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Inverse of an Inverse

Date: 01/20/2002 at 22:19:26
From: Kris
Subject: Modern Algebra (groups)


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.


- Doctor Paul, The Math Forum
Associated Topics:
College Modern Algebra

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