TITLE: Is there a group-theoretic proof of the Riemann rearrangement theorem?
QUESTION [5 upvotes]: The analytic proofs of the Riemann rearrangement theorem are easy to understand but they don't explain why commutativity breaks down when you go from finite sums to infinite sums of real numbers. I suspect that this involves the action of the symmetric group on infinite sets vs finite sets of real numbers. But, so far I haven't come across a proof that only uses tools from Group Theory. Does such a proof exist?

REPLY [4 votes]: First: commutativity of finite sums. $\mathbb{R}$ is a field, and as such, addition is commutative. This means that for two elements $a,b\in\mathbb{R}$, $a+b=b+a$. However, it is possible to extend this result. Consider $a_1,\ldots,a_n\in\mathbb{R}$, and a permutation $\pi$ of $\{1,\ldots,n\}$. Then, by induction, one can easily show that
$$ \sum_{i=1}^n a_i = \sum_{i=1}^n a_{\pi(i)}. $$
Next: infinite sums. While we generally think of an infinite sum as a "sum," it is technically a limit of a sequence of real numbers. In particular, if $a_i\in\mathbb{R}$ for $i\in\mathbb{N}$, then the symbol
$$ \sum_{i=1}^\infty a_i \qquad\text{is defined as}\qquad \lim_{n\to\infty} \sum_{i=1}^n a_i $$
assuming the limit exists. Furthermore, since the limit ignores any finite initial segment of a sequence, we have the result that
$$ \sum_{i=1}^\infty a_i = \sum_{i=1}^\infty a_{\pi(i)} $$ 
whenever $\pi$ is a permutation of $\mathbb{N}$ which fixes all but finitely many elements. You can prove this by applying our result about finite sums being invariant under rearrangement.
Finally: With this in mind, you cannot have a group theoretic proof of the Riemann rearrangement theorem because it is a theorem about infinite sums which are analytic objects in essence. In particular, the group structure of $\mathbb{R}$ (under addition) doesn't encode its analytic structure (completeness).