Finitely Generated Free Groups

sammycaps

So Munkres, on page 424 of Topology (2nd edition) says that "...two free groups are isomorphic if and only if their systems of free generators have the same cardinality (We have proved these facts in the case of finite cardinality)."

Nowhere explicitly does he say this, although it seems that many of the theroems and corollaries allude to it. I've tried a few different things to see a proof that applies only in the finite case, but I'm not sure I have it right.

Any help would be much appreciated.

Vargo

I bet the universal property of free groups would give the simplest proof.

http://en.wikipedia.org/wiki/Free_gr...ersal_property

lavinia

If you think about it you will see that an isomorphism of free groups must be defined from a bijection between generators and any bijection of generators gives an isomorphism by extension as is always true for free groups.

sammycaps

Ok, thanks very much.

mathwonk

if a free group on n generators is isomorphic to a free group on m generators, then their abelianizations are also isomorphic, so the free abelian groups on n and m generators are isomorphic. now if that is true for abelian G and H then their quotients by their subgroups 2G and 2H, i.e. G/2G and H/2H are also isomorphic. Now those two groups are vector spaces over Z/2Z of the same finite dimension, hence the same finite cardinallty. since one has cardinality 2^n and the other has cardinality 2^m, thus n=m.

forgive me if i do not follow, lavinia, but i do not see how your argument proves there cannot be more than one generating set for a free group, of different cardinalities.

I guess it is correct in some abstract sense categorically as Vargo suggests, i.e. there is a functorial equivalence between group maps out of F(n) the free group on {1,2,...,n} and set maps out of {1,2,...,}. but if F(n) and F(m) are isomorphic then they define equivalent functors of group maps, hence set maps out of {1,2,...,n} and {1,.,,.m} are equivalent, hence the two sets are isomorphic. but this requires a bit more of work to prove.

but i believe the proof above showing that if F(n) and F(m) are isomorphic, then 2^n = 2^m is more elementary.

lavinia

i see your point but what i was saying was a weaker statement which just says that an isomorphism requires a bijection between generators and since the groups are free a bijection between sets of generators suffices for an isomorphism. That one can not also have another set of generators of different cardinality is a different question - I think.

lavinia

Some thoughts on whether a free group can have two generating sets of different cardinality

- if the cardinalities of the generating sets are both infinite then they must be the same for otherwise the group would have two cardinalities.

mathwonk

so you are saying that the cardinality of an infinitely generated group would equal the cardinality of the generating set? If so, that makes the infinitely generated case look easier than the finitely generated case. Nice remark.

My point above was that of course as you say, an isomorphism takes a generating set to a generating set of the same cardinality, but that does not prove the statement required. I.e. it only proves that if F(n) and F(m) are isomorphic, then F(m) has a generating set of n elements. But it does not prove no such isomorphism exists when n < m, which I believe is what was asked.

As I understood it the goal was to prove that if F(n) and F(m) are isomorphic, then n = m.

I.e. the statement: "two free groups are isomorphic if and only if they have generating sets of the same cardinality", is true but trivial.

the statement: "two free groups are isomorphic if and only if [all] their generating sets have the same cardinality", is true but not as trivial.

I.e. the second statement is equivalent to: "F(S) is isomorphic to F(T) if and only if S and T have the same cardinality, which is what I believe Munkres meant.

i.e. if two free groups (on given generating sets) are isomorphic, then those generating sets are bijectively equivalent.

if this is obvious, i apologize for belaboring it.

lavinia

Right. I read the statement as assuming that all generating sets in a free group have the same cardinality. The proof assuming this is as you say virtually trivial but not totally. A bijection of generating sets on a non free group may not determine a homomorphism at all.

mathwonk

" I read the statement as assuming that all generating sets in a free group have the same cardinality.":

the statement you are assuming is the statement i read as requiring proof.