When the subgroup generated by all elements of some fixed finite order equals the group Let $G$ be an infinite non-abelian group.
Let $H$ be the subgroup of $G$ generated by all elements of order $n$,
for some fixed $n \geq 2$.
My question: Is it possible to tell when $H=G$? Namely, are there some known groups $G$ that are generated by all elements of some fixed finite order?
(It is easy to see that $H$ is normal in $G$, see this; I do not know if this fact should help in answering my question).
Edit: The motivation for my question is Proposition 5(b), which seems to me too good to be true (perhaps I am missing some implicit assumptions in that proposition).
A relevant question is this question.
 A: You can modify your question a bit: which groups $G$ admit surjection from a free product $\mathbb Z / n\mathbb Z \,* \dots \mathbb * \, Z / n\mathbb Z \twoheadrightarrow^f G$? If you restrict it a bit and assume that $f$ induces isomorphism on abelianisations and $H_2(G, \mathbb Z) = 0$, then (by beautiful Stallings theorem) $f$ induces isomorphisms on all lower central quotients, and this is rather restrictive property.
In other direction, if you know that $G$ is nilpotent, then torsion elements form a normal subgroup, so if your group is simultaneously nilpotent, finitely generated and infinite, it cannot be generated by torsion even normally.
Third aspect is that if your group is finitely generated and equals its $n$-torsion (is subject to relation $x^n = 1)$, then it's finite for $n =$ 2 (obvious), 3 (easy lemma), 4 or 6 (not easy, but can be proven by elementary methods), and it's not known if it must be finite for $n$ = 5, 7, 8 $\dots$, 664 wiki link
Lastly, very plentiful source of natural examples are reflection groups in homogeneous spaces with most well-known case of plane Coxeter groups
