Recently I've been reading J.P. May's [A Concise Course in Algebraic Topology](http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf). 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.)



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


