# Order of Subgroup of Cyclic Group

## Theorem

Let $C_n = \gen g$ be the cyclic group of order $n$ which is generated by $g$ whose identity is $e$.

Let $a \in C_n: a = g^i$.

Let $H = \gen a$.

Then:

$\order H = \dfrac n {\gcd \set {n, i} }$

where:

$\order H$ denotes the order of $H$
$\gcd \set {n, i}$ denotes the greatest common divisor of $n$ and $i$.

## Proof

The fact that $H$ is cyclic follows from Subgroup of Cyclic Group is Cyclic.

We need to show that $H$ has $\dfrac n d$ elements.

Let $\order H = k$.

$k = \order a$

where $\order a$ denotes the order of $a$.

That is:

$a^k = e$

We have that $a = g^i$.

So:

 $\displaystyle a^k$ $=$ $\displaystyle e$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \paren {g^i}^k$ $=$ $\displaystyle e = g^n$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle n$ $\divides$ $\displaystyle i k$ Definition of Order of Group Element

We now need to calculate the smallest $k$ such that:

$n \divides i k$

where $\divides$ denotes divisibility.

That is, the smallest $t \in \N$ such that $n t = i k$.

Let $d = \gcd \set {n, i}$.

Thus:

$t = \dfrac {k \paren {i / d} } {n / d}$

From Integers Divided by GCD are Coprime, $\dfrac n d$ and $\dfrac i d$ are coprime.

Thus from Euclid's Lemma:

$\dfrac n d \divides k$

As $a \divides b \implies a \le b$, the smallest value of $k$ such that $\dfrac k {\paren {n / d} } \in \Z$ is $\dfrac n d$.

Hence the result.

$\blacksquare$