📂Abstract Algebra

Proof 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

With just a little thought this is an obvious fact, but not only is it a fairly important theorem, its proof is not entirely simple either.

Proof

When $H = \left\{ e \right\}$, we have $H = \left< e \right>$, so it is a cyclic group.

When $H \ne \left\{ e \right\}$, for some natural number $n$ we have $a^{n} \in H$; let $m$ be the smallest natural number satisfying this. Setting $c := a^m$, the proof is complete once we show that $H = \left< a^m \right> = \left< c \right>$ holds.

For every $b \in H$ we have $b = a^{n} \in G$, and for some $q , r \in \mathbb{N}$ the relation $n = m q + r$ holds. Here, if we take $ 0 \le r < m$, then $q$ and $r$ are uniquely determined. $$ a^{n} = a^{mq + r} = (a^{m})^{q} a^{r} $$ and rearranging for $a^{r}$, $$ a^{r} = (a^{m})^{-q} a^{n} \\ a^{n} = b \in H $$ and since $a^{m} \in H$ while $H$ is also a group, both $(a^{m})^{-q}$ and $a^{n}$ are contained in $H$. Therefore $$ (a^{m})^{-q} a^{n} = a^{r} \in H $$ On the other hand, $m$ was the smallest natural number satisfying $a^{m} \in H$, and among all cases the only one satisfying $0 \le r < m$ is $r = 0$. In the end $r = 0$ must hold, and $$ b = a^{n} = a^{mq} = (a^{m})^{q} = c^{q} $$ Since every element can be expressed as a power of $c$, $H = \left< c \right>$ is a cyclic group.

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