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:
\(\ds \exists k \in \N: \, \) | \(\ds k n\) | \(=\) | \(\ds r d\) | |||||||||||
\(\ds \) | \(=\) | \(\ds k \paren {\dfrac n d} d\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r\) | \(=\) | \(\ds k \paren {\dfrac n d}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac n d\) | \(\divides\) | \(\ds r\) |
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): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: 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 $4.14$