Inverse map is involutive: Difference between revisions

Latest revision as of 19:39, 31 March 2022

This article gives the statement, and possibly proof, of a basic fact in group theory.
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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

@@ Line 20: / Line 20: @@
 ===Proof of reversal law===
-This follows from associativity. Consider:
+In order to show that the element <math>a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}</math> is a two-sided inverse of <math>a_1a_2 \ldots a_n</math>, it suffices to show that their product both ways is the identity element. Consider first the product:
-<math>a_1a_2 \ldots a_na_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}</math>
+<math>(a_1a_2 \ldots a_{n-1}a_n)(a_n^{-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1})</math>
-We first cancel <math>a_n</math> and <math>a_n^{-1}</math>, and keep proceeding in this way, till the whole word reduces to the identity element.
+Due to associativity, we can drop the parentheses and we get:
-A similar argument follows for the product:
+<math>a_1a_2 \ldots a_{n-1}a_na_n^{-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1}</math>
-<math>a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}a_1a_2 \ldots a_n</math>
+Now, consider the middle product <math>a_na_n^{-1}</math>. This is the identity element, and since the identity element has no effect on the remaining product, it can be removed, giving the product:
+<math>a_1a_2 \ldots a_{n-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1}</math>
+We now repeat the argument with the middle product <math>a_{n-1}a_{n-1}^{-1}</math> 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:
+<math>(a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1})(a_1a_2 \ldots a_n)</math>
 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===