Orders of Cyclic Subgroups

| H | = | x |

when Cyclic Group

If $H = ⟨ x ⟩$ then $| H | = | x |$ . For the finite cases where $| x | = n$ then $H = {1, x, x^{2}, \dots, x^{n - 1}}$ and for the infinite case then $x^{n} \neq 1$ for all $n \neq 0$ and $x^{a} \neq x^{b}$ for all $a \neq b \in Z$ .

Proof

Let $| x | = n$ . The elements $1, x, x^{2}, \dots, x^{n - 1}$ are distinct since if $x^{a} = x^{b}$ with $0 \leq a < b < n$ then $x^{b - a} = x^{0} = 1$ , contradicting that $n$ be the smallest positive power of $x$ giving the identity. Thus $H$ has at least $n$ elements.

To show this is all of them, let $x^{t}$ be any power of $x$ . Using the Division Algorithm write $t = n q + k$ where $0 \leq k < n$ . Then:

x^{t} = x^{n q + k} = (x^{n})^{q} x^{k} = 1^{q} x^{k} = x^{k} \in {1, \dots, x^{n - 1}}

Now for $| x | = \infty$ , no positive power of $x$ is the identity. If $x^{a} = x^{b}$ by assumption then (for clarity say $a < b$ ) we have $x^{b - a} = 1$ . Thus we found a positive power that makes the identity, which is a contradiction.

Since distinct powers of $x$ are distinct elements of $H$ then $| H | = \infty$ .

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