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Section5.1Introduction to cyclic groups

Certain groups and subgroups of groups have particularly nice structures.


A group is cyclic if it is isomorphic to \(\Z_n\) for some \(n\geq 1\text{,}\) or if it is isomorphic to \(\Z\text{.}\)


Examples/nonexamples of cyclic groups.

  1. \(n\Z\) and \(\Z_n\) are cyclic for every \(n\in \Z^+\text{.}\)

  2. \(\R\text{,}\) \(\R^*\text{,}\) \(\M_2(\R)\text{,}\) and \(GL(2,\R)\) are uncountable and hence can't be cyclic.

  3. \(\Z_2^2\) is not cyclic since it would have to be isomorphic to \(\Z_4\) if it were (since it has order 4).

  4. \(\Q\) isn't cyclic. If it were cyclic it would have to be isomorphic to \(\Z\text{,}\) since \(\Q\) is an infinite group (so can't be isomorphic to \(\Z_n\) for any \(n\)). But we showed in Example 3.3.14 that \(\Q \not\simeq \Z\text{.}\)


Clearly, cyclicity is a group invariant.


Let \(a\) be an element of a group \(G\text{.}\) Then

\begin{equation*} \langle a\rangle = \{a^n:n\in \Z\} \end{equation*}

is called the (cyclic) subgroup of \(G\) generated by \(a\). (We will later see why the words “cyclic” and “generated” are used here.)


Note that in the above definition, we were using multiplicative notation. Using additive notation, we have \(\langle a\rangle =\{na:n\in \Z\}=\{\ldots, -2a, -a, 0a, 1a, 2a, \ldots\}.\)


We next prove a lemma to which we will refer when proving the theorem which succeeds it.


Let \(G\) be a group and let \(a\in G\text{.}\) We define the order of \(a\text{,}\) denoted \(o(a)\text{,}\) to be \(|\langle a\rangle |\text{.}\) (Note: If there exists a positive integer \(k\) such that \(a^k=e\text{,}\) then the least such integer is the order of \(a\text{;}\) otherwise, \(o(a)=\infty\text{.}\))


Do not confuse the order of a group with the order of an element of a group. These are related concepts, but they are distinct, and have distinct notations: as we've seen, the order of a group, \(G\text{,}\) is denoted by \(|G|\text{,}\) while the order of an element \(a\) of a group is denoted by \(o(a)\text{.}\)


We have the following handy theorem.


Unfortunately, there's no formula one can simply use to compute the order of an element in an arbitrary group. However, in the special case that the group is cyclic of order \(n\text{,}\) we do have such a formula. We present the following result without proof.


Here are some examples of cyclic subgroups of groups, and orders of group elements.


  1. In \(\Z\text{,}\) \(\langle 2\rangle =\{\ldots,-4,-2,0,2,4,\ldots\}=2\Z\text{.}\) More generally, given any \(n\in \Z\text{,}\) in \(\Z\) we have \(\langle n\rangle =n\Z\text{.}\) For \(a\in \Z\text{,}\) \(o(a)=\infty\) if \(a\neq 0\text{;}\) \(o(0)=1\text{.}\)

  2. In \(\Z_8\text{,}\) we have \(\langle 0\rangle =\{0\}\text{,}\) \(\langle 1\rangle =\langle 3\rangle =\langle 5\rangle =\langle 7\rangle =\Z_8,\) \(\langle 2\rangle =\langle 6\rangle =\{0,2,4,6\},\) and \(\langle 4\rangle =\{0,4\}\text{.}\)

  3. In \(\R\text{,}\) \(\langle \pi\rangle =\pi\Z\text{,}\) so \(o(\pi)=\infty\text{.}\)

  4. In \(\R^*\text{,}\) \(\langle \pi\rangle =\{\pi^n:n\in \Z\}\text{.}\) Again, \(o(\pi)=\infty\text{.}\)

  5. In \(\M_2(\R)\text{,}\)

    \begin{equation*} \left\langle \begin{bmatrix} 1 \amp 1 \\ 0 \amp 1 \end{bmatrix}\right\rangle =\left\{ \begin{bmatrix} c \amp c \\ 0 \amp c \end{bmatrix} : c\in \Z \right\}= \left\{c \begin{bmatrix} 1 \amp 1 \\ 0 \amp 1 \end{bmatrix} : c\in \Z \right\}. \end{equation*}

    The matrix therefore has infinite order.

  6. In \(\M_2(\Z_2)\text{,}\) if \(A= \begin{bmatrix} 1 \amp 1 \\ 0 \amp 1 \end{bmatrix} \text{,}\) then \(\langle A\rangle =\left\{A, \begin{bmatrix}0\amp 0 \\ 0\amp 0\end{bmatrix}\right\}\text{,}\) so \(A\) has order 2.

  7. In \(GL(2,\Z_2)\text{,}\) \(\begin{bmatrix} 1 \amp 1 \\ 0 \amp 1 \end{bmatrix} \) has order 2. (Why?)

  8. In \(GL(2,\R)\text{,}\) \(\begin{bmatrix} 0 \amp -1 \\ 1 \amp \phantom{-}0 \end{bmatrix} \) has order 4. (Why?)

  9. In \(GL(2,\R)\text{,}\) \(\begin{bmatrix} 1 \amp 1 \\ 0 \amp 1 \end{bmatrix} \) has infinite order. (Why?)

  10. In \(GL(4,\Z_2)\text{,}\)\(A=\ \begin{bmatrix} 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \end{bmatrix} \) has order 2. (Why?)


Let \(G\) be a group. An element \(a\in G\) is a generator of \(G\) (equivalently, \(a\) generates \(G\)) if \(\langle a\rangle =G\text{.}\)


  1. Note that if \(G\) has a generator, then it is necessarily a cyclic group.

  2. Note that an element \(a\) of a group \(G\) generates \(G\) if and only if every element of \(G\) is of the form \(a^n\) for some \(n\in \Z\text{.}\)

  3. Generators of groups need not be unique. For instance, we saw in Example 5.1.17 that each of the elements \(1,3,5\) and \(7\) of \(\Z_8\) is a generator for \(\Z_8\text{.}\)


  1. \(\Z\) has generator \(1\) and generator \(-1\text{.}\)

  2. Given \(n\in \Z\text{,}\) \(nZ\) has generator \(n\) and generator \(-n\text{.}\)

  3. Given \(n\geq 2\) in \(\Z\text{,}\) the generators of \(\Z_n\) are exactly the elements \(a\in \Z_n\) such that \(\gcd(n,a)=1\text{.}\) (This follows from Theorem 5.1.14.)


Order of elements provides another group invariant.


Since \(1\) in \(\Z_4\) has order 4 but every element in \(\Z_2 \times \Z_2\) has order less than or equal to 2, these groups cannot be isomorphic.


Note, however, that just because the orders of elements of two groups “match up” the groups need not be isomorphic. For example, every element of \(\Z\) has infinite order, except for its identity element, which has order 1; the same is true for the group \(\Q\text{.}\) However, we have previously proven that these groups are not isomorphic.