16
$\begingroup$

Recently I've been reading J.P. May's A Concise Course in Algebraic Topology. In the section on the classification of covering groupoids, he mentions that sometimes a group G may have two conjugate subgroups H and H' such that H is properly contained in H' (on pp. 26-27, according to his numbering). This seems bizarre to me, and I'm pretty sure I've seen an example before, but I'm having trouble coming up with one now.

Anyways, he continues by saying that it is possible to have an endomorphism of a covering groupoid which is not an isomorphism. I'd like to come up with an example of this, and I'm pretty sure that for me obstruction lies in failing to completely grasp the group-theoretic statement above.

(Of course, when I think of a covering of groupoids I'm secretly thinking about a covering space, partly because this is his motivation for introducing groupoids and partly because it's just easier for me, so ideally but not necessarily the example would really just be a map of covering spaces over the same base space.)

$\endgroup$
1

7 Answers 7

10
$\begingroup$

The subgroup $H=\mathbb Z^\mathbb N$ of $G_1=\mathbb Z^\mathbb Z$ is mapped to a proper subgroup by translation. By considering the semidirect product $G_1\rtimes\mathbb Z$, you can make translation on $G_1$ an inner automorphism.

The first theorem on p.26 tells you that if $G=\pi(B,b)$, $H=p(\pi(E,e))$, $H_1=p'(\pi(E',e'))$, the unique map $g\colon E\to E'$ satisfying $g(e)=e'$ is not an isomorphism. However, by the first proposition on p.23 there is an $e'_1\in E'$ such that $p(\pi(E',e'_1))=p(\pi(E,e))$, and the corresponding $E\to E'$ is an isomorphism.

$\endgroup$
2
  • $\begingroup$ Thanks. Just out of curiosity, does the exponent notation mean direct sum or direct product? The group construction works in either case, of course, and the latter interpretation can evidently arise from some sort of infinite torus -- what about the former? $\endgroup$ Commented Jan 17, 2010 at 18:33
  • $\begingroup$ It means direct product. You get direct sums if you consider the direct limit of finite-dimensional tori. $\endgroup$
    – user2035
    Commented Jan 19, 2010 at 15:30
33
$\begingroup$

The Baumslag-Solitar group $B(1,2) = \langle a,b | bab^{-1} = a^2 \rangle$ has a subgroup $\langle a \rangle \cong \mathbb{Z}$, which is conjugate to $\langle a^2 \rangle$ (which, under the isomorphism, is identified with $2\mathbb{Z}$).

$\endgroup$
14
$\begingroup$

If $G$ is a group and $\sigma$ is an injective endomorphism of $G$, there is a group $K$ containing $G$ such that $\sigma$ extends to an inner automorphism of $K$.

Even more generally, if two subgroups of a group are isomorphic, the group can be embedded in a bigger group where those two subgroups are conjugate via a conjugation map that extends the isomorphism. The construction used is called a HNN-extension, and it basically adjoins elements to the group that act by conjugation as the isomorphism. (This generalizes even further: given any number of isomorphisms between pairs of subgroups of a group, there is a group containing the group such that all these isomorphisms become conjugations in that bigger group.)

Thus, to find examples that answer your question, it is enough to find examples of a group that is isomorphic to a proper subgroup. For instance, if we consider the example of the group of integers isomorphic to the subgroup of even integers, the corresponding HNN-extension is the Baumslag–Solitar group mentioned above.

Incidentally, the statement above (that any two isomorphic subgroups of a group become conjugate in some bigger group) is also true when we restrict to finite groups, though this does not give any examples for the question you are interested in because no finite subgroup can be isomorphic to a proper subgroup.

See isomorphic iff potentially conjugate and isomorphic iff potentially conjugate in finite on the Groupprops wiki for more notes on these.

$\endgroup$
2
  • $\begingroup$ That's pretty damn cool. $\endgroup$ Commented Jan 16, 2010 at 18:28
  • $\begingroup$ Awesome, I like that you generalized the Baumslag-Solitar example that others have already put forward. $\endgroup$ Commented Jan 16, 2010 at 23:22
4
$\begingroup$

Let $V$ be an infinite dimensional vector space, let $T:V\to V\oplus V$ be an isomorphism, and let $G=GL(V\oplus V\oplus V)$. Then $(T^{-1}\oplus T)(I\oplus I\oplus GL(V))(T\oplus T^{-1})=I\oplus GL(V\oplus V)$ gives an example.

$\endgroup$
3
$\begingroup$

To expand on an earlier answer, I will plagiarize an answer provided by Derek Holt in a sci.math thread:

A "standard" example of that is the Baumslag-Solitar group

$$G = \langle x,y \mid y x y^{-1} = x^2 \rangle$$

which is isomorphic to the multiplicative group generated by the $2\times2$ rational matrices

$$x = \left[\begin{array}{cc} 1& 1\\ 0& 1\end{array} \right]\qquad y = \left[\begin{array}{cc} 2& 0\\ 0& 1\end{array} \right]$$

and $N = \langle x \rangle$.

Then $x N x^{-1} = N$ and $y N y^{-1} < N$, but $N$ is not normal in $G$ because $y^{-1} N y$ is not contained in $N$.

End of plagiarism.

