Inverse map is involutive

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Statement

The inverse map in a group, i.e. the map sending any element of the group, to its inverse element, is an involutive map, in the sense that it has the following two properties:

It satisfies the reversal law:

$(a_{1} a_{2} \dots a_{n})^{- 1} = a_{n}^{- 1} a_{n - 1}^{- 1} \dots a_{1}^{- 1}$

Applying it twice sends an element to itself:

$(a^{- 1})^{- 1} = a$

It fact, both these are true in the greater generality of a monoid, under the condition that all the $a_{i}$ s have two-sided inverse $a_{i}^{- 1}$ (note: we still need a monoid to guarantee that two-sided inverses, when they exist, are unique).

Proof

Proof of reversal law

In order to show that the element $a_{n}^{- 1} a_{n - 1}^{- 1} \dots a_{1}^{- 1}$ is a two-sided inverse of $a_{1} a_{2} \dots a_{n}$ , it suffices to show that their product both ways is the identity element. Consider first the product:

$(a_{1} a_{2} \dots a_{n - 1} a_{n}) (a_{n}^{- 1} a_{n - 1}^{- 1} \dots a_{2}^{- 1} a_{1}^{- 1})$

Due to associativity, we can drop the parentheses and we get:

$a_{1} a_{2} \dots a_{n - 1} a_{n} a_{n}^{- 1} a_{n - 1}^{- 1} \dots a_{2}^{- 1} a_{1}^{- 1}$

Now, consider the middle product $a_{n} a_{n}^{- 1}$ . This is the identity element, and since the identity element has no effect on the remaining product, it can be removed, giving the product:

$a_{1} a_{2} \dots a_{n - 1} a_{n - 1}^{- 1} \dots a_{2}^{- 1} a_{1}^{- 1}$

We now repeat the argument with the middle product $a_{n - 1} a_{n - 1}^{- 1}$ 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:

$(a_{n}^{- 1} a_{n - 1}^{- 1} \dots a_{1}^{- 1}) (a_{1} a_{2} \dots a_{n})$

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 $b = a^{- 1}$ . Then, by the inherent symmetry in the definition of inverse element, we see that $a = b^{- 1}$ .

More explicitly, if $b = a^{- 1}$ , that means that $a b = b a = e$ . But this is precisely the condition for stating that $a = b^{- 1}$ .

References

Textbook references

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