Subgroups of Cyclic Groups are Cyclic Groups

Recall from the Cyclic Groups page that a group $G$ is said to be a cyclic group if $G = \langle a \rangle = \{ a^n : n \in \mathbb{Z} \}$ for some $a \in G$, i.e., $G$ can be generated by a single element $a \in G$.

We will now prove an important theorem - every subgroup of a cyclic group is also cyclic.

Theorem 1: Let $G$ be a cyclic group. Then any subgroup $H$ of $G$ is also a cyclic group.

The converse of Theorem 1 is not true in general, i.e., even if every proper subgroup $H$ of a group $G$ is cyclic - it might be the case that $G$ is not cyclic. The simplest example of this is the group $\mathbb{Z}_2 \times \mathbb{Z}_2$. The only proper subgroups of this group are the trivial group and $\mathbb{Z}_2$ - both of which are cyclic. But it is easy to check that $\mathbb{Z}_2 \times \mathbb{Z}_2$ is not cyclic.

• Proof: Let $G$ be a cyclic group. Then there exists an $a \in G$ such that $G = \langle a \rangle$. Let $H$ be a subgroup of $G$.

• If $H$ is the trivial subgroup, then clearly $H = \langle e \rangle$, and so $H$ is a cyclic group.

• If $H$ is not the trivial subgroup then there exists a $b \in H$ such that $b \neq e$. Since $b \in H \subseteq G = \langle a \rangle$, there exists an $n \in \mathbb{Z}$ such that $b = a^n$, and $n \neq 0$. Since $H$ is itself a group and $b \in H$ we have that $b^{-1} = (a^n)^{-1} = a^{-n} \in H$ with $-n \neq 0$.

• So there exists an element $a^k \in H$ with $k > 0$. Consider the set $K = \{ k : a^k \in H, \: k > 0 \}$. This set is a nonempty subset of $\mathbb{N}$, and so by the Well-Ordering principle, it contains a least element, call it $\tilde{k}$.

• Now since $a^{\tilde{k}} \in H$ and $H$ is a group, we have that $\langle a^{\tilde{k}} \rangle \subseteq H$. S let $h \in H$. Since $H \subseteq G$ we have that $h \in G$. So $h = a^m$ for some $m \in \mathbb{Z}$. Using the division algorithm on $m$ and $\tilde{k}$, we see that there exists integers $q, r$ such that:

(1)

\begin{align} \quad m = \tilde{k}q + r \end{align}

• With $0 \leq r < \tilde{k}$. But then:

(2)

\begin{align} \quad h = a^m = a^{\tilde{k}q + r} = a^{\tilde{k}q} a^r = (a^{\tilde{k}})^q a^r = (a^{\tilde{k}})^q a^r \end{align}

• Since $h \in H$ and $(a^{\tilde{k}})^q \in H$ we see that:

(3)

\begin{align} \quad a^r = (a^{\tilde{k}})^{-q} h \in H \end{align}

• If $r > 0$, then $r \in K = \{ k : a^k \in H, k > 0 \}$. But $r < \tilde{k}$, which contradicts the minimality of $\tilde{k}$ in $K$. So we must have that $r = 0$. So:

(4)

\begin{align} \quad a^0 &= (a^{\tilde{k}})^{-q} h \\ \quad e &= (a^{\tilde{k}})^{-q} h \\ \quad h &= (a^{\tilde{k}})^q \end{align}

• Thus $h \in \langle a^{\tilde{k}}$, showing that $H \subseteq \langle a^{\tilde{k}}$. Thus we conclude that $H = \langle a^{\tilde{k}}$, i.e., $H$ is a cyclic group. $\blacksquare$