# Free Groups

1. Sep 5, 2006

### Oxymoron

The only definition of a free group I have is this:

If F is a free group then it must have a subgroup in which every element of F can be written in a unique way as a product of finitely many elements of S and their inverses.

Now, is it possible to form a commutator subgroup of F? That is, the commutator subgroup [F,F] of the free group F is a subgroup generated by all the commutators of F. It will be the smallest normal subgroup so that the quotient group F/[F,F] is abelian.

I think I can define the commutator subgroup of F as

$$[F,F] = \langle g^{-1}h^{-1}gh\,|\,g,h \in F\rangle$$

Does this look right?

Now, what does it mean for a group to be free on more than 1 generator?

Lastly, I want to discover whether or not the commutator subgroup of a free group has infinite index. But I read somewhere that all free groups have infinite index. Then by the Nielsen-Schreier theorem, won't the commutator subgroup be infinite as well, end of story?

Thanks for any help.

Last edited: Sep 5, 2006
2. Sep 5, 2006

### matt grime

The free group on n generators (x_1,..,x_n) is the object whose elements are words in x_i and x_i^-1 with concatenation of strings as the operation.

Of course you can form the commutator, and it is exactly what you wrote. Why would you not be able to do this?

Why should the commutator subgroup not have infinite order? F_n is infinite (the free group on n generators), so is the commutator subgroup (for n>1), and so does the quotient.

3. Sep 5, 2006

### Oxymoron

Ok, but Im not convinced that F_n is infinite.

4. Sep 5, 2006

### Hurkyl

Staff Emeritus
Which of these are equal?

..., g^-2, g^-1, 1, g, g^2, ...

That doesn't look like a definition -- it looks like a theorem. And I know things that aren't free groups for which this holds.

5. Sep 5, 2006

### Oxymoron

None are. Well, g^n = 1 if the generator of the group F has order n. But all I know about F is that it is free on more than one generator, wait a minute, does this simply mean that I can talk about F_n (instead of F) as long as n > 1? Is that all F_n is? ...a free group on n generators?

Hmmm, well upon reviewing my notes I discovered the following theorem:

Every subgroup of a free group is itself free

which sounds a lot like the definition of a free group that I was using.

Last edited: Sep 5, 2006
6. Sep 5, 2006

### matt grime

But by definition free groups do not have any element of finite order. That is a trivial consequence of the definition. Not that you have given a correct definition, so far.

That is what I defined F_n to be.

Last edited: Sep 5, 2006
7. Sep 5, 2006

### mathwonk

a free abelian group generated by a set S, is an abelian group FA(S) plus an in jection from S to F(S), such that given any function g from S to any abelian group G, there is a unique homomorphism F(S)--->G that extends g.

hence a free group on a set S is a group F(S) such that given any function g from S to any group G, there is a unique homomorphism F(S)--->G extending g.

FA(S) consists of funcxtions from S to the integers Z, such that all but a finite number of vbalues are zero.

F(S)( consists of words built from the leters in S and their inverses. the only cancellations possible in a word are to cancel ec = c = ce, and c.c^(-1) = e, where e is the identity c^0 = e.

i hope this is right. michael artin's book algebra has a nice elementary treatment. if this is wrong please tell me as i have to teach this next month.

8. Sep 5, 2006

### AKG

You can always form the commutator subgroup of a given group.

A group G is free on more than one generator if there is some set S such that G is the free group on S and |S| > 1 (possibly infinite). For a definition on free groups, check Mathworld.

There is a difference between the index of a subgroup in another group, and the order of a group. Also, there's no such thing as the index of a group. All free groups (on a non-empty set of generators) are infinite, and by Nielsen-Schreier all subgroups of a free group are free, so all (non-trivial) subgroups of a free group are infinite. The commutator subgroup of a free group on more than one generator is a non-trivial subgroup, hence is an infinite free group. I think it's very easy to prove that the commutator subgroup has infinite index. Why don't you try it and let us know if you get stuck. If you need a clue, recall what you learnt from Hurkyl, namely that ... g-2, g-1, e, g, g2, ... are all distinct.

9. Sep 5, 2006

### Oxymoron

Yeah, that looks right. I liked the definition with the "word" motivation, but my lecturer refuses to teach it with words and letters, etc... for unknown reasons. :(

So I guess I can't simply say "that since F is a free group it is infinite then from the N-S theorem so is any non-trivial subgroup of F, in particular the commutator subgroup. Therefore the commutator subgroup is infinite.".?

