Are there non-trivial superperfect groups with the property that there exists a presentation of the group where the number of generators equals the number of relations?
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2$\begingroup$ Superperfect means 1st and 2nd homology vanish (en.wikipedia.org/wiki/Superperfect_group). In other words, trivial abelianization and trivial Schur multiplier $\endgroup$– YCorCommented Feb 24, 2018 at 21:19
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2 Answers
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Yes, for example ${\rm SL}(2,5) \cong \langle a,b \mid a^2=b^{-3}=(ab)^5 \rangle$.
Furthermore, it is proved in:
C.M. Campbell and E.F. Robertson, ``A deficiency zero presentation for ${\rm SL}(2,p)$'', Bull. Lon. Math. Soc. 12 (1980) 17-20,
that, when $p$ is an odd prime and $k = \lfloor p/3 \rfloor$, then
$$\langle \,x,y \mid x^2(xy)^{-3},\, (xy^4xy^{(p+1)/2})^2y^px^{2k} \,\,\rangle \cong {\rm SL}(2,p).$$
answered Feb 24, 2018 at 22:13
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1$\begingroup$ Just as a consequence, the free product of n copies of any of these ones $\endgroup$– YCorCommented Feb 24, 2018 at 23:03
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For $M$ a closed, connected, orientable 3-manifold, it's well known that the existence of a Heegaard splitting for $M$ implies that the fundamental group $\pi_1M$ admits a balanced presentation (i.e. one with as many generators as relations).
Any such $M$ splits as a connect sum of irreducible pieces $M_i$, meaning that every embedded 2-sphere in the $M_i$ bounds a 3-ball. It follows from the Sphere theorem that $\pi_1M$ is superperfect if and only if $M$ is a homology sphere.
This leads to many examples. The most well known is the Poincare homology sphere (whose fundamental group is $SL(2,5)$, as in Derek Holt's answer), but there are infinitely many others: see Ryan Budney's answer here for a nice summary.
