For more examples see:
C. Campbell, E. Robertson,
A deficiency zero presentation for $SL(2,p)$.
Bull. London Math. Soc. 12 (1980), no. 1, 17–20.
and, more recently,
C. Campbell, G. Havas, C. Ramsay, E. Robertson, Nice efficient presentations for all small simple groups and their covers. LMS J. Comput. Math. 7 (2004), 266–283.
C. Campbell, G. Havas, C. Ramsay, E. Robertson,
All simple groups with order from 1 million to 5 million are efficient. Int. J. Group Theory 3 (2014), no. 1, 17–30.
Here a superperfect group (a perfect group with trivial Schur multiplier) is efficient if and only if it admits a balanced presentation.
The last two papers cover all groups of order up to 5 million.
It can be checked that there is a homomorphism $\phi:B\to {\rm SL}(2,3)$ with $$\phi(r) = \left(\begin{array}{cc}0&1\\2&0\end{array}\right),\ \ \phi(s) = \left(\begin{array}{cc}2&0\\2&2\end{array}\right),\ \ \phi(t) = \left(\begin{array}{cc}0&2\\1&1\end{array}\right),$$
with $\phi(r^2)=\phi(s^3)=\phi(t^3) = \phi(rst) = -I_2$, in which $\phi(s)$ has order $6$, so we just need to check that $s^6=1$ in $B$.
The relations imply that $r=st$, and the element $u=rst$ is central, with $s^3=t^3=(st)^2=u$. Let $x=st$ and $y=ts$.
Then $x^2=u$ and, since $u$ is central, $y^2=tx^2t^{-1}=u$.
Then (using $t^{-1}s^{-1}t^{-1}=su^{-1}$), we have $$[x,y] = x^{-1}y^{-1}xy=u^{-2}xyxy = u^{-2}st^2s^2t^2s = ust^{-1}s^{-1}t^{-1}s = s^3 = u,$$
so, since $[x,y]$ is central, we have $u^2=[x,y]^2=[x^2,y]=[u,y]=1,$ and so $s^6=1$ as claimed.
(Note that $\langle x,y \rangle \cong Q_8$.)
Best Answer
No. Its abelianisation is
$$\langle x, y\mid x^{n-1}=y^{n-1}, xy=yx\rangle,$$
of which $\Bbb Z_{n-1}^2$ is a quotient.