An identity element does not change another element under the binary operation. Applying the operation to the identity element and another element gives that other element.
The video in Figure 13.3.1 defines an identity element for a binary operation. It also gives examples. After the video, we give the formal definition and more examples. Then we give methods for deciding whether a binary operation has an identity element.
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 \(S\) for the operation \(\bullet\) if \(s\bullet e=s\) and \(e\bullet s=s\) for all \(s\in S\text{.}\)
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 \(-\text{.}\)
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 \(S\) for the operation \(\bullet\text{,}\) then it is unique.
Suppose that there are two identity elements \(e\) and \(f\) of the set \(S\) for the operation \(\bullet\text{.}\) Since \(e\) is an identity element, \(s\bullet e=s\) for all \(s \in S\text{.}\) So, in particular, \(f \bullet e = f\text{.}\) But 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 \(f = e\text{.}\) So \(f\) and \(e\) must be the same element. So, there is at most one identity element of \(S\) for \(\bullet\text{.}\)
Example13.3.5.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 13.1.4. 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 \(T\) for the operation \(\star\text{.}\)
When an operation is given by a table, a special row and column show the identity element, if one exists. The row must match the header row at the top of the table. The column must match the header column on the left side of the table.
Example13.3.6.The identity with respect to \(\star:T\times T \to T\).
With the above comment in mind, we revisit Example 13.3.5. Notice that the row matching \(\Ty\) matches the header row at the top of the table and the column matching \(\Ty\) matches the header column on the left side of the table. So, the element \(\Ty\) is the identity element.
Example13.3.7.The identity with respect to \(\oplus:\Z_5\times\Z_5\to\Z_5\).
We consider the binary operation \(\oplus:\Z_5\times\Z_5\to\Z_5\) defined by \(a\oplus b=(a+b)\fmod 5\text{.}\) The identity element for \(\oplus\) is \(0\text{.}\)
Problem13.3.8.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 \(\Z_3\text{.}\)
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 \(\oplus\text{.}\)
