# Inverse Element

Given an element \(a\) in a set with a binary operation, an **inverse element** for \(a\) is an element which gives the identity when composed with \(a.\)

More explicitly, let \(S\) be a set, \(*\) a binary operation on \(S,\) and \(a\in S.\) Suppose that there is an identity element \(e\) for the operation. Then

- an element \(b\) is a
*left inverse*for \(a\) if \(b*a = e;\) - an element \(c\) is a
*right inverse*for \(a\) if \(a*c=e;\) - an element is an
*inverse*(or*two-sided inverse*) for \(a\) if it is both a left and right inverse for \(a.\)

Let \(S = \{a,b,c,d\},\) and consider the binary operation defined by the following table: \[ \begin{array}{|c|cccc|}\hline *&a&b&c&d \\ \hline a&a&a&a&a \\ b&c&b&d&b \\ c&d&c&b&c \\ d&a&b&c&d \\ \hline \end{array} \] The value of \( x * y \) is given by looking up the row with \(x\) and the column with \(y.\)

Which elements have left inverses? Right inverses? Inverses?

Since \(d\) is the identity, and \(b*c=c*a=d*d=d,\) it follows that

- \(a\) has no left or right inverse;
- \(b\) has no left inverse, and has a right inverse \(c;\)
- \(c\) has a left inverse \(b,\) and a right inverse \(a;\)
- \(d\) is its own left and right inverses. \(_\square\)

Consider the set \(\mathbb R\) with the binary operation of addition. The identity element is \(0,\) so the inverse of any element \(a\) is \(-a,\) as \((-a)+a=a+(-a) = 0.\) So every element has a unique left inverse, right inverse, and inverse.

## Existence and Properties of Inverse Elements

An element might have no left or right inverse, or it might have different left and right inverses, or it might have more than one of each.

Let \( S \) be the set of functions \( f\colon {\mathbb R} \to {\mathbb R}.\) There is a binary operation given by composition \( f*g = f \circ g,\) i.e. \( (f*g)(x) = f\big(g(x)\big).\) The (two-sided) identity is the identity function \( i(x) = x.\)

Find a function with more than one left inverse. Find a function with more than one right inverse.

If \(f(x) = e^x,\) then \(f\) has more than one left inverse: let \[ g_1(x) = \begin{cases} \ln(|x|) &\text{if } x \ne 0 \\ 0 &\text{if } x= 0 \end{cases}, \] and let \[ g_2(x) = \begin{cases} \ln(x) &\text{if } x > 0 \\ 0 & \text{if } x \le 0. \end{cases} \] Then \(g_1\big(f(x)\big) = \ln(|e^x|) = \ln(e^x) = x,\) and \( g_2\big(f(x)\big) = \ln(e^x) =x \) because \(e^x \) is always positive. The idea is that \(g_1 \) and \(g_2\) are the same on positive values, which are in the range of \(f,\) but differ on negative values, which are not.

If \[ f(x) = \begin{cases} \tan(x) & \text{if } \sin(x) \ne 0 \\ 0 & \text{if } \sin(x) = 0, \end{cases} \] then \(f\) has more than one right inverse: let \(g_1(x) = \arctan(x)\) and \(g_2(x) = 2\pi + \arctan(x).\) Then \(f\big(g_1(x)\big) = f\big(g_2(x)\big) = x.\)

The first example was injective but not surjective, and the second example was surjective but not injective. \(_\square\)

If the binary operation is associative and has an identity, then left inverses and right inverses coincide:

If \( S\) is a set with an associative binary operation \(*\) with an identity element, and an element \(a\in S\) has a left inverse \( b\) and a right inverse \(c,\) then \(b=c\) and \(a\) has a unique left, right, and two-sided inverse.

Let \(e\) be the identity. Then \[ c = e*c = (b*a)*c = b*(a*c) = b*e = b. \] The same argument shows that any other left inverse \(b'\) must equal \(c,\) and hence \(b.\) Similarly, any other right inverse equals \(b,\) and hence \(c.\) So there is exactly one left inverse and exactly one right inverse, and they coincide, so there is exactly one two-sided inverse. \(_\square\)

## Examples of Inverse Elements

The existence of inverses is an important question for most binary operations. Here are some examples.

Let \(S= \mathbb R\) with \( a*b = ab+a+b.\) It is straightforward to check that this is an associative binary operation with two-sided identity \(0.\) Then the inverse of \(a, \) if it exists, is the solution to \(ab+a+b=0,\) which is \(b = -\frac{a}{a+1},\) but when \(a=-1\) this inverse does not exist; indeed \( (-1)*b = b*(-1) = -1\) for all \(b.\) So every element of \(\mathbb R\) has a two-sided inverse, except for \( -1.\)

Let \({\mathbb R}^{\infty}\) be the set of sequences \( (a_1,a_2,a_3,\ldots) \) where the \(a_i\) are real numbers. Let \(S \) be the set of functions \( f \colon {\mathbb R}^\infty \to {\mathbb R}^\infty.\) Then composition of functions is an associative binary operation on \(S,\) with two-sided identity given by the identity function. Now let \( t \) be the shift operator, \(t(a_1,a_2,a_3) = (0,a_1,a_2,a_3,\ldots).\) Then \(t\) has many left inverses but no right inverses (because \(t\) is injective but not surjective). One of its left inverses is the reverse shift operator \( u(b_1,b_2,b_3,\ldots) = (b_2,b_3,\ldots).\)

Let \(G\) be a group. Then every element of the group has a two-sided inverse, even if the group is nonabelian (i.e. the operation is not commutative).

Let \(R\) be a ring. Then every element of \(R\) has a two-sided additive inverse \((R\) is a group under addition\(),\) but not every element of \(R\) has a multiplicative inverse. In particular, \(0_R\) never has a multiplicative inverse, because \(0 \cdot r = r \cdot 0 = 0\) for all \(r\in R.\) If every other element has a multiplicative inverse, then \(R\) is called a

**division ring**, and if \(R\) is also commutative, then it is called a**field**. In general, the set of elements of \(R\) with two-sided multiplicative inverses is called \(R^*,\) the*group of units*of \(R.\)