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45 votes
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Conjugacy classes of $\mathrm{SL}_2(\mathbb{Z})$

One can proceed as follows for $SL_2(\mathbb{Z})$. First, the trace is a conjugacy invariant. For trace $0$ there are two conjugacy classes represented by $\pmatrix{0 & 1 \\ -1 & 0}$ and $\...
Lee Mosher's user avatar
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23 votes

Conjugacy classes of $\mathrm{SL}_2(\mathbb{Z})$

This is the subject of Gauss' reduction theory, as discussed in Karpenkov's book (among many other places). In this 2007 paper, Karpenkov also extends the method to $SL(n, \mathbb{Z}).$
Igor Rivin's user avatar
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17 votes
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Algebraic structure on conjugacy classes

I claim that there are no natural binary operations (and the same method should apply to any arity $\ge 2$) except those which depend on at most one of the two inputs: Suppose that $f$ is natural. ...
Tom Goodwillie's user avatar
16 votes

Conjugacy classes of $\mathrm{SL}_2(\mathbb{Z})$

The conjugacy classes of elements of ${\rm SL}(2,\mathbb{Z})$ with given trace are counted in: S. Chowla, J. Cowles and M. Cowles: On the number of conjugacy classes in SL(2,Z). Journal of Number ...
Stefan Kohl's user avatar
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14 votes
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Finite groups with lots of conjugacy classes, but only small abelian normal subgroups?

No: there exists a sequence of finite groups with commuting probability bounded away from 0 but with no abelian (normal) subgroup of bounded index. Fix a prime power $q$. Consider the "higher ...
YCor's user avatar
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14 votes

Conjugacy classes as left Kan extension of forgetful functor

$\newcommand{\Gp}{\mathbf{Grp}} \newcommand{\conj}{^{\mathbf{conj}}} \newcommand{\Gpd}{\mathbf{Gpd}} \newcommand{\Set}{\mathbf{Set}} \newcommand{\ho}{\mathrm{ho}} \newcommand{\Fun}{\mathrm{Fun}} \...
Maxime Ramzi's user avatar
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11 votes
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Groups with three conjugacy classes that define an ordering

There are currently no known examples of bi-orderable groups where all positive elements are conjugate. The question of their existence appears as Problem 3.31 of the 2009 problem list Unsolved ...
shane.orourke's user avatar
11 votes

Conjugacy classes in towers of groups

YCor beat me to it, but I will post my answer anyway because it is rather different. I will construct a counterexample to the first question with $\Gamma = F_2 = F\{x,y\}$, the free group on two ...
Sean Eberhard's user avatar
10 votes

Center of a monoid ring

This was originally two questions, one asking about the center of a group ring and one asking about the center of a monoid ring. The answer for groups is quite simple but the answer for monoids is ...
Benjamin Steinberg's user avatar
10 votes
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Are the character degrees determined by the conjugacy class sizes?

SmallGroup(128,227) and SmallGroup(128,731)) are counterexamples. ...
Jeremy Rickard's user avatar
10 votes
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Conjugacy classes in towers of groups

Here is a counterexample, with the Heisenberg group. Define by induction $a_0=1$ and $a_{n+1}=(a_n^2+1)a_n$. Clearly $a_n$ tends to infinity. Let $\Gamma$ be the Heisenberg group, consisting of ...
YCor's user avatar
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9 votes
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Why would dim primitive irrep divide size of some conjugacy class ?

I have not noticed this question before, though it was posted several years ago. As a comment on the question as a whole, and especially Question 1 asked in the text, there are likely to be many such ...
Geoff Robinson's user avatar
9 votes
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A pair of non-conjugate subgroups: a simple proof

I think that the elements $g = \dfrac1{\sqrt2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}^{\oplus3}$ and $h = \dfrac1 2\begin{pmatrix} 1 & 1 & 1 & 0 & 1 & 0 \\ -1 & 1 ...
LSpice's user avatar
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8 votes
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Size of conjugacy classes in infinite groups

The answer is no: there exists a 2-generated group, having finite conjugacy classes of unbounded size. Indeed B.H. Neumann (1937) produced a 2-generated group $G$ with normal subgroups $(H_n)_{n\ge 5}$...
YCor's user avatar
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8 votes

Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$?

