[[Group order]]
# Order of powers of a group element

Given a group element $a \in G$ with $\abs{a} = n$,
then $\langle a^k \rangle = \langle a^{\gcd(n,k)} \rangle$ and $\abs{a^k} = \frac{n}{gcd(n,k)}$. #m/thm/group 

> [!check]- Proof
> Let $d = \gcd(n,k)$ and $k = dr$.
> Since $a^k = (a^d)^r \in \langle a^d \rangle$ by closure $\langle a^k \rangle \sube \langle a^d \rangle$.
> By [[GCD is a linear combination|Bézout's lemma]], there exist $s,t \in \mathbb{Z}$ such that $d = ns + kt$,
> so that $a^d = a^{ns}a^{kt}= (a^n)^s (a^k)^t = (a^k)^t \in \langle a^k \rangle$.
> Hence by closure $\langle a^d\rangle \sube \langle a^k \rangle$
> and therefore $\langle a^k \rangle = \langle a^d \rangle$.
> 
> Keeping in mind $d$ is a divisor of $n$,
> it is clear that $(a^d)^{n/d} = a^n = e$,
> implying $\abs{a^d} \leq \frac{n}{d}$.
> But if $i < \frac{n}{d}$ then $id < n$ and therefore $(a^d)^i = a^{id} \neq e$ by the definition of [[group order|order]].
> Hence $\abs{a^k} = \frac{n}{d}$.
> <span class="QED"/>

Using this technique, computing the cyclic group generated by some power of a basic element becomes simple.[^gallian]

[^gallian]: 2017, [[@gallianContemporaryAbstractAlgebra2017|Contemporary Abstract Algebra]], p. 79

## Corollaries

### Order of elements in finite cyclic groups

It immediately follows that the order of an element in a finite [[Cyclic subgroup|cyclic group]] divides the order of the group. #m/thm/group

### Criterion for ‹𝑎ⁱ› = ‹𝑎ʲ› and |𝑎ⁱ| = |𝑎ʲ| in a group

Given a group element $a \in G$ with $\abs{a} = n$,
then  $\langle{a^i}\rangle = \langle{a^j}\rangle$ iff $\gcd(n,i) = \gcd(n,j)$.
Likewise $\abs{a^i} = \abs{a^j}$ iff $\gcd(n,i) = \gcd(n,j)$. #m/thm/group 

> [!check]- Proof
> From the above theorem, $\langle a^i \rangle = \langle a^j \rangle$ iff $\langle a^{\gcd(n,i)} \rangle = \langle a^{\gcd(n,j)} \rangle$.
> Clearly  $\gcd(n,i) = \gcd(n,j)$ implies $\langle a^{\gcd(n,i)} \rangle = \langle a^{\gcd(n,j)} \rangle$.
> On the other hand, $\abs{a^{\gcd(n,i)}} = \abs{a^{\gcd(n,j)}}$
> implies $n / \gcd({n,i}) = n / \gcd(n,j)$ and thence  $\gcd(n,i) = \gcd(n,j)$.
> 
> It follows immediately that $\gcd(n,i) = \gcd(n,j)$ implies $\abs{a^i} = \abs{a^j}$.
> From the above theorem,  $\abs{a^i} = \abs{a^j} = n / \gcd({n,i}) = n / \gcd(n,j) \iff \gcd(n,i) = \gcd(n,j)$.
> <span class="QED"/>

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