2.6: Examples of Groups/Nongroups, Part II

Example \(\PageIndex{1}\)

Let \(n\in \mathbb{Z}^+\text{.}\) We define \(n\mathbb{Z}\) by

\begin{equation*} n\mathbb{Z}=\{nx: x\in \mathbb{Z}\}: \end{equation*}

that is, \(n\mathbb{Z}\) is the set of all (integer) multiples of \(n\text{.}\)

Theorem \(\PageIndex{1}\)

\(n\mathbb{Z}\) is a group under \(+\) (the usual addition of integers).

Proof

Let \(x,y∈n\mathbb{Z}\). Then there exist \(a,b∈\mathbb{Z}\) such that \(x=na\) and \(y=nb\). Then \(x+y=na+nb=n(a+b)∈n\mathbb{Z}\). So \(\langle n\mathbb{Z},+ \rangle\) \(\langle n\mathbb{Z},+ \rangle\) is a binary structure. The remainder of the proof is left as an exercise for the reader.

Remark

When we are discussing a group \(n\mathbb{Z}\text{,}\) assume that \(n\in \mathbb{Z}^+\text{,}\) unless otherwise noted.

We use an example from our next class of groups all the time; in fact, most six-year-olds do as well, since it is used when telling time! Before we get to the example, we need some more definitions and some notation. Throughout the following discussion, assume \(n\) is a fixed positive integer.

Definition: Congruent modulo (mod)

We say integers \(a\) and \(b\) are congruent modulo [or mod] \(n\) if \(n\) divides \(a-b\text{.}\) If \(a\) and \(b\) are congruent mod \(n\text{,}\) we write \(a \equiv_n b\text{.}\)

Example \(\PageIndex{2}\)

\(1, 7, 13,\) and \(-5\) are all congruent mod \(6\text{.}\)

The following is a profoundly useful theorem; it's so important, it has a special name. We omit the proof of this theorem, but direct interested readers to for, instance, p. 5 in [3].

Theorem \(\PageIndex{2}\): Division Algorithm

Let \(n\in \mathbb{Z}^+\) and let \(a\) be any integer. Then there exist unique integers \(q\) and \(r\text{,}\) with \(0\leq r \lt n\text{,}\) such that \(a=qn+r\text{.}\)

(This is actually a special case of a more general theorem, which states that given any integers \(n\) and \(a\text{,}\) there exist unique integers \(q\) and \(r\text{,}\) with \(0\leq r\lt |n|\text{,}\) such that \(a=qn+r\text{.}\))

It follows that for each positive integer \(n\) and integer \(a\text{,}\) there exists a unique element \(R_n(a)\) (the \(r\) in the above theorem) of the set \(\{0,1,2,\ldots, n-1\}\) such that \(a\) is congruent to \(R_n(a)\) modulo \(n\text{.}\) For example, \(R_3(4)=1\text{,}\) \(R_3(0)=0\text{,}\) \(R_3(17)=2\text{,}\) and \(R_3(-5)=1\text{.}\)

Definition: Remainder

\(R_n(a)\) is the remainder when we divide \(a\) by \(n\text{.}\)

(Note: You were probably already familiar with the remainder when you divide a positive integer by \(n\text{.}\))

Definition: Addition modulo

We define addition modulo \(n\), \(+_n\text{,}\) on \(\mathbb{Z}\) by, for all \(a,b\in \mathbb{Z}\text{,}\)

\begin{equation*} a+_n b=R_n(a+b), \end{equation*}

that is, the unique element of \(\{0,1,\ldots, n-1\}\) that's congruent to the integer \(a+b\) modulo \(n\text{.}\)

Remark

Addition mod \(24\) is what we use to tell time!

The set \(\{0,1,2,\ldots, n-1\}\) of remainders when dividing by \(n\) is so important we give it a special notation.

Definition

We define \(\mathbb{Z}_n\) to be the set \(\{0,1,2,\ldots,n-1\}\text{.}\)

Note

Note that by our definition of \(\mathbb{Z}_n\text{,}\) the integer \(n\) itself is not in \(\mathbb{Z}_n\text{!}\)

We are now ready to consider our next type of group.

Example \(\PageIndex{3}\)

For each \(n\in \mathbb{Z}^+\text{,}\) \(\langle \mathbb{Z}_n,+_n\rangle\) is a group, called the cyclic group of order \(n\) (we will see later why we use the word “cyclic” here). This group is abelian and of order \(n\text{.}\)

Proof

We first check that \(\langle \mathbb{Z}n,+_n\rangle\) is a binary structure. Note that by the definition of \(+_n\), \(a+_nb∈\mathbb{Z}_n\) for each \(a,b∈Z\). Thus, \(a+_nb∈\mathbb{Z}_n\) for each \(a,b∈\mathbb{Z}_n\).