They way I used to work out the index of a subgroup back in the good old days (before all this Free Group stuff!) was to count the number of cosets (and possibly even use Lagrange's theorem if the group was finite). Is this the correct approach?

10. Sep 5, 2006

### AKG

No, you can't say that, but this doesn't have to do with what I said. [F,F] is infinite if F is free on more than one generator. Proof: If F has more than one generator, then [F,F] is not the trivial subgroup. Since F is free and [F,F] is a subgroup, [F,F] is isomorphic to a free group (by Nielsen-Schreier) and since it's non-trivial, it is infinite.
That's what I would do.

Last edited: Sep 5, 2006
11. Sep 7, 2006

### Oxymoron

This is the outline of the proof that I am after, is it not?

12. Sep 7, 2006

### Oxymoron

It seems like the key in proving this is to understand why the commutator subgroup is not trivial.

This the way I picture it: If the commutator subgroup of any group G is the trivial subgroup then the group G is abelian. But surely! ...If F is free on S (and S has more than 1 generator, i.e. $S = \{g_1,g_2,\dots\}$) then F is not abelian! How do I prove this!?

Is the following correct?

Start with a set $S = \{g_1,g_2,g_3,\dots\}$. Then I can construct the free group F containing S. F will not necessarily be abelian in my opinion, but rather the most general group possible that contains S. With S in hand, I adjoin the identity element and relate all elements with the inverse to construct the larger set $\{\dots,g_2^{-1},g_1^{-1},e,g_1,g_2,\dots\}$. Then I can form all finite words from this enlarged set and say that two words are equivalent if and only if inverses and identity can make them so.

This is how I would construct my free group F. Now I can construct the commutator subgroup, [F,F] of F as being a subgroup of F generated by all the commutators of F (that is, containing all $ghg^{-1}h^{-1}$). That is to say that the commutator subgroup of the free group F requires that gh = hg

Last edited: Sep 7, 2006
13. Sep 7, 2006

### Oxymoron

I was thinking that if F is generated by elements of the set S then F is abelian if

$$g_1g_2 = g_2g_1$$

But $g_1g_2$ is a reduced word (is it not?!) and so is $g_2g_1$ - both, of which, have different spellings. Therefore they are distinct and we have $g_1g_2 \neq g_2g_1$. Hence F is not abelian!

Therefore any subgroup of F is non-trivial. If all subgroups of F are non-trivial then [F,F] is non-trivial. Since F is free then every non-trivial proper subgroup of F is free, therefore [F,F] is isomorphic to a free group. Since isomorphisms of groups preserve cardinality, [F,F] is infinite since all free groups are infinite.

How does this sound?

14. Sep 7, 2006

### matt grime

Did you pick a definition of 'free group'? Because from my definition it is trivial that it is not abelian: it is the words in the generators and their inverses with concatenation of strings and the only cancellation is gg^-1=g^-1g=e and eg=ge=g. Then you don't need to write most of the things you did (particularly about isomorphisms preserving cardinalities).

15. Sep 7, 2006

### AKG

What are you trying to prove: that [F,F] is infinite, or that it has infinite index in F?

If h and g are generators of F, then hg and gh are distinct. If F is free with more than one generator, then it is not abelian. The statement:

Therefore any subgroup of F is non-trivial.

is clearly false, and I don't know what you might have meant by it.

N-S Theorem: Every subgroup of a free group is free.
Trivial fact: Every group has a trivial subgroup.
Easily-proved fact: If F is free on more than one generator, then [F,F] is not the trivial subgroup.
Easily-proved fact: If F is a free group and is not the trivial group (i.e. the free group over the empty set of generators is both trivial and free), then it is infinite.

Alternatively, just observe that if F has more than one generator, say g and h are generators, then each of the following are distinct elements of [F,F]:

ghg-1h-1, (ghg-1h-1)2, (ghg-1h-1)3, (ghg-1h-1)4, ...

