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Let $H$ be a group with presentation $\langle h_1, \dots, h_n \mid r_1 = \dots = r_m = 1\rangle$. If there are $g_1, \dots, g_n \in G$ which satisfy the relations $r_1, \dots, r_m$, when is $\varphi : H \to G$ generated by $\varphi(h_i) = g_i$ an isomorphism? In particular, is there an effective way to check that the elements $g_i$ don't satisfy any additional relations?

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  • $\begingroup$ By Von Dyck's theorem, such a map is surjective onto $\langle g_1,\cdots,g_n\rangle$. If $G\neq\langle g_1,\cdots,g_n\rangle$, then certainly $\varphi$ cannot be an isomorphism. On the other hand, if $G=\langle g_1,\cdots,g_n\rangle$, it reduces to checking the kernel of the map $\varphi$. I do not know any efficient ways to do this, and suspect that it's difficult. $\endgroup$ – KReiser Sep 26 '13 at 4:17
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If we could effectively to check that $g_i$ don't satisfy additional relations, then we can decide whether $G\cong H$. But the problem of isomorphism is undecidable.

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  • $\begingroup$ In particular, it is unsolvable when $G$ is the trivial group. $\endgroup$ – Derek Holt Sep 26 '13 at 12:48
  • $\begingroup$ @Derek Holt: Yes, of course. $\endgroup$ – Boris Novikov Sep 26 '13 at 13:17
  • $\begingroup$ Of course, if we know that $H$ and $G$ belong to specific classes of groups, then not all is lost! $\endgroup$ – user1729 Sep 27 '13 at 13:18
  • $\begingroup$ @ user1729: For example, if they are Abelian :-) $\endgroup$ – Boris Novikov Sep 27 '13 at 13:30
  • $\begingroup$ @BorisNovikov: Is there an algorithm in the abelian case or do we just know that it is a decidable problem. $\endgroup$ – Michael Albanese Dec 27 '13 at 23:14

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