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Say that $a, b \in G$ are two elements of a finite group $G$. Is there a structure theorem for the structure of $\langle a,b\rangle$? Is there a way to derive group laws for the group operation in the generated group?

I can think of special cases (the two elements commute, one of the elements is a power of the other, the commutator of the two elements commutes with them, etc.). I am wondering if a general classification results exists, still.

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It is a theorem of Graham Higman, Bernhard Neumann, and Hanna Neumann (Embedding theorems for groups, J. London Math. Society 24 (1949) 247-254) that every countable group can be embedded in a 2-generator group. This was later simplified by Bernhard and Hanna Neumann (Embedding theorems for groups, J. London Math. Soc. 34 (1959) 465-479). Fred Galvin has a paper in the Monthly (Embedding Countable Groups in 2-Generator Groups, Amer. Math. Monthly 100 no. 6 (1993), 578-580; available from JSTOR ) giving a simple proof, showing that in fact:

Theorem. Every countable group is embeddable in a 2-generator group, with one generator of order 11 and the other of order 2.

Given this, it seems hopeless to expect a structure theorem for 2-generator groups.

Arturo Magidin
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  • How about if those two generators come from a well behaved group like $SL(2, F_q)$? –  Jul 27 '15 at 18:31
  • The question is about finite group (not that it makes things much easier). – TT_ stands with Russia Jul 27 '15 at 23:33
  • @TT_: As you say, it's not going to make things any easier; it still tells you that every finite group is embeddable into a 2-generated group (and I would guess one should be able to extend to proving that every finite group can be embedded into a finite 2-generated group). – Arturo Magidin Jul 28 '15 at 03:43
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    The symmetric group on finitely many letters is generated by a `long' cycle and a transposition; therefore every finite group embeds in a 2-generated group. – shane.orourke Jul 28 '15 at 06:36
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    Oops: `...every finite group embeds in a finite 2-generated group.' – shane.orourke Jul 28 '15 at 06:44
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    Not to completely disagree with your heuristic, but every field embeds in an algebraically closed field, but the latter class is significantly easier to classify than the former. – Richard Rast Jul 28 '15 at 12:04
  • @shane.orourke: D'oh. Of course. That's what I get for posting past 11pm... – Arturo Magidin Jul 28 '15 at 16:19
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    According to Theorem 2.1 of F. Levin, Factor Groups of the Modular Group, J. London Math. Soc 43 (1968), 195-203, every countable group is embeddable in a $2$-generator group with generators of prescribed orders $p\ge3$ and $q\ge2.$ – bof Aug 03 '15 at 10:07