= Solution
For the replacement condition, interpret “every element has order $p$” in the usual exponent-$p$ sense: every nonidentity element has order $p$, while the identity has order one. Consider
$$
E=\bigoplus_{j=1}^{\infty}C_p.
$$
Every nonzero element has order $p$. It is infinite, but no finite collection generates it: the supports of finitely many <group generators> cover only finitely many coordinates, so a unit vector outside those coordinates is missing from their generated <subgroup>. Also $E$ is abelian, and one coordinate <subgroup> $C_p$ is a proper nontrivial <normal subgroup>. Therefore
$$
\boxed{\text{the exponent-}p\text{ condition implies neither finite generation nor simplicity}.}
$$
The stronger restriction on all proper <subgroups>, rather than just individual element orders, is what forces the conclusions in parts (i) and (ii).
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