 Solution

Every finite quotient of the abelian group $A_{\neg p}$ is abelian. Part iii excludes elements of prime order $q\ne p$ by the Cauchy theorem for groups, so any finite quotient is a finite abelian $p$-group. If its exponent divides $p^n$, the quotient map kills $p^n{A}_{\neg p}$ and therefore factors throughEvery quotient of a cyclic group is cyclic, so the finite quotient is isomorphic to $\mathbb{Z}/p^k\mathbb{Z}$ for some $k\le n$. Conversely, reduction modulo $p^k$ gives a surjection $A_{\neg p}\to \mathbb{Z}/p^k\mathbb{Z}$. Thus these are exactly the nontrivial finite quotients, together with the trivial case $k=0$.