# Finitely Generated Free Groups

by sammycaps
Tags: finitely, free, generated, groups
 P: 91 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.
 P: 350 I bet the universal property of free groups would give the simplest proof. http://en.wikipedia.org/wiki/Free_gr...ersal_property
P: 1,716
 Quote by 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.
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.

 P: 91 Finitely Generated Free Groups Ok, thanks very much.
 Sci Advisor HW Helper P: 9,453 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.