= Solution
Suppose $1\ne N\lhd G$ is proper. The defining property makes $N$ cyclic of order $p$. Conjugation gives a <homomorphism>
$$
G\longrightarrow\operatorname{Aut}(N),
$$
whose kernel is the <centralizer> $C_G(N)$. The <automorphism> <group> has order $p-1$, so the kernel has finite index in $G$. It contains $N$, because $N$ is abelian. If this kernel were proper, it would be an order-$p$ <subgroup>, and its finite index would force $G$ itself to be finite. Therefore $C_G(N)=G$, so $N$ is central.
Choose $h\notin N$. Part (i) shows that $h$ has order $p$. Its cyclic <subgroup> meets $N$ trivially, since two order-$p$ <subgroups> with nontrivial intersection are equal. Since $N$ is central,
$$
\langle N,h\rangle=N\times\langle h\rangle\cong C_p\times C_p.
$$
This <subgroup> has order $p^2$ and is proper because $G$ is infinite, contradicting the assumed order of every proper nontrivial <subgroup>. Thus no such $N$ exists:
$$
\boxed{G\text{ is simple}.}
$$
Together with part (i), this proves that <Tarski monster groups are two-generated and simple>.
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