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I am learning GAP and would like to ask one question about a command called "TzGoGo":

If $P$ is a finite presentation of a group $G$, then will the eventual result of the command "TzGoGo(P)" be a presentation $Q$ of $G$ such that $Q$ has the same set of generators as $P$ but there does not exist any other presentation of $G$ with the same set of generators but fewer relations?

One example:

gap> F3:=FreeGroup("a","b","c");
gap> B3:=[Comm(Comm(F3.1,F3.2),F3.1),Comm(Comm(F3.1,F3.2),F3.2),
         Comm(Comm(F3.1,F3.3),F3.1),Comm(Comm(F3.1,F3.3),F3.3),
         Comm(Comm(F3.2,F3.1),F3.2),Comm(Comm(F3.2,F3.1),F3.1),
         Comm(Comm(F3.2,F3.3),F3.2),Comm(Comm(F3.2,F3.3),F3.3),
         Comm(Comm(F3.3,F3.1),F3.3),Comm(Comm(F3.3,F3.1),F3.1),
         Comm(Comm(F3.3,F3.2),F3.3),Comm(Comm(F3.3,F3.2),F3.2),
         Comm(Comm(Comm(F3.1,F3.2),F3.3),F3.1),
         Comm(Comm(Comm(F3.1,F3.2),F3.3),F3.2),
         Comm(Comm(Comm(F3.1,F3.2),F3.3),F3.3),
         Comm(Comm(Comm(F3.1,F3.3),F3.2),F3.1),
         Comm(Comm(Comm(F3.1,F3.3),F3.2),F3.2),
         Comm(Comm(Comm(F3.1,F3.3),F3.2),F3.3),
         Comm(Comm(Comm(F3.2,F3.1),F3.3),F3.1),
         Comm(Comm(Comm(F3.2,F3.1),F3.3),F3.2),
         Comm(Comm(Comm(F3.2,F3.1),F3.3),F3.3),
         Comm(Comm(Comm(F3.2,F3.3),F3.1),F3.1),
         Comm(Comm(Comm(F3.2,F3.3),F3.1),F3.2),
         Comm(Comm(Comm(F3.2,F3.3),F3.1),F3.3),
         Comm(Comm(Comm(F3.3,F3.1),F3.2),F3.1),
         Comm(Comm(Comm(F3.3,F3.1),F3.2),F3.2),
         Comm(Comm(Comm(F3.3,F3.1),F3.2),F3.3),
         Comm(Comm(Comm(F3.3,F3.2),F3.1),F3.1),
         Comm(Comm(Comm(F3.3,F3.2),F3.1),F3.2),
         Comm(Comm(Comm(F3.3,F3.2),F3.1),F3.3)];

gap> K3:=F3/B3;
     pre:=PresentationFpGroup(K3);

gap> TzGoGo(pre);

#I  there are 3 generators and 18 relators of total length 204
#I  there are 3 generators and 15 relators of total length 180

Does it mean, in my example, a presentation of $K3$ with the same set of generators has at least 15 relators?

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  • $\begingroup$ Thanks @BabakS. An example has been added. $\endgroup$
    – Zuriel
    Commented Oct 23, 2013 at 10:22
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    $\begingroup$ I am afraid the answer is no. The GAP manual says "Unfortunately there is no algorithm which could be applied to find the shortest presentation which can be obtained by Tietze transformations from a given one. Therefore, what GAP offers are some lower-level Tietze transformation commands and, in addition, some higher-level commands which apply the lower-level ones in a kind of default strategy which of course cannot be the optimal choice for all presentations". It would be good to have an example in GAP demonstrating how this may happen, posted here as a proper answer. $\endgroup$ Commented Nov 15, 2013 at 10:35

1 Answer 1

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To quote my former comment, I am afraid the answer is no.

The GAP manual says here:

"Unfortunately there is no algorithm which could be applied to find the shortest presentation which can be obtained by Tietze transformations from a given one. Therefore, what GAP offers are some lower-level Tietze transformation commands and, in addition, some higher-level commands which apply the lower-level ones in a kind of default strategy which of course cannot be the optimal choice for all presentations".

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