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We have the group $G = \langle a, \,b \; | \; a^2b^2ab^{-1}, \, a^3b^4a^{-2}b^{-3}\rangle$.

Obviously $G/G' = \mathbb{Z}_2$. Is it true that $G = \mathbb{Z}_2$ or equivalently that $G'=1$? It is possible to write down the presentation of $G'$, it has two generators and three relations, however it is not clear from that presentation that $G'=1$. Probably $G=\mathbb{Z}_2$, because there are some coset enumerative algoritms which show that $a$ generates $G$. If this is the case, can we show this by hand?

Another question which immediately comes to mind is the following. What is a probability that randomly chosen group $\langle a, \,b \;| \; r(a,b), \, s(a,b) \rangle$ is finite? Of course this question can be generalized to balanced presentations with any finite number of generators.

Edit: The answer to the original question is yes ($G\simeq\mathbb{Z}_2$), as indicated by Francesco Polizzi, using a computer verification. I would also be interested by a more "explicit" proof, which could be a conceptual argument, or an explicit way of writing, say, $ab$ as a product of conjugates of relators (from which such a writing of $a^2$ and $b^2$ would quickly follow).

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migrated from mathoverflow.net Apr 29 '16 at 7:54

This question came from our site for professional mathematicians.

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    $\begingroup$ I don't believe that we intuitively understand what a randomly chosen presentation is. How are you going to decide how long $r$ and $s$ should be? $\endgroup$ – Derek Holt Apr 28 '16 at 14:16
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    $\begingroup$ "with bare hands, using only pen and paper" is quite self-contradictory :) also I think it does not belong on the title $\endgroup$ – YCor Apr 28 '16 at 14:18
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    $\begingroup$ It's probably not a good idea to alter the question after receiving an answer, in a way that makes the answer no longer relevant. I'm going to revert the tltle and change the body slightly so that Francesco's answer is still relevant. $\endgroup$ – user43208 Apr 28 '16 at 14:21
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    $\begingroup$ I don't see the research angle of the first question (i.e. the one whether the given group $G$ is cyclic of order 2) -- firstly it can be solved by standard methods within milliseconds and secondly there are thousands of similar groups about which one could ask the same question. The second general question is unclear since no choice of probability space is made in the question. For these reasons I have voted to migrate this question to math.SE. -- There it could also be explained in detail that it is not clear what is meant by a "randomly chosen group" of the specified form. $\endgroup$ – Stefan Kohl Apr 28 '16 at 16:39
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    $\begingroup$ I think that it is reasonable to ask about a proof which is not just running an algorithm (e.g., as I mentioned elsewhere, writing explicitly, say, $ab$ as a product of relators, since GAP doesn't provide this); however I agree it was awkward to just discard the answer to the original question because you didn't like it. Recall the original question was "is this group cyclic on 2 elements", and not "I know that this group has 2 elements, can you prove it?" Well, I'll edit, and hope the question will not be migrated since it's not so obvious. $\endgroup$ – YCor Apr 28 '16 at 22:04
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This GAP4 script checks that the group $G$ is actually isomorphic to $\mathbb{Z}/2 \mathbb{Z}$.

f:=FreeGroup("a", "b");
a:=f.1; b:=f.2;
G:=f/[a^2*b^2*a*b^(-1), a^3*b^4*a^(-2)*b^(-3)];
Order(G);
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  • $\begingroup$ Well, I formulated my question incorrectly. I am not interested in the bare fact that $G=\mathbb{Z}_2$. I do not understand how to show this by hand. I amazed with the difficulty which appears when I am trying to deduce some identities. I do not see how to deduce $a^2$. Moreover, at the first look this group looks like a non-abelian one. So, It was a great surprise for me that $G$ is actually $Z_2$ $\endgroup$ – Samarkand Apr 28 '16 at 14:04
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    $\begingroup$ In the reference page of GAP there is a list of books and papers concerning the algorithms used. gap-system.org/Doc/references.html#Si94 For finitely presented groups, Johnson's Presentation of Groups and Sims' Computation with finitely presented groups are especially useful. You can surely show that the group is $\mathbb{Z}/2 \mathbb{Z}$ by hand using the algorithms explained in these books, since they are precisely the algorithms that the software uses. However, I do not know how complicate this can be. $\endgroup$ – Francesco Polizzi Apr 28 '16 at 14:15
  • $\begingroup$ I'd expect that if GAP detects that the group has order 2, it can also explicitly write $a^2$, $b^2$ and $ab$ as an explicit product of conjugates of relators? $\endgroup$ – YCor Apr 28 '16 at 16:43
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    $\begingroup$ @YCor I don't think that is totally straightforward. I believe there is an implementation of coset enumeration that can print out human readable proofs, but I don't know whether it is generally available. $\endgroup$ – Derek Holt Apr 28 '16 at 17:53

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