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 is a

non-trivial word on x,y in the free group , and . We can

carry on now by induction on the length of this word: if , suppose

defining the function the cyclic group of

order , by , we know (the universal property of free groups) that there

exists a unique group homomorphism extending

the function . But this is impossible since then it must be ,

thus getting a straighfroward contradiction. Try now to extend this idea when

Tonio

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

almost trivial.