# Subgroup of Finite Cyclic Group is Determined by Order

## 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.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $7$: Homomorphisms: Exercise $9$ - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 25$: Theorem $25.8$ - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.9$: Theorem $16$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 39$. Cyclic Groups: Worked Example - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.4$: Cyclic groups: Theorem $1$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $4$: Subgroups: Proposition $4.14$