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Representations of Cyclic Groups

When dealing with groups, there are many different ways to give a representation of isomorphic groups. In Subgroups of the Permutations on Three Elements we saw that the groups of permutations on three elements was isomorphic to the group of symmetries of an equilateral triangle. Despite being isomorphic, the actual representation were quite different. It is helpful to have some concrete way of representing a group, so that we can spend more time focused on properties of the group instead of the context explaining what the group is. Here, we will focus on giving a familiar representation of all cyclic groups.

Representation of Cyclic Groups

Let \(G\) be a cyclic group. If \(|G|=k\), then \(G \cong \mathbb{Z}_{k}\). Otherwise, if \(G\) is infinite, \(G \cong \mathbb{Z}\).

Video Explanation

Definition of Cyclic Group

We begin by recalling that a cyclic group is a group which is generated by one element. If we call this generator \(a\), we then have that any cyclic group \(G\) can be written as
G=\langle a \rangle
Note that \(a \in G\) by definition and \(a^{-1} \in G\) since \(G\) is closed under inverses. Furthermore, since \(G\) is closed under its operation (we will denote as a product), we get that \(a^{n} \in G\) for all \(n \in \mathbb{Z}\).

Now, let \(A =\left\{a^{n}: n \in \mathbb{Z}\right\}\) and \(c,d \in A\). Then \(c=a^{j}\) and \(d=a^{k}\). Then, note that

  • \(a^{j}a^{k}=a^{j+k} \in A\). So the \(A\) is closed under products.
  • \(a^{j}a^{0}=a^{j}=a^{0}a^{j}\), so \(A\) has an identity.
  • \(a^{k}a^{-k}=a^{0}=a^{-k}a^{k}\), so \((a^{k})^{-1}=a^{-k} \in A\) and \(A\) is closed for inverses.

Hence, \(A\) is a group containing \(a\).

Now, because the group generated by \(a\) is the smallest group containing \(A\), we know that \(A=\langle a \rangle\), that is our cyclic group \(G\) is of the form
G=\langle a \rangle=\left\{a^{n}: n \in \mathbb{Z}\right\}.

Now, we should note that it is not necessarily the case that each of these \(a^{n}\) is distinct. It is very well possible that \(a^{n}=a^{m}\) for some \(m \neq k\). However, each of these elements will be in the cyclic group.

Can Some Be Equal?

If we do fall into the situation where some \(a^{n}=a^{m}\) with \(n \neq m\), we won’t to know what this cyclic group would look like. We would then see that
a^{n}&=a^{m} \\
a^{-n}a^{n}&=a^{-n}a^{m} \\

We, therefore, see that the identity of the group \(a^{0}\) is equal to some \(a^{k}\) with \(k \neq 0\). Furthermore,
a^{j(m-n)}&=(a^{(m-n)})^{j} \\
&=(a^{0})^{j} \\
Hence, \(a^{0}\) is equal to any integer multiple of \(m-n\). This also tells use that, if \(l_{0}=l_{1}+j(m-n)\) for any integer \(j\), then \(a^{l_{0}}=a^{l_{1}}\). That is, if \(l_{0} \equiv l_{1} \text{ mod } m-n\), then \(a^{l_{0}}=a^{l_{1}}\).

Now, going a step further we can say that \(a^{0}=a^{k}\) for some \(k >0\). We can also assume that this \(k\) is the smallest such \(k\) that this happens for. If there wasn’t a smallest such \(k\) it would imply that there are infinitely many integers smaller than \(m-n\) which would be a contradiction. Now, since \(a^{0}=a^{k}\) we know, from above that \(a^{l_{0}}=a^{l_{1}}\) if \(l_{0} \equiv l_{1} \text{ mod } k\).

Now, suppose that \(a^{l_{0}} \not\equiv a^{l_{1}} \text{ mod } k\). Then, we get that
for any \(j \in \mathbb{Z}\). We can also choose \(j\) so that \(0 < l_{0}-l_{1}+jk < k\) with \(0\) not being an option because the two numbers are not congruent mod \(k\). Hence, if \begin{align*} a^{l_{0}}&=a^{l_{1}}, \text{ then} \\ a^{l_{0}}a^{-l_{1}}&=a_{0} \\ a^{l_{0}-l_{1}+jk}&=a_{0}. \end{align*} This would imply that \(k\) was not the smallest integer with \(a^{0}=a^{k}\), which is a contradiction. Hence \(a^{l_{0}}=a^{l_{1}}\) if and only if \(l_{0} \equiv l_{1} \text{ mod } k\). Therefore, this implies that \(|G|=k\) if and only if \(k\) is the smallest positive integer such that \(a^{0}=a^{k}\).