We next check that \(\mathbb{Z}_n\) under \(+_n\) satisfies the three group axioms. Note that since addition is commutative on \(\mathbb{Z}\),

\(a+_nb=R_n(a+b)=R_n(b+a)=b+_na\)

for all \(a,b∈\mathbb{Z}_n\). Again, a simpler way of stating this is that commutativity of \(+_n\) on \(\mathbb{Z}_n\) is inherited from the commutativity of addition on \(\mathbb{Z}\). One nice result of this is that since \(+_n\) is commutative on \(\mathbb{Z}_n\), we have less to check when verifying group axioms \(\mathcal{G}_2\) and \(\mathcal{G}_3\).

Now, let \(a,b,c∈\mathbb{Z}_n\). We want to show that \((a+_nb)+_nc=a+_n(b+_nc)\). Now,

\(\begin{array}& (a+_nb)+_nc &=R_n(a+b)+_nc \\ &≡_nR_n(a+b)+c \\ &≡_n(a+b)+c \\ &≡_na+(b+c)\\ &≡_na+R_n(b+c)\\ &≡_na+_n(b+_nc).\end{array} \)

So \((a+_nb)+_nc\) and \(a+_n(b+_nc)\) are congruent mod \(n\). Since both of these values are in \(\{0,1,…,n−1\}\), this implies that they are equal, as desired.

Next, clearly, \(0∈\mathbb{Z}_n\) acts as an identity element under \(+_n\), since

\(0+_na=R_n(0+a)=R_n(a)\)

for each \(a∈\mathbb{Z}_n\).

Finally, let \(a∈\mathbb{Z}_n\). If \(a=0\), then clearly aa has inverse \(0∈\mathbb{Z}_n\) since \(0+n_0=0\). If \(a≠0\), then the element \(n−a∈\mathbb{Z}_n\) is an inverse for aa since

\(a+_n(n−a)=R_n(a+(n−a))=R_n(n)=0.\)

Since \(+_n\) is commutative on \(\mathbb{Z}_n\), \(\mathbb{Z}_n\) is an abelian group under \(+_n\).

Finally, we already know that \(|\mathbb{Z}_n|=|\{0,1,2,…,n−1\}|=n\).

Remark

In practice, we often omit the subscript \(n\) and just write \(+\) when discussing addition modulo \(n\) on \(\mathbb{Z}_n\text{.}\)

We first check that \(\langle\Z_n,+_n\rangle\) is a binary structure. Note that by the definition of \(+_n\text{,}\) \(a+_nb \in \Z_n\) for each \(a,b\in \Z\text{.}\) Thus, \(a+_nb \in \Z_n\) for each \(a,b\in \Z_n\text{.}\)

We next check that \(\Z_n\) under \(+_n\) satisfies the three group axioms. Note that since addition is commutative on \(\Z\text{,}\)

\begin{equation*} a+_n b =R_n(a+b)=R_n(b+a)=b+_n a \end{equation*}

for all \(a,b\in \Z_n\text{.}\) Again, a simpler way of stating this is that commutativity of \(+_n\) on \(\Z_n\) is inherited from the commutativity of addition on \(\Z\text{.}\) One nice result of this is that since \(+_n\) is commutative on \(\Z_n\text{,}\) we have less to check when verifying group axioms \(\G_2\) and \(\G_3\text{.}\)

Now, let \(a,b,c\in \Z_n\text{.}\) We want to show that \((a+_n b)+_n c = a +_n(b+_n c)\text{.}\) Now,

\begin{align*} (a+_n b)+_n c\amp =R_n(a+b)+_n c\\ \amp \equiv_n R_n(a+b)+c\\\ \amp \equiv_n (a+b)+c\\\ \amp \equiv_n a+(b+c)\\ \amp \equiv_n a+R_n(b+c)\\ \amp \equiv_n a+_n (b+_n c). \end{align*}

So \((a+_n b)+_n c\) and \(a+_n (b+_n c)\) are congruent mod \(n\text{.}\) Since both of these values are in \(\{0,1,\ldots, n-1\}\text{,}\) this implies that they are equal, as desired.

