Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 3 / 3 / b

The assumed group is a Tarski monster group. For any $g\ne 1$, its cyclic subgroup $⟨g⟩$ must be proper. Otherwise $G$ would be infinite cyclic, and $⟨g^2⟩$ would be a proper nontrivial infinite subgroup, contrary to the hypothesis. Hence $⟨g⟩$ has order $p$.

Choose $h\notin ⟨g⟩$, possible because $G$ is infinite. The subgroup $⟨g,h⟩$ contains $⟨g⟩$ strictly. If it were proper, it would have order $p$ and could not strictly contain the order-$p$ subgroup $⟨g⟩$. ThereforeIn particular, $G$ is a finitely generated group, generated by two elements.

Suppose $1\ne N⊲G$ is proper. The defining property makes $N$ cyclic of order $p$. Conjugation gives a homomorphismwhose 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,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:Together with part (i), this proves that Tarski monster groups are two-generated and simple.

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).