Now that we know what a cyclic group looks like, we are ready to show our theorem that any cyclic group must be isomorphic to \(\mathbb{Z}\) or \(\mathbb{Z}_{k}\) for some \(k \in \mathbb{Z}\). In order to do this, we will need to find an isomorphism between these sets. As we do this, if we recall that all elements of \(\langle a \rangle\) are of the form \(a^{n}\) for some \(n \in \mathbb{Z}\), the natural decision for a mapping will be to send each integer \(n\) to the element \(a^{n}\). However, we will need to be careful about the size of \(G\) in order to ensure we have a one-to-one and onto mapping. Since two groups must have the same cardinality to be isomorphic, we will want to use this mapping from \(\mathbb{Z}_{k} \to G\) if \(|G|=k\) and \(\mathbb{Z} \to G\) if the \(G\) is infinite.

\(G=\left\{a^{n}: n \in \mathbb{Z}\right\}\)

Since we have worked through this above, we will give a formal proof.


Let \(G=\langle a\rangle\), that is \(G\) is the smallest group containing \(a\). First note that \(a^{-1} \in G\), since \(G\) is closed for inverses. Furthermore, \(a^{k}\) and \((a^{-1})^{k}=a^{-k}\) are in \(G\) since \(G\) is closed under multiplication. Finally \(a^{0}=a^{-1}a^{1} \in G\) since \(G\) is closed under multiplication. Hence \(\left\{a^{n}: n \in \mathbb{Z}\right\} \subseteq G\).

Now, let \(A=\left\{a^{n}: n \in \mathbb{Z}\right\}\). Then note that \(a^{n}a^{m}=a^{n+m} \in A\) since \(n+m \in \mathbb{Z}\). Furthermore, \(a^{0} \in A\) and \(a^{-n}=(a^{n})^{-1} \in A\) since \(0,-n \in \mathbb{Z}\). Hence \(A\) is closed under products, identity, and inverses, so \(A\) is a group containing \(a\). Therefore, \(G \subset A\), giving that \(G=\left\{a^{n}:n \in \mathbb{Z}\right\}\)>

\(G\) is infinite

Suppose that \(G\) is infinite. Then, we know that \(a^{n} \neq a^{m}\) for any \(n,m \in \mathbb{Z}\) with \(n \neq m\). If any two such elements were equal, then we would have a finite group, as shown above. Therefore, define \(\alpha: \mathbb{Z} \to G\) as given by \(\alpha(n)=a^{n}\).
Then, we will show that \(\alpha\) is an isomorphism.

First, let \(n \in \mathbb{Z}\), we note that \(a^{n} \in G\), since \(a^{n} \in G\) for all \(n\). Therefore, \(\alpha\) is well-defined. Also, let \(a^{n} \in G\), then \(n \in \mathbb{Z}\), so there exists an \(n \in \mathbb{Z}\) such that \(\alpha (n)=a^{n}\). Hence, \(\alpha\) is onto.

Now, in order to show that \(\alpha\) is one-to-one, we assume that \(\alpha(n)=\alpha(m)\). This means that \(a^{n}=a^{m}\). We know that these can only be equal if \(n=m\), otherwise, we have a contradiction with that fact that \(G\) was infinite.


Let \(G\) be an infinite cyclic group. Then, let \(\alpha: \mathbb{Z} \to G\) be given by \(\alpha(n)=a^{n}\).

We first note that, since \(G\) is closed under inverses, \(a^{-1} \in G\). Then, since \(G\) is closed under products, \(a^{n} \in G\) for all \(n \in G\). Hence, for each \(n \in \mathbb{Z}\), \(a^{n} \in G\), so \(\alpha\) is well-defined.

Next, note that \(b \in G\), then \(b=a^{n}\) for some \(n \in \mathbb{Z}\). Therefore, \(\alpha(n)=a^{n}=b\), so \(\alpha\) is onto.

