0
$\begingroup$

I am trying to show that that $\mathbb{F}_2$ (the free group of rank 2) is CSA. Which means that for every maximal abelian subgroup $H\le \mathbb F_2$, We have that $H\cap gHg^{-1}$ is not trivial iff $g\in H$.

I have showed that in general, a group $G$ is CSA iff it satisfy the two formulas $$\varphi =\forall x,y,x((y\neq1\wedge[x,y]=1\wedge[y,z]=1)\to[x,z]=1)$$ (i.e commutative transitive) and $$\psi=\forall x,y((x\neq1\wedge[x,yxy^{-1}]=1)\to [x,y]=1)$$ While $[x,y]$ is the commutator of $x,y$. I use this characterisation since I want to show the claim for all limit groups and these are universal formulas.

So I have showed that $\mathbb{F}_2$ is commutative transitive i.e satisfies $\varphi$ but I can't manage to show that it satisfy $\psi$ nor to do it directly. I tried to find something that commutes with $x$ and with $yxy^{-1}$ and then use commutative transitivity. And i have tried to use arguments on reduct words but I get stuck in complications

Will appreciate your help!

  • 0
    ${\mathbb F}_2$ usually denotes the finite field of order $2$, so perhaps you should say what you mean by it. – 2017-02-23
  • 0
    Yes I forgot that! – 2017-02-23

1 Answers 1

4

I would argue directly like this. Since subgroups of free groups are free, and abelian free groups are infinite cyclic, a maximal abelian subgroup $H$ is an infinite cyclic group $\langle a \rangle$.

If $g \not\in H$, then by maximality $\langle g,a \rangle$ is not abelian, so it must be free of rank $2$ and freely generted by $g$ and $a$. So we must have $H \cap gHg^{-1} = \{1\}$ or else we would have a nontrivial relation of the form $ga^ig^{-1} = a^j$ for some nonzero $i$ and $j$.

  • 0
    And abelian free groups are infinitely cyclic, since if not, we will have a non trivial relation, ($a\neq b^n$ for all $n\in\mathbb{Z}$ ,the relation $ab=ba$) right? – 2017-02-24
  • 0
    Yes that's right. I am using the general result that if a free group of rank $n$ is generated by $n$ elements, then it is freely generated by those elements. That is usually proved using Nielsen reduction. – 2017-02-24
  • 0
    @DerekHolt: Alternatively, it follows from the fact the free groups are Hopfian, which, is a corollary of their residual finiteness. – 2017-02-24