Thread: free group, generating set

Hi! I have got two questions concerning free groups:
a) If a group is freely generated by infinitely many elements, then it cannot be finitely generated (like the derived subgroup in F(a,b) for example). Is this true? I am pretty sure but cannot show it.

b) Let F(X) be the free group over X and x in X. Then we have a homomorphism $\displaystyle f: F(X) \to \mathbb{Z}$ which for a word w is just the sum of the exponents of x appearing in w. So for example $\displaystyle f(xyzx^2zx^{-1})=2$. Let $\displaystyle K:=\{w \in F(X): f(w)=0\}$. This is the kernel of f, hence a normal subgroup and it of course contains the group generated by X\{x}. But why does X\{x} generate K? For example $\displaystyle xyx^{-1}$ is in K, but how can it be generated from X\{x}?

Thankful for any hints,
Banach

Originally Posted by Banach

Hi! I have got two questions concerning free groups:
a) If a group is freely generated by infinitely many elements, then it cannot be finitely generated (like the derived subgroup in F(a,b) for example). Is this true? I am pretty sure but cannot show it.

b) Let F(X) be the free group over X and x in X. Then we have a homomorphism $\displaystyle f: F(X) \to \mathbb{Z}$ which for a word w is just the sum of the exponents of x appearing in w. So for example $\displaystyle f(xyzx^2zx^{-1})=2$. Let $\displaystyle K:=\{w \in F(X): f(w)=0\}$. This is the kernel of f, hence a normal subgroup and it of course contains the group generated by X\{x}. But why does X\{x} generate K? For example $\displaystyle xyx^{-1}$ is in K, but how can it be generated from X\{x}?

Thankful for any hints,
Banach

For (b), it is the normal closure of the set $\displaystyle X \setminus \{x\}$, not just the group generated by it.

I feel the answer to (a) is quite complicated. However, the fundamental group of the graph consisting of one vertex and infinitely many loops coming out of it (an infinite bouquet) is the infinitely generated free group. Such a graph cannot have a finitely generated infinite group...so that is sort of why, but it does need more of a proof.

You could try looking up either Magnus, Karrass and Solitar, or Lyndon and Schupp. Both books are called Combinatorial Group Theory, and both start with a look at free groups. (The first book is excellent, the second very interesting).

let $\displaystyle Y=X \cup X^{-1}.$ if $\displaystyle F(X)$ was finitely generated, then it would be generated by finitely many words, say $\displaystyle w_1, \cdots , w_n.$ now choose $\displaystyle y \in Y$ which does not appear in $\displaystyle w_j^{\pm 1}, \ j=1, \cdots n.$

then $\displaystyle y \in F(X)$ would not appear in the group generated by $\displaystyle w_1, \cdots, w_n.$

EDIT: ok, i just read part b) of your question more carefully. are you sure that is the question? besides why does $\displaystyle K$ contain the group generated by $\displaystyle X \setminus \{x \}$?

Originally Posted by Banach

Hi! I have got two questions concerning free groups:
a) If a group is freely generated by infinitely many elements, then it cannot be finitely generated (like the derived subgroup in F(a,b) for example). Is this true? I am pretty sure but cannot show it.

If you change a condition a little bit, then the following is true.

"Although a free group H is freely generated by infinitely many elements, it can be a subgroup of a free group G that is freely generated by finitely many elements."

For example, let G=<x, y> and choose generators of H as $\displaystyle g_k=x^kyx^{-k}$ for k=1,2, ..... .

b) Let F(X) be the free group over X and x in X. Then we have a homomorphism $\displaystyle f: F(X) \to \mathbb{Z}$ which for a word w is just the sum of the exponents of x appearing in w. So for example $\displaystyle f(xyzx^2zx^{-1})=2$. Let $\displaystyle K:=\{w \in F(X): f(w)=0\}$. This is the kernel of f, hence a normal subgroup and it of course contains the group generated by X\{x}. But why does X\{x} generate K? For example $\displaystyle xyx^{-1}$ is in K, but how can it be generated from X\{x}?

Thankful for any hints,
Banach

My interpretation is that K is a normal subgroup of F(X) containing a group generated by X\{x}, but it is not equal to the group generated by X\{x}.

Thanks for the quick answers.

b) Ok, with the normal closure of X\{x} the claim makes sense and is easily seen.

a) of course, thank you for helping me over that.

A similar question to a): Let F(X) be freely generated by |X|=n elements. Thinking of vector spaces it will probably have no free generating set consisting of fewer elements. Is that true?

Originally Posted by Banach

Thanks for the quick answers.

b) Ok, with the normal closure of X\{x} the claim makes sense and is easily seen.

a) of course, thank you for helping me over that.

A similar question to a): Let F(X) be freely generated by |X|=n elements. Thinking of vector spaces it will probably have no free generating set consisting of fewer elements. Is that true?

Comparing to a vector space is a bit awkward.

Anyhow, there is a corollary in my topology text book (Massey, "A basic course in algebraic topology, p 78).

"If F and F' are free groups on finite sets S and S', then F and F' are isomorphic if and only if S and S' have the same cardinal number."

Ah, there was a "free" too much in my question . My actual question is:
Can a group (which is freely generated by n elements) have a generating set with fewer elements (it cannot have a FREE generating set according to the corollary from the previous post)?

Originally Posted by Banach

Ah, there was a "free" too much in my question . My actual question is:
Can a group (which is freely generated by n elements) have a generating set with fewer elements (it cannot have a FREE generating set according to the corollary from the previous post)?

Well, know...because every subgroup of a free group is free. So every subset of your free group will freely generate a free group...surely...

True, thank you

Originally Posted by Banach

Ah, there was a "free" too much in my question . My actual question is:
Can a group (which is freely generated by n elements) have a generating set with fewer elements (it cannot have a FREE generating set according to the corollary from the previous post)?

The above corolloary says that if F is a free group on a set S, the cardinal number of S is an invariant of the group, since every group is isomorphic to itself.

If F is a free group on a set S with |S|=n for n>=2 , you have a free subgroup F'' of F with a fewer generating set S'' such that 1<= |S''| < n, but it is not equal to or isomorphic to F.