Now, let \(\alpha(n)=\alpha(m)\), then \(a^{n}=a^{m}\). We, therefore, find that \(a^{n-m}=a^{n}a^{-m}=a^{0}\). This implies that \(a^{l}=a^{l+j(m-n)}\). If \(m-n \neq 0\), there exists a \(j\) such that \(0 \leq l+j(m-n) < m-n\) by the division theorem. Hence \(|G| < m-n\) which is a contradiction to the fact that \(G\) is infinite. Hence \(m-n=0\), so \(m=n\) and \(\alpha\) is one-to-one. Finally note that \begin{align*} \alpha(n+m)&=a^{n+m} \\ &=a^{n}a^{m} \\ &=\alpha(n)\alpha(m). \end{align*} Therefore, \(\alpha\) is a homomorphism and thus is an isomorphism. Hence, every infinite cyclic group is isomorphic to \(\mathbb{Z}\).


The proof is similar to that of the infinite case, so we will give the formal proof.


Let \(G\) be a cyclic group with \(|G|=k\). We then define \(\alpha: \mathbb{Z}_{k} \to G\) as given by \(\alpha(n)=a^{n}\).

Now, since \(a^{n} \in G\) for all \(n \in \mathbb{Z}\), we have that \(\alpha\) is well-defined.

Next, suppose that \(b \in G\). Then \(b=a^{n}\) for some \(n \in \mathbb{Z}\). Now, since \(G\) is not infinite, we may assume that \(a^{j}=a^{m}\) for some \(j, m \in \mathbb{Z}\). Therefore, \(a^{j-m}=a^{0}\) or, if \(m > j\), then \(a^{m-j}=a^{0}\). Hence, \(a^{0}=a^{l}\) for some integer \(l > 0\). There must, therefore, be a smallest integer \(k > 0\) such that \(a^{k}=a^{0}\).

Now, let \(n \equiv l \text{ mod } k\). Then \(n=l+hk\) for some \(h \in \mathbb{Z}\). This gives us that
a^{n}&=a^{l+hk} \\
&=a^{l}a^{hk} \\
&=a^{l}(a^{k})^{h} \\
&=a^{l}(a^{0})^{h} \\
Hence, \(a^{n}=a^{l}\) if \(n \equiv l \text{ mod } k\).

Furthermore, by the division algorithm, for every \(n \in \mathbb{Z}\) there exists a \(l \in \mathbb{Z}\) such that \(b \equiv l \text{ mod } k\) and \(0 \leq l < k\). Hence \(\alpha(l)=a^{l}=a^{n}\), so \(\alpha \) is onto \(G\). Next, suppose that \(\alpha(n)=\alpha(m)\). We may assume that \(n \geq m\). Then \(a^{n}=a^{m}\). This implies that \(a^{n-m}=a^{0}\). Now suppose that \(n \not\equiv m \text{ mod } k\), where \(k\) is as given above, the smallest integer such that \(a^{0}=a^{k}\). Then, by the division algorithm there exists an \(l\) such that \(l \equiv n-m \text{ mod } k\) with \(0 \leq n-m < k\). This implies that for some \(h \in \mathbb{Z}\), we have that \begin{align*} a^{0}&=a^{m-n} \\ &=a^{l+hk} \\ &=a^{l}a^{hk} \\ &=a^{l}(a^{k})^{h} \\ &=a^{l}(a^{0})^{h}\\ &=a^{l}. \end{align*} Since \(l \leq k\) and \(k\) is the smallest positive integer with \(a^{0}=a^{k}\), this implies \(l=0\). Hence, \(n \equiv m \text{ mod } k\), so \(\alpha\) is one-to-one. Finally note that \begin{align*} \alpha(n+m)&=a^{n+m} \\ &=a^{l} \text{ where } l \equiv n+m \text{ mod } k \\ &=\alpha(l) \\ &\equiv \alpha(n)\alpha(m) \text{ mod } k. \end{align*} Therefore, \(\alpha\) is a homomorphism and therefore an isomorphism. Hence \(\mathbb{Z}_{k} \cong G\) if \(G\) is a cyclic group with order \(k\).


We have now given a concrete representation of any cyclic group. This means that instead of dealing with an arbitrary cyclic group, we can work with the groups of the form \(\mathbb{Z}_{k}\) or \(\mathbb{Z}\) to find properties of any cyclic group.

Thank you for reading and, if you found this helpful, please let me know by liking the post and sharing it with others. If you have any questions, feel free to leave a comment below.

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