An identity element with respect to a binary operation is an element such that when a binary operation is performed on it and any other given element, the result is the given element.

In the video in Figure 13.15 we define when an element is the identity with respect to a binary operations and give examples. Following the video we present the formal definition of identity elements, give examples, and discuss methods for determining whether there is an identity element with respect to a given binary operation.

Carefully read the definition.

Definition13.16.

Let \(S\) be a set and \(\bullet:S\times S \to S\) be a binary operation on \(S\text{.}\) An element \(e\in S\) is an identity element of the set \(S\) with respect to the operation \(\bullet\) if \(s\bullet e=s\) and \(e\bullet s=s\) for all \(s\in S\text{.}\)

In Checkpoint 13.17 reproduce the definition by filling in the blanks.

Checkpoint13.17.Identity.

Let S be a set and let * : S \(\times\) S \(\to\) S be a binary operation on S. We read a * b as ’a star b’.

An element e in S is an identity element with respect to * if

select

(a * b) * c = a * (b * c)

a * b = b * a

a * e = a and e * a = a

a * b = e and b * a = e

for

select

all a in S

one a in S

all a in S and all b in S

one a in S and one b in S

all a in S, all b in S, and all c in S

one a in S, one b in S, and one c in S

the identity e with respect to * in S

all e in S

Example13.18.Identities with respect to known binary operations.

We revisit the binary operations from Example 13.3:

Consider the binary operation \(+:\Z\times\Z\to\Z\text{.}\) The sum of 0 and any given integer is the given integer. In other words, for all integers \(s\text{,}\) we have that \(s+0=0+s=s\text{.}\) So, we call the number 0 the additive identity element for the set of integers.

Consider the binary operation \(\cdot:\N\times\N\to\N\text{.}\) The product of 1 and any given natural number is the given natural number. In other words, for all natural numbers \(s\text{,}\) we have that \(s\cdot1=1\cdot s=s\text{.}\) So, we call the number 1 the multiplicative identity element for the set of natural numbers.

Consider the binary operation \(-:\Z\times\Z\to\Z\text{.}\) The only integer \(e\) such that \(a-e=a\) for all integers \(a\) is \(e:=0\text{.}\) For \(0\) to be an identity with respect to the binary operation \(-\) we also need \(0-a=a\) for all \(a\in\Z\text{.}\) We have \(0-1=(-1)\ne 1\text{.}\) So \(0\) is not an identity with respect to the binary operation \(-\text{.}\) As \(0\) was the only candidate for an identity, there is no identity with respect to the binary operation \(-\text{.}\)

Theorem13.19.

Let \(S\) be a set and let \(\bullet:S\times S \to S\) be a binary operation on \(S\text{.}\) Then, there is at most one element \(e\in S\) such that \(s\bullet e=s\) and \(e\bullet s=s\) for all \(s\in S\text{,}\) implying that if there is an identity element of the set \(S\) with respect to the operation \(\bullet\text{,}\) then it is unique.

Proof.

Suppose that there are two identity elements \(e\) and \(f\) of the set \(S\) with respect to the operation \(\bullet\text{.}\) Since \(e\) is an identity element, \(s\bullet e=s\) for all \(s \in S\text{.}\) So, in particular, \(f \bullet e = f\text{.}\) However, since \(f\) is an identity element, \(f\bullet s=s\) for all \(s \in S\text{.}\) Because \(e\in S\text{,}\) then this implies that \(f \bullet e = e\text{.}\) So, \(f \bullet e\) is equal to both \(f\) and to \(e\text{,}\) implying that \(f = e\text{.}\) Therefore \(f\) and \(e\) must be the same element. Thus, there is at most one identity element of \(S\) with respect to \(\bullet\text{.}\)

Since there can be at most one identity element of a set with respect to a binary operation, we call it the identity element, if it exists.

Example13.20.The identity with respect to \(\star:T\times T\to T\).

Let \(T=\{\Tx,\Ty,\Tz\}\text{,}\) and let the binary operation \(\star:T\times T\to T\) be given by the table in Example 13.4. Notice that \(\Tx\star \Ty = \Ty \star \Tx = \Tx\text{,}\)\(\Ty\star \Ty=\Ty\text{,}\) and \(\Ty\star \Tz = \Tz\star \Ty = \Tz\text{.}\) Since \(t\star \Ty=\Ty\star t = t\) for all \(t\in T\text{,}\) we have that \(\Ty\in T\) is the identity element of the set \(T\) with respect to the operation \(\star\text{.}\)

When we have an operation on a set given by an operation table, we can determine the identity element (if there is one) by locating the element corresponding to a special row and special column within the table. That special row within the table would need to match the header row at the top of the table and that special column within the table would need to match the header column on the left side of the table.

Example13.21.The identity with respect to \(\star:T\times T \to T\).

With the above comment in mind, we revisit Example 13.20. Notice that the row corresponding to \(\Ty\) matches the header row at the top of the table and the column corresponding to \(\Ty\) matches the header column on the left side of the table. So, the element \(\Ty\) is the identity element.

