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Section2.6Examples of groups/nongroups, Part II


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

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

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


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


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 \(a \equiv_n b\text{.}\)


\(1, 7, 13,\) 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].

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 \(R_3(-5)=1\text{.}\)


\(R_n(a)\) 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 \(n\text{.}\))


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

\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 \(n\text{.}\)


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.


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


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

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


For each \(n\in \Z^+\text{,}\) \(\langle \Z_n,+_n\rangle\) 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 \(n\text{.}\)


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


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


In the group \(\langle \Z_8,+\rangle\) (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 \(8\not\in \Z_8\text{!}\)).


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


\(\Z_n\) is never a group under \(\cdot_n\) (do you see why?).

But we can consider the following


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

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

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

We end this section by considering a few more examples.


Let \(F\) be the set of all functions from \(\R\) to \(\R\text{,}\) 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 \(x\in \R\text{.}\) We claim that \(F\) is a group under pointwise addition. (For variety, in this proof we don't explicitly refer to \(\G_1\)–\(\G_3\text{,}\) though we certainly do verify they hold.)


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 \(\R\) to \(\R\text{,}\) then \(B\) is a group under function composition. (Prove it!) \(B\) is uncountably infinite and nonabelian.


Let \(\langle G_1,*_1\rangle\text{,}\) \(\langle G_2,*_2\rangle\) , \(\ldots\text{,}\) \(\langle G_n,*_n\rangle\) be groups (\(n\in \Z^+\)). 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 \(\R^*\text{,}\) matrix multiplication on \(GL(2,\R)\text{,}\) and addition modulo \(6\) on \(\Z_6\text{,}\) we have that \(\langle \R^*\times GL(2,\R) \times \Z_6,*\rangle\) is a group in which, for instance,

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

A common example of a group product is the group \(\Z_2^2\text{,}\) equipped with componentwise addition modulo 2.


The group \(\Z_2^2\) 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 \(\Z_2^2\) is sometimes denoted by \(V\text{,}\) which stands for “Vierergruppe,” the German word for “four-group”.