# Free Group

• Dec 3rd 2010, 11:47 PM
Arczi1984
Free Group
Hi,

Prove that every free group is torsion-free and is non-abelian if its rank is $\geq 2$.

The second part is easy.
What about the first part? I think that we should suppose that this free group $F[X]$ is torsion (but I'm not sure).
Thanks for any help.
• Dec 4th 2010, 04:08 AM
tonio
Quote:

Originally Posted by Arczi1984
Hi,

Prove that every free group is torsion-free and is non-abelian if its rank is $\geq 2$.

The second part is easy.
What about the first part? I think that we should suppose that this free group $F[X]$ is torsion (but I'm not sure).
Thanks for any help.

This is an easy task if you know already about the normal form of an element in a free group...besides this, we

can work with F(2) = the free group of rank two, which contains an isomorphic copy of the free

group of any rank between 2 and countably infinite.

If you don't know yet about the normal form, then you can do as follows: suppose $w:=w(x,y)$ is a

non-trivial word on x,y in the free group $F(x,y)$, and $ord(w) =k<\infty$ . We can

carry on now by induction on the length of this word: if $l(w)=1\Longrightarrow w(x,y) =x,y,x^{-1},\,\,or\,\,y^{-1}$ , suppose

$w=x\Longrightarrow$ defining the function $f:\{x,y\}\rightarrow C_{k+1}=\langle c\rangle$ the cyclic group of

order $k+1$, by $f(x):=c\,,\,f(y)=1$ , we know (the universal property of free groups) that there

exists a unique group homomorphism $\phi(F(x,y))\rightarrow C_{k+1}$ extending

the function $f$. But this is impossible since then it must be $\phi(x)=c\Longrightarrow 1\neq c^k=\phi(x)^k=\phi(x^k)=\phi(1)=1$ ,

thus getting a straighfroward contradiction. Try now to extend this idea when $l(w)>1$

Tonio

Ps. Or read about the normal reduced form for an element in a free group. The proof then is

almost trivial.
• Dec 4th 2010, 04:19 AM
Arczi1984
Thanks for help. I'll try to do the rest;)

Could You show how can You do this task using normal reduced form? It will be nice to see the second solution.
• Dec 4th 2010, 04:34 AM
tonio
Quote:

Originally Posted by Arczi1984
Thanks for help. I'll try to do the rest;)

Could You show how can You do this task using normal reduced form? It will be nice to see the second solution.

If $w(x,y)$ is in reduced normal form, then we can assume that its first letter is no the inverse of its

last one (why?), so we get at once that $l(w)=m\Longrightarrow l(w^r)=rm$ , so

if the length of the word is possitive also its multiples also have positive length and are thus

not the unit element.

Tonio
• Dec 5th 2010, 11:53 AM
Arczi1984
We can assume this because this is cyclically reduced word (am I right?).

I'm still do not see how this proof the fact that every free group is torsion-free :(.

Could You show this with details ;), of course if You have time. (I'm thinking with difficulty so this is unclear for me.) I'll be glad for this.
• Dec 5th 2010, 12:37 PM
tonio
Quote:

Originally Posted by Arczi1984
We can assume this because this is cyclically reduced word (am I right?).

I'm still do not see how this proof the fact that every free group is torsion-free :(.

Could You show this with details ;), of course if You have time. (I'm thinking with difficulty so this is unclear for me.) I'll be glad for this.

Well, you know that ONLY the empty word (of length 0 = zero) in free generators represents the unit element, so if ANY non trivial element's

word increases its length as we multiply this element by itself then its length won't ever be zero and this

means the element raised to no (non-zero) power will be the unit element <==> the element has infinite order.

Tonio
• Dec 5th 2010, 01:49 PM
Arczi1984
Ok, I'll look at this tomorrow. Now it's a little late ;)