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Section 13.3 Identity

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 13.3.1 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.

Figure 13.3.1. Identity Elements by Matt Farmer and Stephen Steward.

Carefully read the definition.

Definition 13.3.2.

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 13.3.3 reproduce the definition by filling in the blanks.

Fill in the operation table for the binary operation \(\oplus\) on the set \(\mathbb{Z}_{6}\) defined by \(a \oplus b = (a + b)\bmod 6\) :

\(\oplus\) 0 1 2 3 4 5
0 0 1 2 3 5
1 0
2 4 0 1
3 4 5
4 4 3
5 5 1 4

Complete the following:

In \(\mathbb{Z}_{6}\) with respect to \(\oplus\)

  • select

  • the identity element is 0

  • the identity element is 1

  • there is no identity element

.

Find the inverses of the elements of \(\mathbb{Z}_{6}\) with respect to \(\oplus\) . If an element does not have an inverse answer 'none'.

The inverse of 0 is .

The inverse of 1 is .

The inverse of 2 is .

The inverse of 3 is .

The inverse of 4 is .

The inverse of 5 is .

We revisit the binary operations from Example 13.1.3:

  1. 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.

  2. 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.

  3. 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 \(-\text{.}\)

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.

Let \(T=\{\Tx,\Ty,\Tz\}\text{,}\) and let the binary operation \(\star:T\times T\to T\) be given by the table in Example 13.1.4. 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.

With the above comment in mind, we revisit Example 13.3.6. 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.

\(\star\) \(\color{gray}\Tx\) \(\color{gray}\Ty\) \(\color{gray}\Tz\)
\(\Tx\) \(\Tz\) \(\Tx\) \(\Ty\)
\(\color{red}\Ty\) \(\color{gray}\Tx\) \(\color{gray}\Ty\) \(\color{gray}\Tz\)
\(\Tz\) \(\Ty\) \(\Tz\) \(\Tx\)
\(\star\) \(\Tx\) \(\color{red}\Ty\) \(\Tz\)
\(\color{gray}\Tx\) \(\Tz\) \(\color{gray}\Tx\) \(\Ty\)
\(\color{gray}\Ty\) \(\Tx\) \(\color{gray}\Ty\) \(\Tz\)
\(\color{gray}\Tz\) \(\Ty\) \(\color{gray}\Tz\) \(\Tx\)

The identity element with respect to the operation \(\oplus:\Z_5\times\Z_5\to\Z_5\text{,}\) \(a\oplus b=(a+b)\fmod 5\) is 0. To see this we can either use that \(a+0=0+a=0\) for all integers \(a\) or use the method from the previous example and the operation table from Example 13.1.6.

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

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'.

The operation * is commutative 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

.

Answer 1.

\({\text{a * b = b * a}}\)

Answer 2.

\(\text{all a in S and all b in S}\)