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### 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/
```
Associated Topics:
College Modern Algebra

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