📂Abstract Algebra

Prove that a Subgroup of a Cyclic group is Cyclic

Definition ¹

A subgroup $ H \leqslant G$ of a cyclic group $G$ is a cyclic group.

Explanation

Although it might seem obvious upon a bit of thought, it is a considerably important theorem and its proof is not as straightforward as it might appear.

Proof

If $H = \left\{ e \right\}$, then $H = \left< e \right>$ hence it is a cyclic group.

If $H \ne \left\{ e \right\}$, for some natural number $n$, $a^{n} \in H$ will hold and let’s denote the smallest natural number that satisfies this as $m$. When $c := a^m$, showing that $H = \left< a^m \right> = \left< c \right>$ holds ends the proof.

For all $b \in H$, $b = a^{n} \in G$ will hold, and for some $q , r \in \mathbb{N}$, $n = m q + r$ will be true. Here, if we set $ 0 \le r < m$, $q$ and $r$ are uniquely determined. $$ a^{n} = a^{mq + r} = (a^{m})^{q} a^{r} $$ and organizing it for $a^{r}$ gives $$ a^{r} = (a^{m})^{-q} a^{n} \\ a^{n} = b \in H $$ and if $a^{m} \in H$ while $H$ is also a group, then $(a^{m})^{-q}$ and $a^{n}$ are included in $H$. Hence $$ (a^{m})^{-q} a^{n} = a^{r} \in H $$ On the other hand, since $m$ was the smallest natural number satisfying $a^{m} \in H$, the only case that satisfies $0 \le r < m$ for all cases is $r = 0$. Eventually, it must be $r = 0$, $$ b = a^{n} = a^{mq} = (a^{m})^{q} = c^{q} $$ Since every element can be expressed as a power of $c$, $H = \left< c \right>$ is a cyclic group.

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