Oh, I did not know about this ongoing discussion on math overflow. Amri pointed out to me about this discussion today morning only. As I discussed with you in a private communication, I don't know how ...
Pooja Singla's user avatar
8 votes
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Number of conjugacy classes of pairs of commuting elements II

Revised answer, to cover more of question: The proof that $r_{G} \geq \frac{8p}{3}$ if $G$ is non-Abelian is tedious, but not too difficult, I think, and this is attained, (as you say),for $p = 3$ by ...
Geoff Robinson's user avatar
7 votes
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Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$?

I would have preferred to not answer my own question, but here it goes. Yes, the two groups have the same number of conjugacy classes and in fact, the groups $\mathrm{SL}_{n}(W_{2}(\mathbb{F}_{q}))$ ...
A Stasinski's user avatar
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7 votes
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Constructing the largest finite group with a fixed number of conjugacy classes

Following @verret comment, here are the list of the largest finite groups $G_k$ with a fixed class number $k\le 14$ (i.e. the number of conjugacy classes of elements, or the number of irreducible ...
Sebastien Palcoux's user avatar
7 votes

Conjugacy classes as left Kan extension of forgetful functor

An alternative to Maxime's sophisticated proof is to observe that $X$ and $L$ are both corepresented by the free group, because a group homomorphism from a free group (up to conjugacy) is the same as ...
Shay Ben Moshe's user avatar
6 votes
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Variety of conjugacy classes

As already noted, the quotient need not be even $T_1$. For example, $SL_2(\mathbb{C})/SL_2(\mathbb{C})$ is homeomorphic to $\mathbb{C}$ with double points at $\pm 2$. This quotient is not $T_1$ ...
Sean Lawton's user avatar
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6 votes

Are the character degrees determined by the conjugacy class sizes?

Here is a general comment related to my answer to a previous MO question. If $\chi$ is a complex irreducible character of a finite group $G$, and $\chi$ takes a root of unity value at $x \in G$, then $...
Geoff Robinson's user avatar
6 votes

What are the conjugacy classes of the category of ($\kappa$-small) sets?

So for endonorphisms up to isomorphisms, you're asking for a description of endomorphisms of sets. It sort of depends what kind of description you're looking for, but you could imagine something like &...
Maxime Ramzi's user avatar
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5 votes

Size of conjugacy classes in infinite groups

The size of all conjugacy class of a group $G$ is bounded if and only if the derived subgroup $G^\prime$ of $G$ is finite. This is a celebrated theorem of Neumann proved in the following paper: B.H. ...
Salvatore Siciliano's user avatar
5 votes
Accepted

A probability problem in the conjugacy classes of symmetric group

Let $kp$ be the size of the support of $\sigma$. Let $1,2,3,4$ be four points of the ground set. The probability that $\sigma_1$ maps $1 \mapsto 2$ is $kp/n(n-1)$, because there is a $kp/n$ chance ...
Sean Eberhard's user avatar
5 votes

Number of conjugacy classes of a semi-direct product of two finite groups

For an example with equality, consider the order-$16$ central product of $D_4$ and $C_4$, i.e., the group (of order $16$) $G=NK$ with $N\cong D_4$, $K\cong C_4$, $|N\cap K|=2$ and $C_G(N)=K$. This can ...
Joachim König's user avatar
5 votes
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Number of conjugacy classes of pairs of commuting elements

I think it is false in general that $r_{G} \geq p^{\frac{3}{2}},$ where $p$ is the largest prime divisor of the order of $G$. If we take a Frobenius group $G$ of order $pq,$ where $p,q$ are primes ...
Geoff Robinson's user avatar
5 votes
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Is there a way to study the relationship between the category of finite groups and their conjugacy classes categorically?

This question will likely be closed, but I'll try to give the OP some references before it is. First, let me answer the OP's question in the comments, regarding why this is vague and why the question ...
David White - gone from MO's user avatar
4 votes

Groups with three conjugacy classes that define an ordering

I have proposed a positive solution to this problem in a preprint entitled Hyperexponentially closed fields, to be found here, more precisely in Sections 10.1 and 10.2. The solution is based on work ...
nombre's user avatar
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4 votes
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Conjugacy classes in the automorphism group of a simple Lie algebra

If $\mathfrak{s}$ is $K$-anisotropic, where $K$ is a real or $p$-adic field (this is equivalent to $\mathfrak{s}$ not containing $\mathfrak{sl}_2$, and also to the corresponding group be compact), ...
YCor's user avatar
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