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Let $G$ be an abelian group and $| S |$ is infinite with $S\subset G$.

Then $\langle S\rangle = \bigcup \langle T\rangle$ where $T \subset S$ and $T$ is finite.

My attempt is below.

$\langle S \rangle \supset \bigcup\langle T\rangle$ is clear, since every $T$ is contained in $S$ and then each $\langle T \rangle$ is contained in $\langle S\rangle $.

Remains to show is $\langle S \rangle \subset \bigcup \langle T\rangle$.

Take any $ x \in \langle S\rangle$.

Case I. $ x \in S$. In this case, there is $ T' $ containing $x$. so it is clear.

Case II. $ x \notin S $. In this case, there are $a, b \in \langle S \rangle $ such that $ ab=x $. and then there is $ T'' $ containing $ a $ and $ b$. Then $\langle T'' \rangle$ contains $ x $.

But I'm not sure the case II. Could I say there are such $ a $ and $ b $ ? Does $ x \in \langle S \rangle$ guarantee it?

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Hint (which, observed in a proper way, it's practically a complete answer):

If $\,x\in\langle S\rangle\,$ , then $\;x\;$ is a finite combination of elements in $\,S\,$ , meaning:

$$\exists\,s_1,...s_k\in S\;\;s.t.\;\;x=s_1^{\epsilon_1}\cdot\ldots\cdot s_k^{\epsilon_k}\;,\;\;\epsilon_i=\pm 1\;\;\forall\,i$$

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This also works with non-abelian groups (using finite words instead of finite power products). – Hagen von Eitzen Apr 20 '13 at 15:59
Words, products, sums: it's all the same, depending on how we call the operation that forms the group. Thanks. – DonAntonio Apr 20 '13 at 16:00

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