# How to show this obvious and basic property of abelian groups?

I have a question that is probably very silly, but let's go. Let $(G,+)$ be an abelian group. In that case we know that $+$ is associative and commutative. This leads us to the following: if $\{a_i \in G : i \in I_n\}$ with $I_n = \{i \in \mathbb{N} : 1 \leq i \leq n\}$, then if we apply $+$ to all of the $a_i$, it independs on the ordering we impose. In truth, if we want to define the sum of all of those elements as:

$$\sum_{i \in I_n}a_i = a_1+\cdots+a_n,$$

we should first know what means $a_1 + \cdots + a_n$. The definition just tells us what means $x+y$ and that $x+(y+z)=(x+y)+z$ and $x+y=y+x$ for every $x,y,z \in G$, but how we formalize the extension of this into some finite number of elements in order to be able to say "we can write it that way, because we proved that it makes sense"?

Indeed this is something very obvious, and I've never seem someone giving great arguments about it. Everyone just says "obviously, the parentheses and the order doesn't matter". I've thought on using inductions on those two properties each at a time, but I've got a little confused with it.

How is this really done? Is a proof necessary? Or we just leave it there without proof?

Thanks very much in advance, and sorry if this is not the place for this kind of question.

-
Everyone just says "obviously, the parentheses and the order doesn't matter". Not everybody. Sometimes it's proved. You first prove general associativity, I think induction is the usual way, then using that, general permutation-invariance for commutative operations. –  Daniel Fischer Aug 12 '13 at 2:32
Thanks for the help @DanielFischer! I've said everybody meaning "everybody I've seem until now". I'll try doing it in this way you've said. Thanks for the hint. –  user1620696 Aug 12 '13 at 2:35

To prove general associativity, define $$\begin{cases}\prod_{i=1}^1 a_i=a_1\\\prod_{i=1}^n a_i=\prod_{i=1}^{n-1}a_i\cdot a_{n}\end{cases}$$

The claim is that for any $m$, we have that $$\prod_{i=1}^n a_i\prod_{i=1}^m a_{n+i}=\prod_{i=1}^{n+m}a_i$$

By definition we have the truth of $m=1$. Thus assume true for $m=r$, and consider $r+1$. Then \begin{align}\prod_{i=1}^n a_i\prod_{i=1}^{r+1} a_{n+i}&=\prod_{i=1}^na_i \left(\prod_{i=1}^{r} a_{n+i}a_{n+r+1}\right)\\&=\left(\prod_{i=1}^na_i \prod_{i=1}^{r} a_{n+i}\right)a_{n+r+1}\\&=\prod_{i=1}^{n+r}a_i a_{n+r+1}\\&=\prod_{i=1}^{n+r+1}a_i\end{align}

General commutativity can be proven as follows. Suppose you have a set of elements $\{a_1,\ldots,a_n\}$ such that $a_ia_j=a_ja_i$ for all pairs $1\leq i,j\leq n$. Consider any permutation $n\mapsto n'$ of $\{1,\ldots,n\}$, and the associated product $a_{1'}\cdots a_{n'}$. Suppose that the term $a_n$ occurs it the place $h'=n$. Then me way write by the commutativity hypothesis $$a_{1'}\cdots a_{n'}=a_{1'}\cdots a_{(h-1)'}a_{(h+1)'}\cdots a_{(n-1)'}a_{n'}a_n$$

and then induction does the rest, since we have stepped down to the case $n-1$

-
As Daniel notes it is generally first proved (with no assumption of commutativity) that generalized associativity holds. (There are a number of questions about that on MSE.) Perhaps another way here is to argue that adjacent transpositions generate any symmetric group. –  anon Aug 12 '13 at 3:30
@anon Agreed. ${}{}{}$ –  Pedro Tamaroff Aug 12 '13 at 4:13

As the others point out, you can prove that parentheses and order don't matter via induction. Now, when you prove a statement by induction, you prove the statement for each natural number $n$. In other words, the statement applies only to finite sums.

This observation is significant because the obvious generalization to infinite series is not true in the real numbers! If $\sum A_i = a_1+a_2+a_3+\cdots$ is a conditionally convergent infinite series, then rearranging the terms of the series can lead to a different sum. In fact, you can get any sum you want!

(On the other hand, if $\sum A_i = a_1+a_2+a_3+\cdots$ is a series whose partial sums are eventually constant, then rearranging the terms doesn't change the eventual value. Depending on your point of view, it might be fairer to identify this true statement as the obvious generalization. After all, it doesn't depend on the topology of our abelian group. Or, rather, it requires convergence in the discrete topology.)

-
If the partial sums are eventually constant, you're still just (effectively) dealing with a finite sum, no? –  Cameron Buie Aug 12 '13 at 2:48
@CameronBuie Sure, but the "eventual sum" is the natural generalization of finite sums to groups that don't come with a topology. I edited the answer to explain this point a little better. –  Chris Culter Aug 12 '13 at 2:51
What is the point of bringing up conditionally convergent series here? This would better suit as a (rather long) comment. –  Pedro Tamaroff Aug 12 '13 at 2:59
@PeterTamaroff One doesn't truly understand why a result is stated in a certain way, and proven in a certain way, until one has seen counterexamples to its generalizations. Wouldn't you agree? –  Chris Culter Aug 12 '13 at 3:04
@ChrisCulter "...the obvious generalization to infinite series..." This is really digressing! –  Pedro Tamaroff Aug 12 '13 at 3:10