Free group over a set of two elements is abelian

Question

Let $A=\{a,b\}$ and $a\ne b$. Let $F(A)$ be the free group constructed on $A$. Let $f_a,f_b$ be the canonical homomorphisms of $\mathbb{Z}$ into $F(A)$. Let $g:F(A)\rightarrow\mathbb{Z}\times\mathbb{Z}$ be the unique homomorphism for which $g(f_a(1))=(1,0)$ and $g(f_b(1))=(0,1)$. Let $h:\mathbb{Z}\times\mathbb{Z}\rightarrow F(A)$ be the mapping defined by $h(n,m)=f_a(n)f_b(m)$. Then $h\circ g=id_{F(A)}$ and $g\circ h=id_{\mathbb{Z}\times\mathbb{Z}}$. This means that $g$ is an isomorphism. Since $\mathbb{Z}\times\mathbb{Z}$ is abelian, $F(A)$ is also abelian.

This is surely wrong, right? Where is the mistake?

Answer 1

The map $h$ is not a homomorphism. Indeed, if you try and prove it is a homomorphism, you will find that you need to know that $f_a(1)$ and $f_b(1)$ commute in $F(A)$, which you do not know to be true (and it is in fact false).

Answer 2

Mainly I think your maps are backwards. Well the word "canonical" is tricky it interpret but there isn't a map $\mathbb{Z}\to F(A)$ that is unique in general, some would want $1\mapsto a$, others $1\mapsto b$, etc. But there are maps $\hat{\pi}_a,\hat{\pi}_b:F\langle a,b\rangle\to \mathbb{Z}$. Why? Take $\pi_a(a)=1$ and $\pi_a(a)=0$, $\pi_b(a)=0$ and $\pi_b(b)=1$. That gives you (by the universal mapping property) the homomorphism $\hat{\pi}_a$ and $\hat{\pi}_b$. Taken together you will make an amalgamated product, not a direct product. In this case you get $F\langle a,b\rangle\to \mathbb{Z}*\mathbb{Z}$ because the kernels of $\hat{\pi}_a$ and $\hat{\pi}_b$ intersect trivially.

Answer 3

The free group on two generators is not abelian. If we have $ab=ba$, then it is not free. A free group satisfies no relations on its generators, hence the terminology.

Answer 4

For $h$ to be a homomorphism, you need $h(1,0)h(0,1)$ to be the same as $h(0,1)h(1,0)$, because in your domain $(1,0)+(0,1)=(0,1)+(1,0)$. But in fact, that requires $ab=ba$, which you don’t know yet.

So your $h$ is not known to be a homomorphism. That means that your claim that $hg=\mathrm{id}$ is unjustified: it is implicitly predicated on the assumption that you have a homomorphism $F(A)\to F(A)$ that maps the generators to themselves identically. But because $h$ is not a homomorphism, it is not enough to check what $hg$ does to $a$ and $b$ in order to conclude the composition is the identity.

In fact, you have problems. For example, $h(g(aba)) = h(2,1) = a^2b$. So for $hg$ to be the identity, you would need $aba=a^2b$, or equivalently, $ba=ab$. That is, you need the very commutativity you are trying to prove. By asserting $hg$ is the identity, you are implicitly assuming your groups is commutative; and of course, from the assumption that $F(A)$ is commutative you can conclude that $F(A)$ is commutative...

Your $g$ is not a injective: indeed, $g(ab) = (1,1) = g(ba)$; but you don’t know that $ab=ba$. In fact, they are not equal. The kernel of $g$ consists of all products of elements of the form $xyx^{-1}y^{-1}$ with $x,y\in F(A)$; this is called the commutator subgroup of $F(A)$, and it is the smallest normal subgroup $N$ of $F(A)$ such that $F(A)/N$ is abelian.