Abstract Algebra/Group Theory/Subgroup/Cyclic Subgroup/Order of a Cyclic Subgroup

= Theorem = Define Order of an Element g of Finite Group G:
 * o(g) = the least positive integer n such that gn = e

Define Order of a Cyclic Subgroup generated by g:
 * $$o(\langle g \rangle)$$ = # elements in $$\langle g \rangle$$


 * o(g) = $$o(\langle g \rangle)$$

= Proof = [[File:cyclic group order.png|thumb|200px| Since $$\langle g \rangle$$ is cyclic, and has $$o(\langle g \rangle)$$ elements.

$$g^{o(\langle g \rangle)}=e$$]]

By diagram,
 * 0. $$g^{o(\langle g \rangle)}= e = g^{o(g)}$$.


 * 1. Let n = o(g), and m = $$o(\langle g \rangle)$$


 * 2. gn = gm


 * 3. gn – m = e


 * 4. Let n – m = sn + r where r, n, s are integers and 0 ≤ s < n.


 * 5. gsn + r = e


 * 6. $$ [g^{n}]^{s} \ast g^{r} = e $$

By definition of n = o(g)
 * 7. gr = e

As n is the least that makes gn = e and 0 ≤ r < n.
 * 8. r = 0

Lemma: Let $$k,m \in \mathbb{Z}$$.

$$g^k = g^m$$ if and only if $$k \equiv m (\,\bmod\ o(g))$$.

Proof: Let $$n = o(g)$$.

$$g^k = g^m$$ if and only if $$g^{k-m} = e$$.

By Euclidean division: $$k - m = qn + r$$, some integers $$q, r$$ with $$0 \leq r < n$$.

We have $$g^r = g^{k - m - qn} = g^{k-m} {(g^n)}^{-q} = g^{k-m}$$, hence $$g^r = e$$ if and only if $$g^k = g^m$$.

But $$g^r = e$$ if and only if $$r = 0$$ (i.e. if and only if $$n \mid k - n$$),

since, by definition, $$n = o(g)$$ is the least positive integer satisfying $$g^n = e$$.

Hence the result. $$\square$$

By definition: $$\langle g \rangle = \{ g^r : r \in \mathbb{Z} \}$$.

Therefore, $$g, g^2, \ldots, g^n$$ (where $$n = o(g)$$) all lie in $$\langle g \rangle$$ – furthermore, by lemma above, these are pairwise distinct.

Finally, any element of the form $$g^r$$, $$r \in \mathbb{Z}$$ equals one of $$g, g^2, \ldots, g^n$$ (again by lemma).

We conclude that $$g, g^2, \ldots, g^n$$ are precisely the elements of $$\langle g \rangle$$,

so $$o(\langle g \rangle)=n=o(g)$$, as required.

- Q.E.D. -