Find a finite subgroup of order at least $6$ from an infinite non-abelian group which every element is of order less than $6$. Let $G$ be an infinite non-abelian group such that for all $g\in G$, $1\leq |g|\leq 5$.
Prove or disprove that there exists a finite subgroup $H$ of $G$ such that $|H|\geq 6$.
By the assumption, first we know that if $H$ exists, $H$ must not be cyclic.
Secondly, if $g^2=1$ for every $g\in G$, then $G$ is abelian; a contradiction.
Hence there exists $g\in G$ such that $3\leq|g|\leq 5$.
Yet the two information are not sufficient to prove or disprove the statement.
 A: (I'm posting cw since I partly copy feedback from Derek Holt)

It is a notoriously hard old problem whether the Burnside group $B(2,5)$ is infinite (equivalently, whether there exists an infinite 2-generated group of exponent 5).
It is also open whether there exists a Tarski monster of exponent 5, that is, an infinite 2-generated group of exponent 5 satisfying the additional condition: every pair either is contained in a cyclic subgroup, or generates the whole group. This question was asked here at Mathoverflow in 2013. If the OP's question has a positive answer (that is every infinite torsion group with all elements of order $<6$ has a finite subgroup of order $|6|$), then a negative answer follows (which is weaker than proving that $B(2,5)$ is finite). Hence it seems hard to prove a positive answer.
And it would be definitely hard to prove a negative answer, as it would imply that the Burnside group $B(2,60)$ is infinite (indeed let $G$ be a counterexample: take two non-commuting elements, so they generate a subgroup of order $\ge 6$, hence infinite, and the exponent divides $\operatorname{lcm}(2,3,4,5)=60$). Because of the additional constraints, it's even much more than just proving $B(2,60)$ is infinite.
