Inverse map is involutive

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:

$$(a_1a_2 \ldots a_n)^{-1} = a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}$$


 * 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 of reversal law
In order to show that the element $$a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}$$ is a two-sided inverse of $$a_1a_2 \ldots a_n$$, it suffices to show that their product both ways is the identity element. Consider first the product:

$$(a_1a_2 \ldots a_{n-1}a_n)(a_n^{-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1})$$

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

$$a_1a_2 \ldots a_{n-1}a_na_n^{-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1}$$

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

$$a_1a_2 \ldots a_{n-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1}$$

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:

$$(a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1})(a_1a_2 \ldots a_n)$$

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 $$ab = ba = e$$. But this is precisely the condition for stating that $$a = b^{-1}$$.

Textbook references

 * , Page 18, Proposition 1