Inverse of an Inverse
Date: 01/20/2002 at 22:19:26 From: Kris Subject: Modern Algebra (groups) Question: In a group, prove that (a^-1)^-1=a for all a. Is this question asking me to show that this is a group? If that is the case, then I must show associativity, existence of an identity, and existence of inverse. How do I go about starting this problem? I am not even sure that's exactly what it's asking. I would very much appreciate some help on this. Thank you in advance.
Date: 01/21/2002 at 13:53:28 From: Doctor Paul Subject: Re: Modern Algebra (groups) The question is not asking you to show that this is a group. It is saying that given a group, pick an arbitrary element in the group and prove that the inverse of the inverse of this element is actually the element itself. To prove that x^(-1) = y, you need to show that x*y = e (where e is the identity element in the group). So in this problem, you need to show that a^(-1) * a = e. But this is certainly the case since a^(-1) * a = a^(-1 + 1) = a^0 = e. I hope this helps. Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.