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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?


A group $G$ is said to be locally finite iff every finitely generated subgroup is finite.

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up vote 4 down vote accepted

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.

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What do you mean by "there are only finitely many possibilities for $(H_\alpha, g_{1\alpha},\dots, g_{n\alpha})$"? After all this is a direct product, not a direct sum, so we can have non-trivial projections in all $H_\alpha$. Am I missing something? – Christoph Feb 27 '14 at 9:14
Ah, I get it now. @ChristophPegel There is a map which takes in as input an index $\alpha$ and another $1\le i\le n$ and spits out the tuple $(H_\alpha,\pi_\alpha(g_1),\cdots,\pi_\alpha(g_n))$. For each $\alpha$, there are only finitely many values this tuple can take. Moreover, there are only finitely many values $H_\alpha$ can take, because there are only finitely many groups of order $n$. Try imagining an $n\times\aleph_0$ matrix whose $i$th row gives the coordinates of $g_i$. This matrix will have a lot of redundant columns that can be deleted: in fact, only finitely many distinct columns. – anon Feb 27 '14 at 9:22
In fact, the cardinality $\aleph_0$ is irrelevant, this argument shows an arbitrary direct product of finite groups of bounded order will be locally finite. – anon Feb 27 '14 at 9:29
I'll do a stupid question! Are there finitely many values for $H_\alpha$? If we take infinitely many isomorphic $H_\alpha$ but with different underlying sets can we consider them as the same on the same underlying set? – W4cc0 Feb 27 '14 at 9:35
Oh the emphasis here is really distinct projections, I get it. — @W4cc0 You get an isomorphism to a product where you don't have non-equal isomorphic factors immediatly, that's no big deal. – Christoph Feb 27 '14 at 9:36

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.

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