Products of torsion groups

Question

Given an infinite family of non-zero torsion groups $G_i$ (not necessarily commutative). Prove that their Cartesian product is a torsion group iff all but finitely many (i.e. "almost all") of the groups have a finite uniform bound (i.e., there is a positive natural number n with $G_i^n=1$, for all i, except a measly few that may be torsion, unbounded groups). I am pretty sure this is true in the category of Abelian groups, but what about non-commutative groups? Added: $G_i^n=1$ means that orders of all elements in $G_i$ divide $n$.

Answer 1

Let $G = \prod_\alpha G_\alpha$, where $G_\alpha$ are torsion subgroups. It is obvious that if all but finitely many of the $G_\alpha$ have a uniform bound, then $G$ is torsion.

For the other direction, it suffices to show that the product of a countable collection $\{G_i\}_{i\in \mathbb N}$ without a uniform bound is not a torsion group. Let $\{n_i\}_{i\in \mathbb N}$ be an unbounded sequence of natural numbers such that $G_i^{n_i} \neq 1$ for all $i$. For each $i\in \mathbb N$, choose $g_i \in G_i$ such that $g_i^{n_i} \neq 1$. Then the element $$ (g_1, g_2, \dots) \in \prod_{i \in \mathbb N} G_i $$ is not torsion.