Cartesian Product of Boundedly Finite Groups Let $G$ be the cartesian product of countably many finite groups $H_\alpha$,  $\alpha\in \omega$. Assume also that there exists an $ n\in\mathbb N$ such that $|H_\alpha|\leq n,\, \forall \alpha\in \omega$. Is $G$ locally finite? Is there an easy way to prove it?
Notation
A group $G$ is said to be locally finite iff every finitely generated subgroup is finite.
 A: I think so. The isomorphism type of $G$ does not change if we replace a group $H_\alpha$ by an isomorphic group, so we can replace each $H_\alpha$ by a canonical copy for its isomorphism type, and hence assume that $H_\alpha = H_\beta$ whenever $H_\alpha \cong H_\beta$. So, since each $|H_\alpha| \le n$, there are only finitely many $H_\alpha$ that occur.
Let $K$ of $G$ be a subgroup of generated by $g_1,\ldots,g_n$, with projections $g_{i\alpha} \in H_\alpha$ onto the components. Then there are only finitely many distinct possibilities for the $(n+1)$-tuple $(H_\alpha,g_{1\alpha},\ldots,g_{n\alpha})$.
Suppose that these finitely many distinct projections occur at $\alpha = \alpha_1,\ldots,\alpha_m$. Then $K$ is a subgroup of the direct product of the $m$ finite groups $K_1,\ldots,K_m$, where $K_i := \langle g_{1\alpha_i},\ldots g_{n\alpha_i} \rangle$, so $K$ is finite.
A: There are only finitely many groups of order at most $n$, let $H$ be the direct product of all of these, then $H$ is still finite and we have embeddings $H_\alpha\hookrightarrow H$ for all $\alpha\in\omega$. Let $X = \prod_{\alpha\in\omega} H$, then $G\hookrightarrow X$, so it suffices to show that the $X$ is locally finite.
Thus we have reduced the claim to the special case of all $H_\alpha$ being the same finite group.
Let me know if this helps.