Next, clearly, \(0\in \Z_n\) acts as an identity element under \(+_n\text{,}\) since

\begin{equation*} 0+_n a =R_n(0+a)=R_n(a) \end{equation*}

for each \(a\in \Z_n\text{.}\)

Finally, let \(a\in \Z_n\text{.}\) If \(a=0\text{,}\) then clearly \(a\) has inverse \(0\in \Z_n\) since \(0+_n 0 = 0\text{.}\) If \(a\neq 0\text{,}\) then the element \(n-a\in \Z_n\) is an inverse for \(a\) since

\begin{equation*} a+_n(n-a)=R_n(a+(n-a))=R_n(n)=0. \end{equation*}

Since \(+_n\) is commutative on \(\Z_n\text{,}\) \(\Z_n\) is an abelian group under \(+_n\text{.}\)

Finally, we already know that \(|\Z_n|=|\{0,1,2,\ldots,n-1\}|=n\text{.}\)

Note

Do not confuse \(n\mathbb{Z}\) and \(\mathbb{Z}_n\text{!}\) They are very different as sets and as groups.

Example \(\PageIndex{4}\)

In the group \(\langle \mathbb{Z}_8,+\rangle\) (where, as indicated by our above remark, \(+\) means addition modulo \(8\)), we have, for instance, \(3+7=2\) and \(7+7=6\text{.}\) The numbers 2 and 6 are each other's inverse, and \(7^{-1}=1\text{.}\) The number \(0\) has inverse \(0\) (it can't be \(8\text{,}\) since \(8\not\in \mathbb{Z}_8\text{!}\)).

Definition: Multiplication modulo

For \(n\in \mathbb{Z}^+\text{,}\) we define multiplication modulo \(n\), denoted \(\cdot_n\text{,}\) on \(\mathbb{Z}_n\) by \(a\cdot_n b = R_n(ab)\text{,}\) the remainder when \(ab\) is divided by \(n\text{.}\)

Remark

\(\mathbb{Z}_n\) is never a group under \(\cdot_n\) (do you see why?).

But we can consider the following

Definition

For \(n\in \mathbb{Z}^+\text{,}\) we define \(\mathbb{Z}_n^{\times}\) to be the set

\begin{equation*} \{a\in \mathbb{Z}_n\,:\,\gcd(a,n)=1\}\text{.} \end{equation*}

Example \(\PageIndex{5}\)

\(\langle \mathbb{Z}_n^{\times},\,\cdot_n\, \rangle\) is a group under multiplication. We omit the proof.

We end this section by considering a few more examples. 

Example \(\PageIndex{6}\)

Let \(F\) be the set of all functions from \(\mathbb{R}\) to \(\mathbb{R}\text{,}\) and define pointwise addition \(+\) on \(F\) by

\begin{equation*} (f+g)(x)=f(x)+g(x) \end{equation*}

for all \(f,g\in F\) and \(x\in \mathbb{R}\text{.}\) We claim that \(F\) is a group under pointwise addition. (For variety, in this proof we don't explicitly refer to \(\mathcal{G}_1\)–\(\mathcal{G}_3\text{,}\) though we certainly do verify they hold.)

Proof

If \(f,g∈F\) then clearly \(f+g\) is also a function from \(\mathbb{R}\) to \(\mathbb{R}\), so \(\langle F,+ \rangle\) is a binary structure.

Next, let \(f,g,h∈F\). Then for all \(x∈\mathbb{R}\),

\( \begin{array} &((f+g)+h)(x) &=(f+g)(x)+h(x)& \\&=((f(x)+g(x))+h(x)&\\ &=f(x)+(g(x)+h(x)) &\text{(since addition is associative on \(\mathbb{R}\))} \\ &=f(x)+(g+h)(x)& \\&=(f+(g+h))(x).& \end{array}\)

Note that the key fact used in this argument is that \(f(x)\), \(g(x)\) and \(h(x)\) all lie in \(\mathbb{R}\), and addition is associative on \(\mathbb{R}\). When you get used to such arguments, it is sufficient to say that associativity of \(+\) on \(F\) is inherited from the associativity of addition on \(\mathbb{R}\).

Next, let \(z:\mathbb{R}→\mathbb{R}\) be the function \(z(x)=0\) for all \(x\). Then for all \(f∈F\) and \(x∈R\),

\(\begin{array}& (f+z)(x) &=f(x)+z(x)\\ &=f(x)+0\\ &=f(x)\\ &=0+f(x)\\ &=z(x)+f(x)\\ &=(z+f)(x). \end{array}\)

So \(z\) is an identity element of \(\langle F,+ \rangle\).

Finally, let \(f∈F\), and define \(g∈F\) by \(g(x)=−f(x)\) for all \(x∈\mathbb{R}\). It is easy then to see that \(g\) is an inverse for \(f\) in \(F\).

