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Prove that the each element of a group G has a unique inverse. That is, if a, b, W E G satisfy then b

User Robaker
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Explanation:

Say
a is an element of
G which might have more than 1 inverse. Let's call them
b, and
c. So that
a has apparently two inverses,
b and
c.

This means that
a*b = e and that
a*c=e(where
e is the identity element of the group, and * is the operation of the group)

But so we could merge those two equations into a single one, getting


a*b=a*c

And operating both sides by b by the left, we'd get:


b*(a*b)=b*(a*c)

Now, remember the operation on any group is associative, meaning we can rearrange the parenthesis to our liking, gettting then:


(b*a)*b=(b*a)*c

And since b is the inverse of a,
b*a=e, and so:


(e)*b=(e)*c


b=c (since e is the identity of the group)

So turns out that b and c, which we thought might be two different inverses of a, HAVE to be the same element. Therefore every element of a group has a unique inverse.

User Rodrigo Zepeda
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