Perhaps you can enhance this example to your purpose. Gerhard "Ask Me About System Design" Paseman, 2009.01.15

$\endgroup$
2
  • $\begingroup$ (2009 ended a couple weeks ago.) $\endgroup$
    – S. Carnahan
    Commented Jan 15, 2010 at 16:23
  • $\begingroup$ Clearly I should stop hanging on to the past. $\endgroup$ Commented Jan 15, 2010 at 16:27
3
$\begingroup$

(Old question and many answers already, but adding a very natural and elementary example not yet mentioned.)

Take the permutation group $\newcommand{\Z}{\mathbb{Z}}\newcommand{\Sym}{\operatorname{Sym}}\newcommand{\Stab}{\operatorname{Stab}_{\mathrm{pw}}}\Sym(\Z)$, with the subgroups of permutations acting trivially on the non-negative and the strictly positive integers respectively, $\Stab(\Z_{\geq 0}) < \Stab(\Z_{> 0}) < \Sym(\Z)$. These are evidently conjugate via the permutation “+1”.

Of course, nothing here is specific to $\Z$: it just needed a set $X$ (necessarily infinite) with an automorphism $s : X \to X$ and a subset $Y$ such that $s(Y) \subsetneq Y$; then we have $\Stab(s(Y)) < \Stab(Y) < \Sym(X)$, and they are conjugate via $s$.

More generally, $X$ can be an object of any category, $s$ an automorphism, and $Y$ a subobject of $X$ with $s Y \lneq Y$; then $\Stab(s(Y)) < \Stab(Y) < \operatorname{Aut}(X)$, conjugate via $s$ as before. Jonas Meyer’s answer is a special case of this.

$\endgroup$
2
$\begingroup$

Old question, but it seems like the right place to add the following example:

  • Let $G$ be the free group on the generators $g_{0},g_{1}\in G$. Recall that the collection $\left(g_{1}^{n}g_{0}g^{-n}\right)_{n\in\mathbb{Z}}$ of elements of $G$ freely generates$^{\text{1}}$ a subgroup of $G$ (so its elements are in particular independent therein). Thus $$G':=\left\langle g_{1}^{n}g_{0}g_{1}^{-n}\right\rangle_{n\in\mathbb{Z}_{\geq 0}}\ \subseteq\ G$$ (note the altered index) is a subgroup of $G$ with the property that $$g_{1}G'g_{1}^{-1}\ \subsetneq \ G'\ \subsetneq\ g_{1}^{-1}G'g_{1}.$$

  • Let $G/G'$ be the standard transitive (left) $G$ action on the set of left cosets of $G'$. Remark that $$\varphi\ \colon\ \left[x\right]\in G/G'\ \mapsto\ \left[xg_{1}^{-1}\right]\in G/G'$$ defines a (left) $G$-set endomorphism$^{\text{2}}$ of $G/G'$, where $\left[x\right]$ denotes the coset to which each $x\in G$ pertains. In fact, $$\left[g_{1}\right]\ \neq\ \left[g_{0}g_{1}\right]$$ but \begin{align*}\varphi\left(\left[g_{1}\right]\right)\ &=\ \left[1\right]\\ &=\ \left[g_{0}\right]\\ &=\ \varphi\left(\left[g_{0}g_{1}\right]\right).\end{align*} Thus $\varphi$ is a noninjective, hence noninvertible, (left) $G$-set endomorphism of $G/G’$.

  • In particular, there is an equivalence$^{\text{3}}$ from the category of transitive (left) $G$-sets to the category of (un-based) connected covers of $S^{1}\vee S^{1}$. Thus there exists a connected cover of $S^{1}\vee S^{1}$ admitting a noninvertible endomorphism; working through the (not too bad) details produces about as tractable of a "concrete" such example as one could reasonably$^{\text{4}}$ hope for.


  1. This can itself be motivated by studying covering spaces; cf. diagram 11 on page 58 of Hatcher.

  2. The well-definition of the underlying map at the level of elements is by that if $\left[x\right]=\left[x'\right]$, then $x^{-1}x' \in G'$, so $g_{1}x^{-1}x'g_{1}^{-1} \in G'$, so \begin{align*}\left[xg_{1}^{-1}\right]\ &=\ \left[xg_{1}^{-1}g_{1}x^{-1}x'g_{1}^{-1}\right]\\ &=\ \left[x'g_{1}^{-1}\right].\end{align*} Equivariance is by the commutativity of action on the left with action on the right.

  3. Said equivalence is given, of course, by the fundamental theorem of covering spaces, but can be established just as easily by direct argument due to the simplicity of the specific base at hand.

  4. It should be noted, albeit entirely trivially, that the connectedness of the covering space is essential to making the problem interesting; already the point has finite covers with noninjective automorphisms. In the other direction, it's not hard to show that an endomorphism of a connected cover of a locally connected space is necessarily surjective and (even more easily) that a bijective endomorphism of a cover of a locally connected space is invertible. So the failure of an endomorphism of a connected cover of a locally connected space to be an automorphism is precisely its failure to be injective, just as the failure of an endomorphism of a field extension to be an automorphism is precisely its failure to be surjective.

$\endgroup$
1
  • $\begingroup$ Ps.: Let X be the genus 2 surface. There is an obvious continuous map f : X → S¹ ∨ S¹ that induces a surjection on π₁ (for a suitable choice of basepoint). Pulling back along f then gives us an example of a noninvertible endomorphism of a connected cover of a compact smooth surface. So such (endomorphisms of) covers exist for even the most "real world" examples of spaces. $\endgroup$
    – Rafi
    Commented Mar 30, 2023 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.