Hence, \(F\) is a group under pointwise addition. Note that it is uncountably infinite and abelian.

Example \(\PageIndex{7}\)

The set \(F\) is not a group under function composition (do you see why?). But if we define \(B\) to be the set of all bijections from \(\mathbb{R}\) to \(\mathbb{R}\text{,}\) then \(B\) is a group under function composition. (Prove it!) \(B\) is uncountably infinite and nonabelian.

Example \(\PageIndex{8}\)

Let \(\langle G_1,*_1\rangle\text{,}\) \(\langle G_2,*_2\rangle\) , \(\ldots\text{,}\) \(\langle G_n,*_n\rangle\) be groups (\(n\in \mathbb{Z}^+\)). Then the group product

\begin{equation*} G=G_1\times G_2\times \cdots \times G_n \end{equation*}

is a group under the componentwise operation \(*\) defined by

\begin{equation*} (g_1,g_2,\ldots, g_n)*(h_1,h_2,\ldots,h_n)=(g_1*_1h_1, g_2*_2h_2,\ldots, g_n*_nh_n) \end{equation*}

for all \((g_1,g_2,\ldots, g_n),(h_1,h_2,\ldots,h_n)\in G\text{.}\)

For instance, considering multiplication on \(\mathbb{R}^*\text{,}\) matrix multiplication on \(GL(2,\mathbb{R})\text{,}\) and addition modulo \(6\) on \(\mathbb{Z}_6\text{,}\) we have that \(\langle \mathbb{R}^*\times GL(2,\mathbb{R}) \times \mathbb{Z}_6,*\rangle\) is a group in which, for instance,

\begin{equation*} \left(-1, \begin{bmatrix} 1 &3 \\ 0 &-1 \end{bmatrix}, 3\right) *\left(\pi, \begin{bmatrix} 2 &1 \\ 1 &1 \end{bmatrix}, 4\right)=\left(-\pi, \begin{bmatrix} 5 &4 \\ -1 &-1 \end{bmatrix} ,1\right). \end{equation*}

If \(f,g\in F\) then clearly \(f+g\) is also a function from \(\R\) to \(\R\text{,}\) so \(\langle F,+\rangle\) is a binary structure.

Next, let \(f,g,h\in F\text{.}\) Then for all \(x\in \R\text{,}\)

\begin{align*} ((f+g)+h)(x)\amp =(f+g)(x)+h(x)\amp \amp\\ \amp =((f(x)+g(x))+h(x)\amp \amp\\ \amp =f(x)+(g(x)+h(x))\amp \amp \text{ (since addition is associative on \(\R\)) }\\ \amp =f(x)+(g+h)(x)\amp \amp\\ \amp =(f+(g+h))(x)\amp \amp . \end{align*}

Note that the key fact used in this argument is that \(f(x)\text{,}\) \(g(x)\) and \(h(x)\) all lie in \(\R\text{,}\) and addition is associative on \(\R\text{.}\) When you get used to such arguments, it is sufficient to say that associativity of \(+\) on \(F\) is inherited from the associativity of addition on \(\R\text{.}\)

Next, let \(z:\R\to\R\) be the function \(z(x)=0\) for all \(x\text{.}\) Then for all \(f\in F\) and \(x\in \R\text{,}\)

\begin{align*} (f+z)(x)\amp =f(x)+z(x)\amp \amp\\ \amp =f(x)+0\amp \amp\\ \amp =f(x)\amp \amp \\ \amp =0+f(x)\amp \amp\\ \amp =z(x)+f(x)\amp \amp\\ \amp = (z+f)(x). \end{align*}

So \(z\) is an identity element of \(\langle F,+\rangle\text{.}\)

Finally, let \(f\in F\text{,}\) and define \(g\in F\) by \(g(x)=-f(x)\) for all \(x\in \R\text{.}\) It is easy then to see that \(g\) is an inverse for \(f\) in \(F\text{.}\)

Hence, \(F\) is a group under pointwise addition. Note that it is uncountably infinite and abelian.

Example \(\PageIndex{9}\)

A common example of a group product is the group \(\mathbb{Z}_2^2\text{,}\) equipped with componentwise addition modulo \(2\).

Definition: Klein-4-group

The group \(\mathbb{Z}_2^2\) is known as the Klein 4-group. (Felix Klein was a German mathematician; you may have heard of him in relation to the Klein Bottle.) The group \(\mathbb{Z}_2^2\) is sometimes denoted by \(V\text{,}\) which stands for “Vierergruppe,” the German word for “four-group”.