Subgroup of Finite Cyclic Group is Determined by Order

Theorem
Let $$G = \left \langle {g} \right \rangle$$ be a cyclic group whose order is $$n$$ and whose identity is $$e$$.

Let $$d \backslash n$$, where $$d \backslash n$$ in this context means "$$d$$ is a divisor of $$n$$."

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

Proof

 * Let $$G$$ be generated by $$g$$, such that $$\left|{g}\right| = n$$.

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

Thus $$\left \langle {g^{n / d}} \right \rangle$$ has $$d$$ elements.


 * Now suppose $$H$$ is another subgroup of $$G$$ of order $$d$$. Then by Subgroup of a Cyclic Group is Cyclic, $$H$$ is cyclic.

Let $$H = \left \langle {y} \right \rangle$$ where $$y \in G$$. Thus $$\left|{y}\right| = d$$.

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

Since $$\left|{y}\right| = d$$, it follows that $$y^d = x^{r d} = e$$.

From Equal Powers of Finite Order Element, $$n \backslash r d$$.

Thus $$\exists k \in \N: k n = r d = k \left({n / d}\right) d \implies r = k \left({n / d}\right)$$.

So $$n / d \backslash r$$.

Thus $$y$$ is a power of $$g^{n/d}$$, so $$H$$ is a subgroup of $$\left \langle {g^{n / d}} \right \rangle$$.

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