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Is there an example of an infinite group G acting on a set X such that each non-identity element of G fixes exactly two elements of X, and some two elements of G have in common exactly one fixed point in X.
(This is impossible if G is finite.)

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If this is impossible when G is finite, can you tell us where in that proof the result possibly fails for infinite groups? – Ketil Tveiten Jun 26 '10 at 15:10
When a finite group G acts transitively on a set, the average over all g in G of the number of elements fixed by G is always 1. – Tom Goodwillie Jun 26 '10 at 15:39
Is the action transitive? Otherwise there is a finite such group and action (see Tom's answer). – Victor Protsak Jun 27 '10 at 2:06
@Victor: No, we overlooked something. See my edited answer. – Tom Goodwillie Jun 27 '10 at 11:18
Ketil, we can show the result is impossible if G is finite by using Burnside's counting formula. This only applies when G is finite. – Martin Erickson Jun 28 '10 at 16:38

Find a nontrivial group that acts on a set in such a way that each nontrivial group element fixes exactly one point but they don't all fix the same one -- for example, SO(3) acting on the projective plane in the usual way.* Now add one more point fixed by everybody.

Now I suppose you will ask for a transitive action.

  • This cannot be done if G is finite.

EDIT: It can be done if G is finite, as Victor Protsak points out in a comment.

NEW EDIT: This answer was nonsense. Each element of order $2$ in SO(3) fixes infinitely many lines through the origin.

In fact, I claim that a finite group cannot act on a set in such a way that every nontrivial element of $G$ has exactly one fixed point, unless either the group or the set has exactly one element.

FINAL (?) EDIT: After several false starts, here is what I have learned about finite group actions in response to this question.

Recall the following useful fact:

(0) When a finite group $G$ acts on a finite set $X$, then the number of $x\in X$ fixed by $g\in G$, averaged over all $g$, is the number of orbits of the action.

This can be proved by comparing the two ways of counting the set of pairs $(g,x)$ such that $gx=x$. An immediate consequence of (0) is:

(1) If a finite group acts transitively but not trivially on a set, then some element of the group has no fixed points.

You can also use (0) to show:

(2) When a nontrivial finite group acts on a set in such a way that every $g\ne 1$ has exactly one fixed point, then apart from free orbits there must be exactly one orbit, of size $1$.

With a little more work than that, you can show:

(3) When a nontrivial finite group acts on a set in such a way that every $g\ne 1$ has exactly two fixed points, then apart from free orbits there must be either exactly two orbits, both of size $1$, or else exactly three orbits. In the latter case, either

(a) The group has order $12$ and the orbit sizes are $(6,4,4)$, or

(b) The group has order $24$ and the orbit sizes are $(12,8,6)$, or

(c) The group has order $60$ and the orbit sizes are $(30,20,12)$, or

(b) The group has order $2k$ and the orbit sizes are $(k,k,2)$.

By thinking about groups of order $12$, $24$, and $60$ and noting that the hypotheses rule out nontrivial normal cyclic subgroups, you can show:

(4) In each of cases (a), (b), and (c), the only possibility is the one you are already thinking of, corresponding to the action on the $2$-sphere of the symmetries of the tetrahedron, cube/octahedron, or dodecahedron/icosahedron.

(5) In case (d) there is the dihedral group acting on the $2$-sphere, but there are also plenty of other examples, all of them semidirect products based on actions of a group of order $2$ on a group of order $k$. If $k$ is odd then the group of fixed elements must be trivial; if $k$ is even, it must have order two.

(6) I believe that in all of these cases it is impossible for two elements of the group to have only one common fixed point, so I agree with the OP that the kind of example asked for in the question would have to involve an infinite group. Is there an easier way to see this?

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Do you mean $SO(3)$ acting on $\mathbb{RP}^2?$ If $G$ is abelian (e.g. $SO(2)$) and $g\in G$ fixes $x$ then it fixes the full $G$-orbit of $x$ so "each nontrivial group element fixes exactly one point but they don't all fix the same one" is impossible. – Victor Protsak Jun 27 '10 at 0:13
Thanks, yes, I meant SO(3) acting on the set of lines through the origin in $\mathbb R^3$ -- now corrected. – Tom Goodwillie Jun 27 '10 at 0:43
Ah, and instead of full $SO(3)$ you can consider its irreducible finite subgroup (tetrahedral, octahedral, or iscosahedral), which gives counterexamples to the OP's claim that its impossible to achieve with $G$ finite (without transitivity assumption). – Victor Protsak Jun 27 '10 at 2:01
-1: It seems you have not touched the question. – Dror Speiser Jun 27 '10 at 18:46
True. In its final form, my "answer" was not an answer to the question. I wrote it because I was exploring the OP's statement that no finite group action could have the properties asked for. This leads me to a question, which I will post as such. – Tom Goodwillie Jun 27 '10 at 19:02

One way to get an example is to let $F_2$ be the free group on two generators, then consider the collection of all maximal cyclic subgroups and choose one representative $H_n$ for each conjugacy class. $F_2$ then acts by left multiplication on the disjoint union of the left coset spaces $F_2/H_n$. Since maximal cyclic subgroups in $F_2$ are malnormal, each non-trivial element will have exactly one fixed point. Hence we can do as Tom Goodwillie suggests and add a global fixed point.

