7.39 (1980)

Let $G = \langle a, b \mid a^p = (ab)^3 = b^2 = (a^\sigma ba^{2/\sigma}b)^2 = 1 \rangle$, where $p$ is a prime, $\sigma$ is an integer not divisible by $p$. The group $PSL_2(p)$ is a factor-group of $G$ so that there is a short exact sequence $1 \to N \to G \to PSL_2(p) \to 1$. For each $p > 2$ there is $\sigma$ such that $N = 1$, for example, $\sigma = 4$. Let $N^{ab}$ denote the factor-group of $N$ by its commutator subgroup. It is known that for some $p$ there is $\sigma$ such that $N^{ab}$ is infinite (for example, for $p = 41$ one can take $\sigma^2 \equiv 2 \pmod{41}$), whereas for some other $p$ (for example, for $p = 43$) the group $N^{ab}$ is finite for every $\sigma$.
$\qquad$ a) Is the set of primes $p$ for which $N^{ab}$ is finite for every $\sigma$ infinite?
$\qquad$ b) Is there an arithmetic condition on $\sigma$ which ensures that $N^{ab}$ is finite?