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informally speaking, what algebraic structure does the set of conjugacy classes of a group carry?

Formally, I'm interested in natural operations on conjugacy classes. Let $\mathsf{Grp}$ be the category of groups and $\mathsf{CC} : \mathsf{Grp} \to \mathsf{Set}$ the functor mapping every group to its set of conjugacy classes. Then a natural operation of arity $n \in \mathbb{N}$ is a natural transformations of the form $$\mathsf{CC}^{\times n} \longrightarrow \mathsf{CC}.$$

Here are the ones that I know of (hat tip to YCor):

  • Projecting onto one of the factors.
  • In arity $n = 1$, taking a power $x \mapsto x^k$ for fixed $k \in \mathbb{Z}$, which descends to conjugacy classes.
  • Composing the first kind with the second.
  • In arity $n = 0$, returning the class of the neutral element.

Are there any other ones? In particular, are there nontrivial operations of arity $> 1$?

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    $\begingroup$ Arity 1: for each $d$, the map $x\mapsto x^d$ factors modulo conjugation. $\endgroup$
    – YCor
    Commented May 25 at 11:46
  • $\begingroup$ @YCor, d'oh, of course...! I'll add that to the question. $\endgroup$
    – Tobias Fritz
    Commented May 25 at 11:47
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    $\begingroup$ For power maps, you don’t need $k\in\mathbb{N}$. Negative $k$ works as well. $\endgroup$
    – Jeremy Rickard
    Commented May 25 at 12:03
  • $\begingroup$ thanks @JeremyRickard, corrected. $\endgroup$
    – Tobias Fritz
    Commented May 25 at 12:05
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    $\begingroup$ Given a group $G$, there are also left/right $G$-actions $$\begin{align*}\lambda_G&\colon\mathrm{Z}(G)\times\mathrm{CC}(G)\to\mathrm{CC}(G),\\\rho_G&\colon\mathrm{CC}(G)\times\mathrm{Z}(G)\to\mathrm{CC}(G)\end{align*}$$ given by left/right multiplication. It would be interesting to explore how the natural operations $\mathrm{CC}^{\times n}\to\mathrm{CC}$ interact with these, e.g. which ones would be maps of left/right $G$-sets componentwise? Alternatively, we could also study natural operations $\mathrm{CC}^{\boxtimes n}\to\mathrm{CC}$, where $\boxtimes$ is the tensor product of $G$-sets. $\endgroup$
    – Emily
    Commented May 25 at 15:19

1 Answer 1

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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. Write $[a]$ for the conjugacy class of $a$. Then, taking $G$ to be free on generators $x$ and $y$ and choosing a representative for the class $f([x],[y])$, i.e. a word $w(x,y)$ such that $[w(x,y)]=f([x],[y])$, by naturality we have $[w(a,b)]=f([a],[b])$ in general. In the free group on generators $x,y,z$ the words $w(zxz^{-1},y)$ and $w(x,y)$ are then conjugate. It is clear then that in the (reduced) word $w(x,y)$ either the letter $x$ does not occur or the letter $y$ does not.

Indeed, unless $w$ is of the form $x^k$ or $y^k$, by replacing $w$ with a conjugate we may assume $w(x,y) = x^{a_1} y^{b_1} \cdots x^{a_n} y^{b_n}$ where $a_1, b_1, \dots, a_n, b_n$ are all nonzero. Then $w(zxz^{-1}, y) = z x^{a_1} z^{-1} y^{b_1} \cdots z x^{a_n} z^{-1} y^{b_n}$. This expression is cyclically reduced and contains a $z$, so $w(zxz^{-1}, y)$ is not conjugate to $w(x, y)$.

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    $\begingroup$ Not so fast. $w$ could be $x^k$ or $y^k$. $\endgroup$
    – Sean Eberhard
    Commented May 25 at 18:44
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    $\begingroup$ @SeanEberhard, if one of the letters doesn’t appear then this is really arity 1, and we know this case. I think the point is if both letters appear, then these won’t be conjugate $\endgroup$
    – Benjamin Steinberg
    Commented May 25 at 19:25
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    $\begingroup$ @Sean Eberhard: Yes, I was too hasty. I've edited accordingly. $\endgroup$
    – Tom Goodwillie
    Commented May 25 at 22:01
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    $\begingroup$ Nice! If someone can add a bit more detail on the "I believe it is clear" part, then I'll accept this answer. $\endgroup$
    – Tobias Fritz
    Commented May 26 at 8:35
  • $\begingroup$ Each conjugacy class is represented by a cyclically reduced word w. If w contains both x,y then you can cyclically conjugate it to make sure that w doesn't both begin and end with x or it's inverse. Wlog assume w begins with x or x^-1 and ends with y or y^-1. Then replacing x by zxz^-1 and reducing will end up with a cyclically word beginning with z and ending with y or y^-1 and this is not conjugate to w $\endgroup$
    – Benjamin Steinberg
    Commented May 26 at 12:28

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