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The defining equations of a group are a set of equations involving the group's generators that determine the group's multiplication table completely.

What I want to know is: Is the least number of distinct defining equations for a group determined entirely by the number of generators of that group? If so, how do we calculate that number if we know the number of generators? (By 'distinct', I mean that each equation represents a separate piece of information that could not have been inferred from the other given equations. So, for example, the equations $a^2=e$ and $ab=ba^2$ are distinct, while the equations $a^2=b$ and $a=ba^{-1}$ would not be distinct.)

I'm interested in this mainly because I want to see if there is a way to find all the defining equations of a group without having to write down the entire group's table.

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  • $\begingroup$ There seems to be an implicit assumption here that all sets of 'distinct' defining equations contain the same number of equations. This is not the case; e.g. in a cyclic group generated by $a$, $a^4=1$ and $a^6=1$ are 'distinct' equations which together are equivalent to the single equation $a^2=1$. $\endgroup$ – joriki Aug 18 '15 at 19:29
  • $\begingroup$ @joriki Oops, good point - I'll edit the question now $\endgroup$ – Asker Aug 18 '15 at 19:33
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    $\begingroup$ Are you talking about group presentations? en.wikipedia.org/wiki/Presentation_of_a_group $\endgroup$ – Cheerful Parsnip Aug 18 '15 at 19:44
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    $\begingroup$ @GrumpyParsnip Yes, that's it. I didn't know that's what they were called though. Thanks for the link $\endgroup$ – Asker Aug 18 '15 at 19:55
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The answer to your question is no. The Klein $4$-group $C_2 \times C_2$ is a $2$-generator group with presentation $\langle x,y \mid x^2=y^2=1, xy=yx \rangle$, and it is not too hard to show that it cannot be defined with less than $3$ relations.

However, the quaternion group $Q_8$ is also a $2$-generator group with presentation $\langle x,y \mid x^2=y^2, y^{-1}xy=x^{-1} \rangle$ that can be defined with $2$ relations.

Finite groups require at least as many relations as generators, but that is not true for infinite groups. For any $n$ the free group on $n$ generators requires no relations, and that is the only group with that property.

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