Find a group G that contains elements a and b such that a^2 = e, b^2 = e, but the order of the element ab is infinite. Can G be abelian?

Obviously, G can't be Abelian, or else (ab)(ab) = (aa)(bb) = ee = e, and thus ab has order 2 (or 1 if ab=e).

Let G be the free group on a and b, with a^2 = b^2 = e. In a free group, there are no relations other than what's given for a and b. In G, the elements are simply strings of alternating a's and b's. You can begin with a or b, and end with a or b.

If you multiply two elements g and h, you just concatenate the expressions if g ends with a different letter than h begins; otherwise, you get a string of a^2 and b^2 terms that collapse the string until you run out of terms in one side:
(ababab)(abab) = ababababab
(abababab)(babab) = aba

Inverses are found by just reversing the terms:
(abababab)^-1 = babababa

Hghbbbhhg