Results concerning Order of Element

Theorem
Let $$G$$ be a group whose identity is $$e$$.

Let $$a \in G$$ have finite order such that $$\left|{a}\right| = k$$.

Then the following results apply:

Theorem 1
$$\forall n \in \mathbb{Z}: n = q k + r: 0 \le r < k \Longrightarrow a^n = a^r$$

Theorem 2
$$\forall n \in \mathbb{Z}: a^n = e \iff k \backslash n$$

Theorem 3
$$\left\{{a^0, a^1, a^2, \ldots, a^{k - 1}}\right\}$$ is a complete repetition-free list of the elements of $$\left \langle {a} \right \rangle$$.

Theorem 4
$$\left|{\left \langle {a} \right \rangle}\right| = k$$, where $$\left \langle {a} \right \rangle$$ is the smallest subgroup of $$G$$ containing $$a$$.

That is, the order of the subgroup generated by $a$ is equal to the order of $$a$$.

(Some sources use this as the definition of the order of an element, and from it derive Order of an Element.)

Proof of Theorem 1
Let $$n \in \mathbb{Z}$$.

Then from the Division Theorem, $$\exists q, r \in \mathbb{Z}: n = q k + r, 0 \le r < k$$.

From the index laws and the fact that $$a^k = e$$, we have:

$$ $$ $$ $$

Proof of Theorem 2
Let $$k \in \mathbb{N}$$ be the smallest such that $$a^k = e$$ as per the hypothesis.


 * Let $$a^n = e$$.

Let $$n = q k + r, 0 \le r < k$$.

By Theorem 1 above, $$a^r = a^n = e$$.

But $$0 \le r < k$$.

Since $$k$$ is the smallest such that $$a^k = e$$, $$1 \le s < k \Longrightarrow a^s \ne e$$.

Thus $$r = 0$$, i.e. $$k \backslash n$$.


 * Now suppose $$k \backslash n$$.

Then $$\exists s \in \mathbb{Z}: n = s k$$.

So:

$$ $$ $$ $$

Proof of Theorem 3
By Theorem 1 above, every power of $$a$$ is equal to one appearing in the list $$a^0, a^1, a^2, \ldots, a^{k - 1}$$.

This list has to be repetition free, otherwise it would contain $$a^m = a^n$$ with $$0 \le m < n < k$$ which violates the hypothesis.

Proof of Theorem 4
It follows straight away from Theorem 3 above that $$\left|{\left \langle {a} \right \rangle}\right| = k$$.