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I have group elements, that I'd like to condense down according to group rules. Take the modular group. This has generators $S,T$ that satisfy the groups relations $S^2 = I, (ST)^3 = I$.

So given a group element as a list of generators (for instance, $S T^2 S$ I'd write as {S,T,T,S} in Mathematica), I'd like to collapse the expression down according to group rules.

For instance, in our example, we'd have 

Collapse[{S,T,S,S,T}] = {S,T,T}

I've tried using ReplaceAll, but can't get it to match substrings. StringReplace seems to have the correct functionality, but only works on actual strings.

How would one go about doing this?

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    $\begingroup$ In general, this is sort of a hard problem. My take on it in the past has been to choose a "normal order", since writing group elements in terms of generators is not unique in general. Maybe start with this question and answer. I know there are many more on this site. The real question is, what do you want to do with this? Do you want to have a list of all the group elements in terms of the generators? How do you decide when you're done? $\endgroup$
    – march
    Jan 31, 2016 at 18:52
  • $\begingroup$ By the way, does this particular presentation lead to a finite group? $\endgroup$
    – march
    Jan 31, 2016 at 19:04

1 Answer 1

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Using replacement Rules

collapse[lst_?VectorQ] := lst //. {
   {b___, S, S, e___} :> {b, e},
   {b___, S, T, S, T, S, T, e___} :> {b, e}}

collapse[{S, T, S, S, T}]

(*  {S, T, T}  *)

collapse[{S, S, T, S, T, S, T, S, T, S, S, T}]

(*  {T, T}  *)
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    $\begingroup$ I think I would go with {b___, T, S, T, S, T, e___} :> {b, S, e} instead of the second one. It along with the first rule makes your second rule, but this one will get more cases. +1 $\endgroup$
    – march
    Jan 31, 2016 at 19:21

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