either $G_1$ or $G_2$. In other words, if $\varphi$ is good, then $(G_1\rtimes_\varphi G_2)_A\cong (G_1/A)\times (G_2/A)$. In particular, if you can find two good twisting homomorphisms $\varphi_1,\varphi_2$, then the category invariant you consider will not be able to differentiate between $G_1\rtimes_{\varphi_1} G_2$ and $G_1 \rtimes_{\varphi_2}G_2$. If $\varphi_1,\varphi_2$ can be chosen to produce non-isomorpihc semi-direct products, then you're done. This latter property is easy to achieve (generally, semi-direct products on the same factors are highly sensitive to the twisting homomorphism, even for very small groups). I suspect making sure the twisting homomorphisms are also good is possible, though I don't have a construction right now.

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