Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 3 / 3 / b / 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. ConsiderEvery 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. ThereforeThe stronger restriction on all proper subgroups, rather than just individual element orders, is what forces the conclusions in parts (i) and (ii).