Is there any concept in monoids that is similar to the concept "conjugate class" in groups? For example, are there any such similar concept in symmetric inverse monoids? Thank you very much.


There are several different definitions of conjugacy for semigroups. For inverse semigroups the best, in my opinion, definition is this: $a$ is conjugate to $b$ if there exists $t$ such that $t^{-1}at=b$, $tt^{-1}=e, t^{-1}t=f$, $ae=ea=a$, $bf=fb=b$. In this case the conjugacy relation is an equivalence relation and the concept of conjugacy classes is well defined, and easy to describe in the case of symmetric inverse monoids. See, for example, the following article by Akihiro Yamamura, http://journals.cambridge.org/download.php?file=%2FJAZ%2FJAZ70_02%2FS1446788700002639a.pdf&code=213e793755d6e9249ea5704c1f57a0b3 .

  • $\begingroup$ Mark, are Green's (semigroup) relations any help in determining conjugacy or notions similar to conjugacy? Gerhard "Ask Someone Else About Semigroups" Paseman, 2011.03.05 $\endgroup$ – Gerhard Paseman Mar 6 '11 at 6:27
  • $\begingroup$ @Gerhard: Green relations are relevant but a connection is not direct. Certainly two conjugate elements must be in the same $D$-class but the converse is not true (as the example of an Abelian group shows). Still sometimes conjugacy classes can be described by looking at the conjugacy classes of the maximal subgroups and on the structure of Green relations. For example, I believe it is so for the Brandt semigroup and completely 0-simple semigroups in general. $\endgroup$ – user6976 Mar 6 '11 at 19:42
  • $\begingroup$ Actually I am not completely certain that conjugate elements in an inverse semigroup are $D$-related. They are clearly $J$-related. $\endgroup$ – user6976 Mar 6 '11 at 20:14

As so often happens when you generalize from groups to monoids, it gets more complicated.

Conjugacy is, of course, an equivalence relation on any group X. Two elements x, y are conjugate if and only if there exists g in X with gx = yg. This last equation makes sense in a monoid, but does not define an equivalence relation: it's not symmetric.

Here's one way to think about it. With every monoid X there is associated a category C(X). The objects of C(X) are the elements of X, and a map x --> y in C(X) is an element g of X such that gx = yg. Composition is defined in the obvious way. When X is a group, C(X) is a groupoid (i.e. a category in which every map is invertible).

Now, in a category there are at least two natural equivalence relations on the collection of objects: isomorphism, and being in the same connected-component. When the category is a groupoid, these are the same.

Apply this observation to C(X), where X is your monoid. It tells us that there are two natural equivalence relations on the elements of X:

  • "isomorphism", which identifies x and y when there exists an invertible g with gx = yg
  • "being in the same connected component", which is the equivalence relation generated by identifying x and y whenever there exists a (not necessarily invertible) g with gx = yg.

When X is a group, these two equivalence relations are the same thing---conjugacy.

Postscript The category C(X) can be understood abstractly as follows. A one-object category is the same thing as a monoid. Functors between one-object categories are the same thing as monoid homomorphims. A homomorphism from N (the additive monoid of natural numbers) to a monoid X is the same thing as an element of X. Thus, an element of X is the same thing as a functor N --> X. In other words, the elements of X are the objects of the functor category X^N = [N, X]. And this functor category is exactly C(X).


This seems to be a relevant paper: Kudryavtseva and Mazorchuk - On three approaches to conjugacy in semigroups.

  • $\begingroup$ Link broken; can you provide the name of the article please? $\endgroup$ – Morgan Rogers Feb 27 '19 at 10:43
  • 2
    $\begingroup$ @MorganRogers, I fixed it. $\endgroup$ – Benjamin Steinberg Jan 28 at 21:41

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