It is not too difficult to show that if $X$ is an infinite set, then there exists a two-element subset of the group $\operatorname{Sym}(X)$ with trivial centralizer iff $\lvert X\rvert \leq \lvert\mathbb{R}\rvert$.

My question is if this is true if we replace $\operatorname{Sym}(X)$ with $\operatorname{End}(X)$.

I.e., for what infinite sets $X$ do there exist functions $f,g: X \rightarrow X$, such that if $h:X \rightarrow X $ satisfies $fh = hf$ and $gh = hg$, then $h = I$? The same argument from the $\operatorname{Sym}(X)$ case shows that it is true when $|X| \leq \mathbb{R}$ (and was given as a problem in the 6th Romanian Masters of Mathematics competition). But is it false for $|X| > |\mathbb{R}|$?

  • $\begingroup$ I think $h = I$ is too much to ask for in general, as you can always set $g = f^{\circ k}$ for some $k\in\mathbb{N}$. Then by re-associating you have that any $h = f^{\circ n}$ works, but may not be the identity. $\endgroup$
    – Mark
    Apr 20, 2020 at 19:07
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    $\begingroup$ @Mark OP is right that there exists a pair in $\mathrm{Sym}(c)$ whose centralizer is trivial. If the idea you have in mind contradicts this, it means it does not work. $\endgroup$
    – YCor
    Apr 20, 2020 at 19:08
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    $\begingroup$ Don't use Sym$(X)$ Sym$(X)$ for upright text in math mode. $\operatorname{Sym}(X)$ $\operatorname{Sym}(X)$ is best. I have edited accordingly. $\endgroup$
    – LSpice
    Apr 20, 2020 at 19:10
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    $\begingroup$ It's very interesting; I'm curious what your motivation is. One is that the sentence, in a group "there exists a pair with trivial centralizer" can be written as a first-order formula, and for the symmetric groups $\mathrm{Sym}(X)$, it indeed characterizes $|X|\le c$ (while for $|X|>c$ every countable subset has centralizer of cardinal $|X|$). $\endgroup$
    – YCor
    Apr 20, 2020 at 19:19

1 Answer 1


The answer is no: for every set $X$ there exists a pair in the monoid $X^X$ of self-maps of $X$, with centralizer reduced to $\{\mathrm{id}\}$.

(I first left my original "groupwise" answer because it's easier and because it has other follow-up questions. It's now deleted and copied as an answer to another question).

For $X$ empty take $(\mathrm{id},\mathrm{id})$. For $X$ finite nonempty, take a constant, and a cycle. So henceforth I assume that $X$ is infinite.

(a) First I use Sierpiński-Banach theorem [cf. here and here] that every countable subset (here just finite is fine) of $X^X$ is contained in the subsemigroup generated by a 2-element subset. This reduces to proving that there is a finite (actually 6-element) subset $\Sigma\subset X^X$ with trivial centralizer.

(b) Next I split $X$ as union of two subsets $Y,Z$ of the same cardinal. Let $f,g\in X^X$ have image equal to $Y$ and $Z$ respectively. If $u$ commutes to $f$, then $u$ stabilizes $\mathrm{Im}(Y)$, and similarly with $g$, $Z$. I'll therefore assume $f,g\in\Sigma$, and hence every $u$ in the centralizer of $\Sigma$ stabilizes both $Y$ and $Z$.

(c) It was proved in [VPH] that there exists a "strongly rigid" binary relation on $Y$: a subset $R\subset Y^2$ (actually, $R$ being subset of a well-ordering) such that the only endomorphism $u$ of $(Y,R)$ is the identity. (Here endomorphism means that $u\times u:Y^2\to Y^2$ maps $R$ into itself.) Clearly the cardinal of $R$ is that of $|Y|=|X|$.

Choose a partition $Z=Z'\sqcup Z''$ of $Z$ in subsets of the same cardinal. Choose a bijection $i$ from $R$ to $Z'$. Define self-maps $p,q$ of $X$ as follows. On $Y$, $p$ and $q$ are chosen as injective maps into $Z''$. Also $p$ and $q$ are defined on $Z'$ by: for $(y,y')\in Y^2$ and $z=i(y,y')$, $q(z)=p(y)$ and $p(z)=q(y')$. Finally, extend $p,q$ arbitrarily choosing maps $Z''\to Y$.

Then, for $(y,y')\in Y^2$, we have $(y,y')\in R$ if and only if there exists $z_1,z,z_2\in Z$ such that $p(y)=z_1$, $q(z)=z_1$, $p(z)=z_2$, $q(y')=z_2$. [Intuition: this is a "$\stackrel{p}\to\stackrel{q}\leftarrow\stackrel{p}\to\stackrel{q}\leftarrow$ path" from $y$ to $y'$]

Indeed $\Rightarrow$ works by construction with $z_1=p(y)$, $z=i(y,y')$, $z_2=q(y')$. Conversely, suppose that such elements exist; write $(Y,Y')=i^{-1}(z)$, so $(Y,Y')\in R$. By definition $p(z)=q(Y')$ and $q(z)=p(Y)$. So $q(Y')=q(y')$ and $p(Y)=p(y)$. By injectivity of $p$ and $q$ on $Y$, we have $(y,y')=(Y,Y')\in R$.

As a consequence, if $u$ stabilizes $Z$ and $Y$ and commutes with $p$ and $q$, then $u$ preserves $R$ on $Y$.

Next we define similarly $p',q'$ from a strongly rigid binary relation on $Z$.

Then the above proves that the centralizer of $\{f,g,p,q,p',q'\}$ in $X^X$ is reduced to $\{\mathrm{id}\}$.

[VPH] Vopěnka, P.; Pultr, A.; Hedrlín, Z. A rigid relation exists on any set. Comment. Math. Univ. Carolinae 6 (1965), 149–155.

Informal outline: the hard step is the above reference (existence of a strongly rigid binary relation). Then, the 0th step is Sierpinski-Banach (which is not hard) to pass from 6 to 2. The second is quite trivial: there exists a pair such that centralizing this pair implies preserving each component of a partition into two moieties. The third step is to encode a binary relation into a pair of self-maps using such a $\stackrel{p}\to\stackrel{q}\leftarrow\stackrel{p}\to\stackrel{q}\leftarrow$ path and the "coloring" by the 2-component partition.

  • $\begingroup$ One related question I couldn't answer so far is whether there exists $f\in X^X$ whose centralizer is reduced to $\{f^n:n\ge 0\}$. Or if at the opposite, is it true that for $|X|>c$ every $f\in X^X$ has a centralizer cardinal $2^{|X|}$. $\endgroup$
    – YCor
    Apr 28, 2020 at 10:40
  • $\begingroup$ Eventually I made it a separate question: mathoverflow.net/questions/359660 $\endgroup$
    – YCor
    May 7, 2020 at 19:09
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    $\begingroup$ I used above the existence of a strongly rigid binary relation on every set (in ZFC). Let me mention that Hamkins–Palumbo proved that it is consistent with ZF that there exists a set with no rigid binary relation, that is, in which every binary relation has a nontrivial automorphism group, see this post. $\endgroup$
    – YCor
    May 8, 2020 at 8:31
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    $\begingroup$ I took the liberty to add references to the result by Sierpiński and Banach (which I did not know). $\endgroup$
    – Gro-Tsen
    Dec 2, 2021 at 22:52

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