This is not homework. Self-study. And I'm really enjoying it. But, as I'm going through this book ("A Book of Abstract Algebra" by Charles C. Pinter) every so often I run into a problem or concept I don't understand. Let G be a finite abelian group, say G = (e,a1, a2, a3,...,an). Prove that (a1*a2*...an)^2 = e. So, it has a finite number of elements and it's a group. So it's associative, has an identity element and an inverse as elements of G, and as it's abelian so it's also commutative. But I don't see how squaring the product of its elements leads to the identity element e. Wait. Writing this has me thinking that each element might be being 'multiplied' by it's inverse yielding e for every pair, which when all mutiplied together still yields e, even when ultimately squared. Could that be the answer, even though I may not have stated it elegantly? There's no one I can ask so I brought it to this forum.
You're on the right track. But the product is squared for a reason. What if your group has elements of order 2? This won't cause problems, but it's necessary to consider it.
'If the group has elements of order 2' I don't really understand that. The terminology in this book I understand (so far) is if the group is of order 2 that means it is a finite group with two elements. Things are sometimes squared to get rid of a negative sign. But if the elements are numbers I would think that multiplying a negative number by its inverse (which would also be negative so the outcome is 1) would take care of that. But perhaps not so I'll go with squaring would knock a negative out of the final e. Is that it?
What spamiam means is that there might be an element [itex]a_i[/itex] such that [itex]a_i=a_i^{-1}[/itex]. In that case, your proof would not hold anymore. Indeed, its inverse does not occur in the list since it equals [itex]a_i[/itex].
So there could be an element of G that equals its own inverse. So squaring the product insures that this is reduced to e as well? Since they equal each other they should square to identity I think. Is this it?