Does associativity imply commutativity? I used to think that commutativity and associativity are two distinct properties. But recently, I started thinking of something which has troubled this idea:
$$(1+1)+1 = 1+ (1+1)\implies 2+1=1+2$$ Here using associativity of addition operation, we've shown commutativity. 
In general, $$\underbrace{(1+1+\dots+1)}_{a \, 1\text{'s }}\ \ + \ \ \underbrace{(1+1+\dots+1)}_{b \, 1\text{'s }}=\underbrace{(1+1+\dots+1)}_{b \, 1\text{'s }}\ \ + \ \ \underbrace{(1+1+\dots+1)}_{a \, 1\text{'s }} \\ \implies a+b=b+a$$
For any natural $a,b$.
Hence using only associativity we prove commutativity. That this can be done, is disturbing me too much. Is this really correct? If yes, then are associativity and commutativity closely related? Or is it because of some other property of natural numbers? If yes, then can it be done for other structures as well?

Clever observation. This is called a cyclic group (or the generalized version, free object). $ G $ is said to be cyclic, if there is some set $ s $, called the "generating element", such that
$$ G = \langle s \rangle = \{ n \in \mathbb{Z} : s^n \} = \{e, s, s^2, s^3, \ldots s^{-1}, s^{-2}, s^{-3}, \ldots \} $$
That is, all elements can be "generated" from a single element.
In this case $ s = 1 $, but let's forget that for a moment, and consider the general case:


*

*Are cyclic objects always abelian?

*Are all groups cyclic?
To 1., the answer is relatively simple: Yes. To verify this, we simply observe that $$ s^n s^m = \underbrace{sss\ldots s}_{n + m} = s^{n + m} = s^{m + n} = s^m s^n $$
Thus $ ab = ba $ in $ G $.
Now, for 2., the easiest way to prove that not all algebraic objects are cyclic, is to construct an object that isn't cyclic nor abelian. The simplest example is the group, $ G $, of bijective functions, $ \mathbb{R} \to \mathbb{R} $, under composition.
You can easily see that it is associative, at the same time it is relatively easy to verify that this is in fact not abelian (I will let you do that). Therefore it cannot be cyclic.
Another (more funky) way we can see that it is not cyclic is by Cantor's diagonal argument: Define $ \mathbb{R} $ as $ \{ f \in G : f(0) \} $. If $ G $ was cyclic, we would be able to enumerate the real numbers, by $ \mathbb{R} = \{ n \in \mathbb{Z} : s^n(0) \} $, which is not possible, as shown by the diagonal argument.
A bonus: Often people mistakenly believe that commutativity implies associativity, but that's in fact not true! Consider $ a * b = ab + 1 $. This is obviously commutative, but it isn't associative.
