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My question is:

  1. Are there $G$, which is a fundamental group of a surface or a 2-orbifal, and its two finite index subgroups $H_1$, $H_2$ satisfying the following condition: $H_1$ is isomorphic to $H_2$, but for any automorphism $\varphi$ of $G$, $\varphi(H_1)$ is not $H_2$?

  2. If the answer to question 1 is Yes, can $H_1$ and $H_2$ be normal subgroups of $G$?

This question arise from the extension of the automorphism of a surface or an orbifal to its covering space.

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closed as too localized by Fernando Muro, Mark Sapir, Anton Petrunin, Benoît Kloeckner, Igor Rivin Oct 1 '12 at 15:02

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Yes to all questions: just take $G$ equals the integers and $H_1$, $H_2$ be the subgroups generated by $p$ and $q$ with $|p|$ different from $|q|$. All the groups involved are infinite cyclic and hence isomorphic to the fundamental group of the annulus. –  Bruno Martelli Oct 1 '12 at 12:39
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Do you really mean 'homeomorphic'? `Isomorphic' would seem to make more sense (and the answer to 1 and 2 would then be YES.) If you really mean 'homeomorphic' you'd better give us more information about what this group G is, e.g. it is, at least, a topological group. –  Nick Gill Oct 1 '12 at 12:42
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This question doesn't look like a research level question to me. –  Fernando Muro Oct 1 '12 at 13:18
    
Sorry for the first two stupid questions. I originally think of the covering space (hence the subgroup) of some surfaces. May the title should be "Are there two homeomorphic subgroups of a surface group which cannot be transferred to each other under any automorphism of the mother group?" –  X.M. Du Oct 1 '12 at 14:54
    
@X.M. Du Please edit your question so that it asks what you want in a clear and precise way. The guidelines here are useful mathoverflow.net/howtoask . Then it has a chance of being reopened. –  j.c. Oct 1 '12 at 17:09

1 Answer 1

Let me answer this question under the assumption that we are talking about `isomorphism' of groups. So I have a group $G$ containing two isomorphic subgroups $H$ and $K$. When can I extend the isomorphism between $H$ and $K$ so that it becomes an automorphism of $G$? The answer is that usually you can't.

In particular the model theoretic concept of homogeneity is relevant here: roughly speaking (because I am no model theorist) a class of structures is homogeneous if whenever you have three structures $H$, $K$ and $G$ in the class such that (a) $H$ and $K$ are substructures of $G$ and (b) $H$ is isomorphic to $K$, then an isomorphism between them can be extended to an automorphism of $G$.

An example of a class of groups that is homogeneous would be $\mathcal{E}_p$, the class of elementary abelian $p$-groups. That the isomorphism is always extendible follows from the fact that any linearly independent set of vectors in a vector space can be extended to a basis.

To return to the original question: in addition to Bruno's example given above, here is another: $G=C_2\times C_4$ and $H$ and $K$ are cyclic subgroups of order $2$. Take $H$ to lie in a subgroup $C_4$ and $K$ to not lie in such a subgroup. Note that $G$ is abelian so $H$ and $K$ are normal, which deals with your second question.

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