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Let $G$ be any countable discrete group, and $H_1, H_2$ be two subgroups of $G$ with some fixed finite index $m>1$.

My question is:

Is it always possible to find some automorphism $\phi$ of $G$, i.e., $\phi\in Aut(G)$, such that $\phi(H_1)=H_2$?

If not, could any one give some counterexample to illustrate this? Thanks in advance!

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Nope: let $G=C_4\times V_4\times\mathbb Z$, where $C_4$ is the cyclic group of order $4$ and $V_4$ is the Klein $4$-group. Let $H_1=C_4\times\{e\}\times\mathbb Z$ and $H_2=\{e\}\times V_4\times\mathbb Z$. Both subgroups have index $4$, but they aren't isomorphic.

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  • $\begingroup$ So we can take two non-isomorphic finite groups $G_1, G_2$, of the same order $n$S and a take an infinite group $H$, and consider $G=H\times G_1\times G_2$. Then $H_1=G\times G_1, H_2=G\times G_2$ may work generalizing your idea. $\endgroup$ – P Vanchinathan Mar 21 '14 at 1:15
  • $\begingroup$ @PVanchinathan It depends. Using your notation, if $H$ is a direct sum of countably many copies of $G_1\times G_2$, then $G_1\times H$ is isomorphic to $G_2\times H$ after all. In my example, there is no such accidental isomorphism, since $C_4\times\{e\}\times\mathbb Z$ has an element of order $4$, and $\{e\}\times V_4\times\mathbb Z$ doesn't. $\endgroup$ – Chris Culter Mar 21 '14 at 1:25
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The dihedral group of order 8 has a subgroup isomorphic to the cyclic group of order 4, and has a subroup isomorphic to the Klein group. Both subgroups have index 2 but are non-isomorphic since the first has an element of order 4 but the second does not. No smaller order group G can be found as a counterexample since a nontrivial subgroup of such a G would have order 2 or order 3.

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    $\begingroup$ Dihedral group G of order 8 also has two subgroups of order 2 with no automorphism of G taking one to the other (take one central, and one not). $\endgroup$ – Jack Schmidt Mar 21 '14 at 1:57
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Take $m\mathbf{Z}, n\mathbf{Z}\subset \mathbf{Z}$. As isomorphisms have to send generator to generator we get conflicting requirement $\phi(1)=\pm1, \phi(m)= \pm n $. So your question has a negative answer.

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    $\begingroup$ They don't have the same index though $\endgroup$ – Bruno Joyal Mar 21 '14 at 1:05
  • $\begingroup$ OOps, I looked at the title and was misled. My answer is incorrect. $\endgroup$ – P Vanchinathan Mar 21 '14 at 1:11

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