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For the finite cyclic group of order $n$, there is the standard presentation $\langle a \mid a^n\rangle$. Also for $S_n$ (symmetric group), I know a few presentations where the number of relations is strictly bigger than the number of generators.

I want to get examples of $G$, where $G$ is a finite group with the property that for any presentation of $G$ the number of relations is strictly bigger than the number of generators.

More generally, can one put some condition on a finite group $G$, so that for any presentation the number of relations is strictly bigger than the number of generators?

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    $\begingroup$ TeX note: please use $\langle\rangle$ \langle\rangle instead of $<>$ <>. It is also often preferable (though, unlike \langle\rangle, probably not universally agreed) to use \mid for a divider rather than |. (The results are $\langle a \mid a^n\rangle$ vs. $<a|a^n>$.) I have edited accordingly. $\endgroup$ Commented Sep 29, 2022 at 22:45
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    $\begingroup$ $(\mathbb Z/p)^n$ is an example for all $n>1$, which can be proven using mod $p$ cohomology. $\endgroup$ Commented Sep 30, 2022 at 0:32
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    $\begingroup$ The lingo for this kind of thing is deficiency. $\endgroup$ Commented Sep 30, 2022 at 0:40
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    $\begingroup$ In general the "quest" is rather to find groups of deficiency zero among finite groups. See for instance this 1996 paper by Havas, Newman, and O'Brien. They mention in particular that a theorem of Golod-Shafarevich says that a finite $p$-group with deficiency zero is 3-generated (whether this holds for all finite groups is mentioned there as open). $\endgroup$ Commented Sep 30, 2022 at 6:42
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    $\begingroup$ And in general (with the |relators|$-$|generators| sign convention) the deficiency of a finite group is $\ge$ that of its Schur multiplier. See also this paper by Gardam (publ. Bull LMS 2017) for examples of finite groups with arbitrary positive deficiency. $\endgroup$ Commented Sep 30, 2022 at 6:57

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