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What I want is to give the generators and relations, as well as an integer $n$, then tell the program to compute the number of elements having word length $\leq n$.

Can someone please show me how to do this? I have very little knowledge of group theory software.

Thanks!

Chris Sanders
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    This problem is not solvable in general. If the group has a shortlex automatic structure that can be computed then it is easy to use this to compute the growth function as a rational function, and there is functionality available in GAP or Magma (or the standalone KBMAG package) for doing this. There may be other particular classes of groups for which it can be computed. – Derek Holt Feb 17 '22 at 17:13
  • I see. Is there some sort of database where I can look at, say, the first 20 terms of the growth sequence, for many different examples of groups? This is motivated by the following question that I found: https://math.stackexchange.com/questions/4383792/are-jumps-in-the-growth-function-of-an-infinite-group-increasing – Chris Sanders Feb 17 '22 at 17:29
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    You may be interested in the Wikipedia article word problem for groups. – Somos Feb 17 '22 at 17:29
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    I don't know of any such database. – Derek Holt Feb 17 '22 at 19:18

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For a given f.g. recursively presented group $G$ and finite generating family, this is algorithmically solvable iff the word problem is solvable.

Indeed, if the word problem is solvable, consider all group words of length $\le n$, compare any two to check what is the "equality in $G$" equivalence relation is, and output the number of classes.

Conversely, consider two group words, say both of length $\le n$. Compute the cardinal $c_n$ of the $n$-ball. Enumerating consequences of relators within the $n$-ball eventually reaches $c_n$ equivalence classes within the $n$-ball. Once reached, just check whether your two elements lie in the same class.

YCor
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