16. Sep 7, 2006

### Oxymoron

I want to prove it is infinite and it has infinite index

17. Sep 7, 2006

### AKG

Well, does post #15 not make it clear how to prove that [F,F] is infinite when F has more than one generator? I've also mentioned how to prove that it has infinite index. To reiterate, first prove that [F,F] is not all of F, then find infinitely many cosets for [F,F]. If you're proven that [F,F] is not all of F, then there is some generator g not in [F,F]. Consider the cosets gk[F,F]. If they aren't all distinct, then there are distinct j, k such that gj[F,F] = gk[F,F], implying that gj-k is in [F,F], with j-k non-zero. So to neaten, there is some non-zero m such that gm is in [F,F]. Prove this to be impossible. Hint: if you write out any x in [F,F] as a product of generators and their inverses, what can you say about the sum of the exponents? For example, the sum of the exponents in ghg-1h-1 is 1 + 1 + (-1) + (-1) = 0. The sum of the exponents in gm, on the other hand, is m. Okay, at this point I've given everything away. :(

18. Sep 7, 2006

### Oxymoron

Oh it certainly does! :) EDIT: Because you can keep the exponent of the commutators increasing forever and they will all be distinct in [F,F].

Don't be sad AKG. I still have to completely understand every little bit of what you wrote, down to the finest detail, and no-one can give that away. So I still have some work to do.

Well I figured that $[F,F] < F$ because the commutator subgroup is generated by elements of F that commute (basically). What about the non-commutative elements of F? They wont have anything to do with [F,F] will they? Therefore there will be elements in F that aren't in [F,F] and therefore [F,F] is not all of F.

How can I find infinitely many cosets for [F,F] if all I know how to do is count? Surely there is something else about the commutator subgroup that lets me "count" the number of cosets in this case?

Let me guess...this is how you count all the cosets?

So you are saying that if [F,F] is not all of F then there exists a generator $g\in F$ such that $g \notin F$. Then using these particular generators, left-multiply elements in the commutator by them, i.e. $g^k[F,F]$. Ok, well I can see that if they are not all distinct then what you wrote in the above quote makes sense. But what if they are distinct? Does that mean that my g (which was supposedly not in [F,F]) does actually belong in [F,F]?

Now this is interesting. The sum of the exponents for any string in [F,F] is zero because all you have are generators (+1) and their inverses (-1) and they always commute, therefore their multiplications always equal the identity (0). But If I multiply my generator $g^m$ on the left of the commutator: $g^m[F,F]$ then all of a sudden the sum of the exponents is m. Hence $g^m$ is not in [F,F].

Unfortunately I still do not see how this lets me say that there are an infinite number of cosets...yet

Last edited: Sep 7, 2006
19. Sep 8, 2006

### AKG

Commute with what? With everything else in the group? No, that's the center of the group, i.e. Z(F) consists of all elements that commute with everything else in F. [F,F] is the subgroup of commutators. If F is a free group of more then one generator, then the only thing in Z(F) is the identity. In light of what I just said, the following questions don't make sense:
Moving on:
No, it means you have infinitely many cosets of [F,F], meaning that it has infinite index, and you're done!
I don't know what you mean by "they always commute", but yes, the sum of the exponents is always 0 in the commutator subgroup.
You need simply to say that every element of [F,F] has a zero exponent sum, and gm has a non-zero exponent sum, hence gm is not in [F,F].
Recall g is a generator not in [F,F] (in fact, no generator is in [F,F], but you only need one). To prove [F,F] is infinite, you need to find infinitely many cosets. I propose the gk[F,F] to be such an infinite bunch of cosets. But what if the list:

[F,F], g[F,F], g2[F,F], ...

only "looks" infinite, whereas in fact it repeats itself? Well, this is in fact not the case, as all the cosets listed above are in fact distinct. That is, there are no distinct j and k such that gj[F,F] = gk[F,F]. If there were, then gj-k[F,F] would equal [F,F], implying that gj-k is in [F,F]. This would mean that some non-zero integer m existed such that gm was in [F,F]. But this can't be the case, as proven above. So there is no non-zero integer m such that gm is in [F,F]. Hence there are no distinct integers j, k such that gj[F,F] = gk[F,F]. So the list:

[F,F], g[F,F], g2[F,F], ...

is indeed infinite. So [F,F] does indeed have infinite index.

20. Sep 8, 2006

### AKG

Note, this also holds in the case where F is free over 1 generator, as [F,F] = {e} is the trivial subgroup of size 1, whereas F is infinite. It's easy to see that its cosets are ... g-2{e}, g-1{e}, {e}, g{e}, g2{e}, ... where g is the generator for F.