Subgroup of Finite Cyclic Group is Determined by Order

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Theorem

Let $G = \gen g$ be a cyclic group whose order is $n$ and whose identity is $e$.

Let $d \divides n$, where $\divides$ denotes divisibility.


Then there exists exactly one subgroup $G_d = \gen {g^{n / d} }$ of $G$ with $d$ elements.


Proof

Let $G$ be generated by $g$, such that $\order g = n$.

From Number of Powers of Cyclic Group Element, $g^{n/d}$ has $d$ distinct powers.

Thus $\gen {g^{n / d} }$ has $d$ elements.


Now suppose $H$ is another subgroup of $G$ of order $d$.

Then by Subgroup of Cyclic Group is Cyclic, $H$ is cyclic.

Let $H = \gen y$ where $y \in G$.

Thus $\order y = d$.

Thus $\exists r \in \Z: y = g^r$.

Since $\order y = d$, it follows that $y^d = g^{r d} = e$.

From Equal Powers of Finite Order Element:

$n \divides r d$

Thus:

\(\, \displaystyle \exists k \in \N: \, \) \(\displaystyle k n\) \(=\) \(\displaystyle r d\)
\(\displaystyle \) \(=\) \(\displaystyle k \paren {\dfrac n d} d\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r\) \(=\) \(\displaystyle k \paren {\dfrac n d}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac n d\) \(\divides\) \(\displaystyle r\)


Thus $y$ is a power of $g^{n / d}$.

Hence $H$ is a subgroup of $\gen {g^{n / d} }$.

Since both $H$ and $\gen {g^{n / d} }$ have order $d$, they must be equal.

$\blacksquare$


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