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I am trying to show that a group defined by the generators $a$, $b$, $c$, and $d$ and relations $adb=b^2a$, $a^3=c$, $b^2=c$, and $d^2=c$ is infinite and non-commutative.

I'm not really sure how to start on this problem- am I meant to assume that $c$ is the identity? Any help would be appreciated.

Sarah
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  • Since there's no general method for solving such problems, you have to think that some person contrived the question, so look for "coincidences/secrets", rather than having any systematic approach. E.g., in the first relation, if the group were commutative, things would cancel and you'd have $d=b$. Whoa! What are the odds of that? :) – paul garrett Jan 05 '20 at 22:38
  • @paulgarrett: Looking at the equations, I see that the $b^2$ in the first equation can be replaced by $c$ (due to the third equation) and then by $a^3$ (as of the second equation), which allows to cancel one $a$. There's still an $a^3$ left for which we can do the substitutions backwards, so that now $b$ can be cancelled out. And thus, unless I made an error on the way, you'll get $d=b$ even without commutativity. – celtschk Jan 05 '20 at 22:46
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    @celt Yes, so it just reduces to two generators $a, b$ with $a^3=b^2$. – Matt Samuel Jan 05 '20 at 23:06
  • Indeed, there is no general method for solving a problem like this in a very strong sense--it is a theorem that there is no algorithm that takes a finite presentation for a group and decides whether the group is infinite, or non-commutative. – Eric Wofsey Jan 05 '20 at 23:06
  • That is really interesting! Thanks – Sarah Jan 05 '20 at 23:38

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One way to try to solve this type of problem would be to add in extra conditions to produce a group or groups that you know. For example:

A group defined by these generators must be infinite because if we add in the conditions $$a=f^2,b=d=f^3,c=f^6$$ we obtain the infinite cyclic group $<f>$.

A group defined by these generators must be non-commutative because if we add in the conditions $$a^3=b^2=c=e,d=b,a^b=a^{-1}$$ we obtain the group $S_3$.