Consider the following: For a group $G$ with identity $e$, define $s: G \to \mathbb{N} \cup \{ \infty \}$ by $s(g) = \min \{ k \in \mathbb{N} : g^{k} = e \}$, where $ \min \emptyset = \infty$. Moreover, let $\Theta(G) = \sup \{ s(g) : g \in G \}$. Refer to $G$ as bounded if there exists $N \in \mathbb{N}$ such that $g^{N} = e$ for all $g \in G$.
We see that $G$ is bounded iff $\Theta(G) < \infty$. If $\Theta(G) $ is finite, then $g^{\Theta(G) !} = e$; if $\Theta(G) = \infty$, then for every $N$ there exists $g \in G$ for which $s(g) > N$, so $g^{N} \neq e$.
Moreover, any group of order $n$ is bounded. To see this, the claim is made that $s(g) \leq n$ for all $g \in G$. To see this, consider the set $ \{ g, g^{2}, \ldots, g^{n + 1} \}$. By pigeonhole, there exist $1 \leq i < j \leq n + 1$ such that $g^{i} = g^{j}$, so $g^{j - i} = e$; then $s(g) \leq j - i \leq n$. Thus $\Theta(G) \leq n$.
Initially, I conjectured that a set was bounded off it was finite, considering the multiplicative group $\{ e^{iq} : q \in \mathbb{Q} \}$, where $s(g)$ is always finite, but unbounded. Then I dismissed this considering $G = \mathbb{Z}_{2}^{I}$, where $I$ is an infinite indexing set and $\Theta(G) = 2$. I revised this to the claim that a group $G$ is bounded if and if it can be written as a product $G = \prod_{i \in I} H_{i}$ of finite groups for some index set $I$, where $\sup \{ \# H_{i} : i \in I \} < \infty$. I believe I know how to show any factoring of $G$ into finite groups would satisfy this, but I can't show that every bounded group is factorable into finite groups, i.e. can't show there doesn't exist a bounded group which cannot be expressed as a product of finite groups.
Is this correct? If so, how might I show it? If not, is there a counterexample? Thanks.
The Tarski monster group has elements all of finite order $p$ for some large prime, but since it is simple it cannot be written as a direct product of groups.