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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.

For a Tarski monster group, any nonidentity element generates a proper subgroup of order $p$. Adjoining an element outside that subgroup generates the whole group. A proper nontrivial normal subgroup would have order $p$; conjugation into its finite automorphism group forces it to be central. Combining it with another order-$p$ cyclic subgroup would then give a proper subgroup of order $p^2$, a contradiction.

// Target: algebra.bigb