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Let $F$ be the free group on the generating set $A$. Note that $F$ is not abelian if $A$ contains more than one element.

How am I suppose to note this? Doesn't $Z\times Z$ has 2 elements in it's generating set (more than one element) but abelian?

How?

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    $\begingroup$ $\mathbb{Z}\times \mathbb{Z}$ is a not a free group, but $\mathbb{Z}\ast \mathbb{Z}$ is! $\endgroup$ – Dietrich Burde Jun 1 '17 at 11:38
  • $\begingroup$ "Free" means that relations never hold unless they have to (from the group axioms). Powers of one element have to commute among each other, but beyond that elements do not have to commute; hence free groups on more than one generator cannot be Abelian. $\endgroup$ – Marc van Leeuwen Jun 1 '17 at 19:35
  • $\begingroup$ What's the difference between $\mathbb{Z} \times \mathbb{Z}$ and $\mathbb{Z} * \mathbb{Z}$? $\endgroup$ – Obinna Nwakwue Jun 1 '17 at 20:17
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    $\begingroup$ In an unfortunate turn of nomenclature, a "free abelian group" is not free unless it is $\mathbf Z$ or the trivial group. $\endgroup$ – bertram Jun 1 '17 at 20:38
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$\mathbb Z\times \mathbb Z$ is the free abelian group of two generators, but it is not the free group of two generators. The free group of two generators $F(a,b)$ is the set of all (reduced) finite strings containing the letters $a, b, a^{-1},$ and $b^{-1}$, whose group operation is concatenation-and-reduction. It, then, should be clear that $ab$ and $ba$ are two different elements of $F(a,b)$, and so this group is not abelian.

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By definition of a free group, all words in the alphabet $a,b,\ldots ,$ of $A$ are different if they are not identical after cancelling all occurences of $xx^{-1}$, etc. In particualr, the words $w_1=ab$ and $w_2=ba$ are different. Hence the group is non-abelian for rank at least $2$.

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Let $u,v$ be any two distinct elements of $A$. Suppose that $F$ is commutative.

Let $G$ be any group, and $x,y$ be any two elements of $G$.

There is a group homomorphism $\varphi:F \to G$ such that $\varphi(u) = x$ and $\varphi(v) = y$. Therefore,

$$ xy = \varphi(uv) = \varphi(vu) = yx $$

Thus, we've proven that every group is abelian.

Since this is not true, our assumption that $F$ is commutative must be false.

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  • $\begingroup$ But how would you show that another group is non-abelian? Probalby by taking two elements and explicitely calculating their compositions (especially if we a priori cannot rely on the fact that free groups are non-abelian). But it doesn't get much easier as in free groups, where $ab\neq ba$ simply by definition. $\endgroup$ – klirk Jul 12 '18 at 21:02
  • $\begingroup$ @klirk: Explicit calculation, as you suggest; e.g. the symmetric group on 3 objects is an easy group to work with. Regarding your suggestion $ab \neq ba$ does not appear anywhere in the definition of free group; I imagine you're referring to the group of words over an alphabet. If you've already made that comparatively complicated construction and proven a theorem about normal forms or other statement implying $ab \neq ba$, then that also serves as a good example. The method of my post is a rather general argument form for reasoning about free structures, however, which is why I posted it. $\endgroup$ – Hurkyl Jul 12 '18 at 21:11
  • $\begingroup$ I agree. If the free group is introduced by its universal property, then your proof is even more from the definition. $\endgroup$ – klirk Jul 13 '18 at 9:14

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