[GAP Forum] Checking equality of elements of words in a finitely presented group

[James Mitchell]

Dear Edwin,

Probably someone else will reply with a more elaborate answer, but you
might want to consult:

https://en.wikipedia.org/wiki/Word_problem_for_groups#Unsolvability_of_the_uniform_word_problem

A very short answer is that the word problem for finitely presented groups
is unsolvable, i.e. it is known that there is no single algorithm that can
for every finitely presented group say whether two words represent the same
element.

Best wishes,

James

On Thu, 28 Apr 2016 at 20:03 W. Edwin Clark <wclark at mail.usf.edu> wrote:

> I am trying to use GAP to establish equality of two words in a finitely
> presented group and am having problems.
>
> First here is a successful test case:
>
> f:=FreeGroup(5);;
> rel:=[(f.(5)*f.(1))^5];;
> g:=f/rel;;
> u:=GeneratorsOfGroup(g);;
> (u[5]*u[1])^5 = Identity(g);  #testing this identity which is actually a
> given relation works successfully
>
>        true
>
> But if I add some more relators to rel, I cannot even get
> equality to the identity of the first relator.  Here's what I tried:
>
> f:=FreeGroup(5);;
> rel:=[(f.(5)*f.(1))^5];;
> for i in [1..4] do
>  Add(rel,(f.(i)*f.(i+1))*(f.(5)*f.(1))^(-1));
> od;
> rel;;
> g:=f/rel;;
> u:=GeneratorsOfGroup(g);
>
> (u[5]*u[1])^5 = Identity(g);  #I gave up waiting for an answer
>
> Why is this so difficult?
>
> What I really want in this group is to show that
>
> u[1]^5*u[2]*u[4]*u[1]*u[3]*u[5] = Identity(g)
>
> Any help would be appreciated.
>
> --Edwin Clark
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