Inverse map is involutive
This article gives the statement, and possibly proof, of a basic fact in group theory.
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- It satisfies the reversal law:
- Applying it twice sends an element to itself:
It fact, both these are true in the greater generality of a monoid, under the condition that all the s have two-sided inverse (note: we still need a monoid to guarantee that two-sided inverses, when they exist, are unique).
Proof of reversal law
In order to show that the element is a two-sided inverse of , it suffices to show that their product both ways is the identity element. Consider first the product:
Due to associativity, we can drop the parentheses and we get:
Now, consider the middle product . This is the identity element, and since the identity element has no effect on the remaining product, it can be removed, giving the product:
We now repeat the argument with the middle product and cancel them. Proceeding this way, we are able to cancel all terms and eventually get the identity element.
A similar argument follows for the product the other way around:
Thus, the elements are two-sided inverses of each other.
Note: In fact, it suffices to check only one of the two inverse conditions, i.e., check only that the first product is the identity element. This is because, in a group, every element has a two-sided inverse. Further, equality of left and right inverses in monoid forces any one-sided (left or right) inverse to be equal to the two-sided inverse.
Proof for applying it twice
This is direct from the definition. let . Then, by the inherent symmetry in the definition of inverse element, we see that .
More explicitly, if , that means that . But this is precisely the condition for stating that .
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 18, Proposition 1