two conjugate subgroups and one is a proper subset of the other? plus, a covering space interpretation.

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.)

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.

• 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? – Aaron Mazel-Gee Jan 17 '10 at 18:33
• It means direct product. You get direct sums if you consider the direct limit of finite-dimensional tori. – user2035 Jan 19 '10 at 15:30

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 $2\mathbb{Z}$).

If there is a group G and an injective endomorphism $\sigma$ 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 this and http://groupprops.subwiki.org/wiki/Isomorphic_iff_potentially_conjugate_in"finite">this for more notes on these.

• That's pretty damn cool. – Harry Gindi Jan 16 '10 at 18:28
• Awesome, I like that you generalized the Baumslag-Solitar example that others have already put forward. – Aaron Mazel-Gee Jan 16 '10 at 23:22

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.

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 = < x,y | y x y^-1 = x^2 >

which is isomorphic to the multiplicative group generated by the 2x2 rational matrices

x = [ 1 1; 0 1 ], y = [ 2 0; 0 1 ],

and N = < x >.

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

• (2009 ended a couple weeks ago.) – S. Carnahan Jan 15 '10 at 16:23
• Clearly I should stop hanging on to the past. – Gerhard Paseman Jan 15 '10 at 16:27