\(\star\)

\(\color{gray}\Tx\)

\(\color{gray}\Ty\)

\(\color{gray}\Tz\)

\(\Tx\)

\(\Tz\)

\(\Tx\)

\(\Ty\)

\(\color{red}\Ty\)

\(\color{gray}\Tx\)

\(\color{gray}\Ty\)

\(\color{gray}\Tz\)

\(\Tz\)

\(\Ty\)

\(\Tz\)

\(\Tx\)

\(\star\)

\(\Tx\)

\(\color{red}\Ty\)

\(\Tz\)

\(\color{gray}\Tx\)

\(\Tz\)

\(\color{gray}\Tx\)

\(\Ty\)

\(\color{gray}\Ty\)

\(\Tx\)

\(\color{gray}\Ty\)

\(\Tz\)

\(\color{gray}\Tz\)

\(\Ty\)

\(\color{gray}\Tz\)

\(\Tx\)

Example13.22.The identity with respect to \(\oplus:\Z_5\times\Z_5\to\Z_5\).

We consider the binary operation \(\oplus:\Z_5\times\Z_5\to\Z_5\) definied by \(a\oplus b=(a+b)\fmod 5\text{.}\) The identity element with respect to \(\oplus\) is \(0\text{.}\)

To see this note that for every \(a\in\Z_5\) we have

\begin{equation*}
0\oplus a = (0+a) \fmod 5 = a\fmod 5= a.
\end{equation*}

Thus \(0\) is the identity with repect to \(\oplus\text{.}\)

We can also arrive at this conclusion by applying the method from the previous example to the operation table from Example 13.6.

Problem13.23.Identity with respect to \(\oplus:\Z_3\times\Z_3\to\Z_3\).

Consider the binary operation \(\oplus:\Z_3\times\Z_3\to\Z_3\) given by \(a\oplus b=(a+b)\fmod 3\text{.}\) Find the identity with respect to \(\oplus\) in \(\Z_3\text{.}\)

Solution.

Recall that \(\Z_3=\{0,1,2\}\text{.}\)

The binary operation \(\oplus\) is based on the addition of integers and the identity with respect to the addition of integers is 0. So we check whether 0 is also the identity with respect to \(\oplus\text{.}\)

For all \(a\in\Z_3\) we have

\begin{equation*}
a\oplus 0 = (a+0) \fmod 3 = a \fmod 3 = a
\end{equation*}

where the last equality holds because \(a\in\Z_3\text{.}\) Also for all \(a\in\Z_3\) we have

\begin{equation*}
0\oplus a = (0+a) \fmod 3 = a \fmod 3 = a\text{.}
\end{equation*}

Thus \(0\) is the identity with respect to \(\oplus\) in \(\Z_3\text{.}\)

In Checkpoint 13.24 decide whether a given element is the identity with respect to a binary operation. If the element is not the identity, give a counterexample.

Checkpoint13.24.Is there an identity ?

Decide whether the following statements are true or false. If the statement is false give a counterexample, otherwise leave the field empty.

(1) Let the binary operation \(\otimes:\mathbb{Z}_{13}\times \mathbb{Z}_{13}\to \mathbb{Z}_{13}\) be given by \(a \otimes b = \left(a\cdot b\right) \bmod 13\text{.}\)

select

The statement is true.

The statement is false.

The identity element with respect to \(\otimes\) is \(7\text{.}\)

Counterexample: The statement is false, because for \(b:=\)\(\in \mathbb{Z}_{13}\) we have \(7 \otimes b\ne b\text{.}\)

(2) Let the binary operation \(\ominus:\mathbb{Z}_{12}\times \mathbb{Z}_{12}\to \mathbb{Z}_{12}\) be given by \(a \ominus b = \left(a-b\right) \bmod 12\text{.}\)

select

The statement is true.

The statement is false.

The identity element with respect to \(\ominus\) is \(11\text{.}\)

Counterexample: The statement is false, because for \(b:=\)\(\in \mathbb{Z}_{12}\) we have \(11 \ominus b\ne b\text{.}\)

(3) Let the binary operation \(\ominus:\mathbb{Z}_{8}\times \mathbb{Z}_{8}\to \mathbb{Z}_{8}\) be given by \(a \ominus b = \left(a-b\right) \bmod 8\text{.}\)

select

The statement is true.

The statement is false.

The identity element with respect to \(\ominus\) is \(1\text{.}\)

Counterexample: The statement is false, because for \(b:=\)\(\in \mathbb{Z}_{8}\) we have \(1 \ominus b\ne b\text{.}\)

Hint.

An element \(e\in \mathbb{Z}_{13}\) is the identity with respect to \(\otimes\) if \(e \otimes b=b\) and \(b \otimes e=b\) for all \(b\in \mathbb{Z}_{13}\text{.}\)

An element \(e\in \mathbb{Z}_{12}\) is the identity with respect to \(\ominus\) if \(e \ominus b=b\) and \(b \ominus e=b\) for all \(b\in \mathbb{Z}_{12}\text{.}\)

An element \(e\in \mathbb{Z}_{8}\) is the identity with respect to \(\ominus\) if \(e \ominus b=b\) and \(b \ominus e=b\) for all \(b\in \mathbb{Z}_{8}\text{.}\)