I'm not sure how one would get a transitive action, but one thing which might help in this direction is that Ashot Minasyan, building on the work of Denis Osin, shows in Lemma 3.4 of that there is a countable torsion free group containing a malnormal copy of $\mathbb Z$ and having every element being conjugate to an element in $\mathbb Z$. It follows that this group acts transitively on a set such that each non-trivial element has exactly one fixed point.

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I don't understand your description of Minasyan's result. A generator $x$ of the malnormal copy of $\mathbb Z$ must be conjugate to $x^{-1}$ as there are only two conjugacy classes and then it wouldn't be malnormal would it? – Torsten Ekedahl Jun 27 '10 at 18:48
Thanks for the comment. I meant to say that every element is conjugate to an element in $\mathbb Z$. I edited my answer accordingly. – Jesse Peterson Jun 27 '10 at 18:59
Jesse's example could also be described more canonically by saying let the group act by conjugation on its set of maximal cyclic subgroups. – Tom Goodwillie Jun 29 '10 at 10:14
Good point Tom. In fact, by looking at the point stabilizers, a group will act non-trivially on a set such that each non-identity element has exactly one fixed point if and only if the group is a non-trivial disjoint (except for $e$) union of a family of malnormal subgroups, which is invariant under conjugation. Non-trivial free products (here $\mathbb Z_2 * \mathbb Z_2$ is a trivial free product), torsion free Gromov hyperbolic groups, and torsion free groups which have deficiency at least 2 all have such a family. On the other hand, non-trivial direct products have no such family. – Jesse Peterson Jun 29 '10 at 18:07
The torsion free Gromov hyperbolic groups should not be $\{ e \}$ or $\mathbb Z$ of course. – Jesse Peterson Jun 29 '10 at 18:19

The trivial group acting on a one-point set. Ha, ha. Or maybe you meant the two elements in the last clause to be distinct; it's ambiguous.

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This is silly... – Douglas S. Stones Jun 27 '10 at 9:19

Consider the action of $\mathrm{PU}_2$ of projective transformations of the complex projective line $\mathbb P^1(\mathbb C)$ represented by unitary matrices. The fixed points of a non-identity element in its action on $\mathbb P^1(\mathbb C)$ are its two eigenspaces (there are exactly two as we have identified all scalar matrices with the identity element and any element of $\mathrm{PU}_2$ is diagonalisable). Furthermore, there are two elements that have exactly one eigenspace in common.

Addendum: I just now realised that this is essentially Tom's example with a twist that makes it work: Look at the action of $\mathrm{SO}_3$ on the set of oriented lines through the origin. We have $\mathrm{PU}_2=\mathrm{SO}_3$ and the action on $\mathbb P^1(\mathbb C)$ is just this action.

Failure: Having realised this I also see that this example fails. The last condition is not fullfilled, for any non-identity element $g$, an element that fixes one of the fixed points of $g$ also fixes the other. This is clear in the $\mathrm{SO}_3$-model. In the $\mathrm{PU}_2$-model it follows because the second eigenspace is the orthogonal complement of the first.

Addendum: The double covering situation with oriented lines covering non-oriented ones turns out to be the reason why the last condition is not fulfilled. In fact asssume that we have a $G$-set $X$ fulfilling the first but not necessarily the last property. By throwing away the orbits with trivial stabiliser we may also assume that every point has a non-trivial stabiliser. Now define a $G$-graph $\Xi$ with vertex set $X$ where $x$ and $y$ are connected when they are both fixed by the same non-identity element of $G$. The last condition is now equivalent to some connected component of $\Xi$ containing more than two elements, hence if it is not fulfilled we get a double cover $G$-map $X\to\pi_0(\Xi)$.

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The following was inspired by Jesse Peterson's answer.

Let $M^n$ be a closed hyperbolic manifold, eg a surface, and consider the action of $\Gamma=\pi_1M$ on the boundary at infinity of hyperbolic $n$-space. Then each $\gamma\in\Gamma\smallsetminus 1$ has precisely two fixed points, namely the end points of its axis of translation.

Any non-commuting pair of elements whose axes share an end point now provide an example of the kind required. It isn't too difficult to construct explicit examples.


As Torsten Ekedahl points out, the last sentence should have read 'It's impossible to construct such examples'! In particular, this doesn't work.

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I don't think this works at least for $n=3$ (or $2$). We should have two elements with exactly one fixed point in common. We may assume that the common fixed point is $\infty$ and one element has $0$ as the other fixed point. Then they have the form $\mu x$ and $\lambda x+a$ (as Möbius transformations) with $a\ne0$ and $\lambda\ne 1$. Their commutator is then $x+(\mu-1)a$ which has exactly one fixed point. – Torsten Ekedahl Jun 28 '10 at 5:22
Huh, good point. Now that I think harder about it, if two elements of a hyperbolic group have axes that remain uniformly close for an infinite distance then of course they commute. Silly me. – HJRW Jun 28 '10 at 21:30

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