Cyclic Groups

Let G be a group. G is cyclic if there exists a $g \in G$ such that $G = \{g^n : n \in \mathbb{Z}\}$. This property is denoted $G = <g>$ and we say that G is generated by g.

For example, $(\mathbb{Z}, +) = <1>$ ($\mathbb{Z}$ is the group of all integers with addition as the group operation) and $(\mathbb{Z}/n\mathbb{Z}, +) = <1>$ ($\mathbb{Z}/n\mathbb{Z}$ is the group of integers modulo n, ie integers from 1 to n-1, with addition as the group operation).

A note on notation: when we are working with $\mathbb{Z}/n\mathbb{Z}$ all integers (and variables) are modulo n. This is comonly denoted by placing a bar on top of numbers and variables (eg, $\bar{x}, \bar{2}$ are both modulo n) but for simplicity I will not use this notation unless not using it would result in ambiguity $a\vert b$ means that a divides b, and (a, b) denotes the greater common divisor of a and b.

Proposition: the order of an element is equal to the size of the group it generates. If $H = <x>$ is infinite then $x^i = x^j$ if and only if i = j, hence x has infinite order. If $H = <x>$ is finite and $|H| = n$ then $H = \{1, x, ..., x^{n-1}\}$, so x has order n.

Theorem: let H be cyclic generated by x, then (1) if the order of x is infinite then $H \cong \mathbb{Z}$ (2) if $\vert x\vert = n \in \mathbb{Z}_{> 0}$ then $H \cong \mathbb{Z}/n\mathbb{Z}$.

Proof: (1) Let $f: \mathbb{Z} \rightarrow H$ be a homomorphism that sends $i \mapsto x^i$. Since $f(i+j) = x^{i+j} = x^i \cdot x^j = f(i)\cdot f(j)$ f is well defined. Since $f(i) = 1$ if and only if $x^i = 0$ if and only if i = 0 then f is injective. Finally since $H = \{x^n : n \in \mathbb{Z}\} = f(\mathbb{Z})$, f is surjective. This completes the proof for part 1. (2) let $f : \mathbb{Z}/n\mathbb{Z} \rightarrow H$ be a homomorphism that sends $i \mapsto x^i$. If i = j then $x^i = x^j$ since if i = j then $n | i-j = n \cdot k, x^{i+j} = x^ix^j = x^n = 1$. We can also see that $f(i+j) = x^{i+j} = x^ix^j = f(i)+f(j)$ so this homomorphism is well defined. $f(i) = 1$ if and only if $x^i = 1$ where i = qn+r and $0 \leq r < n$. Hence $x^i = x^{qn+r} = (x^n)^q\cdot x^r = 1$ so r = 0 and i = 0, hence f is injective. Since $H = \{1, x, ..., x^{n-1}\} = f(\mathbb{Z}/n\mathbb{Z})$ f is surjective. Thiscompletes the proof for part 2.

Remark: if H and G are cyclic, $H \cong G$ if and only if $\vert H\vert = \vert G\vert$

Proposition: (1) Let $H = <x>$ be finite, then for all $a \in \mathbb{Z}-{0}$ (all integers except 0) $x^a$ has finite order. (2) Let $H = <x>$ be finite with $|H| = n$ then for all $a \in \mathbb{Z}$ $x^a$ has order $\frac{n}{(n,a)}$

Proof: (1) $(x^a)^m = x^{am} = 1$ if and only if am = 0 if and only if m = 0, so $x^a$ has infinite order. (2) $(x^a)^m = x^{am} = 1$ if and only if am = 0 mod n if and only if $n| am$. Let d = (n,a) where $n = d \cdot k, a = d \cdot l$ and (k,l) = 1. From that it follows that $n | am$ if $dk | dlm$ if and only if $k | lm$ if and only if $k | m$, hence the order of $x^a$ is k = $\frac{n}{(n, a)}$.

Proposition: (1) if $H = <x>$ is finite, with $a, b \in \mathbb{Z}$, $<x^a> = <x^b$ if and only if $|a| = |b|$ (2) If $H = <x>$ is finite and $|H| = n$ with $a, b \in \mathbb{Z}$, $<x^a> = <x^b>$ if and only if (a,n) = (b,n).

Proof: (1) $<x^a> = \{ x^n : n \in \mathbb{Z}\}$. If $x^b \in \{ x^n :n \in \mathbb{Z}\}, x^b = x^{am} = b = am$ so $a\vert b$. Similarly $b \vert a$ so $\vert b \vert = \vert a\vert$. Conversely if $\vert a\vert = \vert b\vert$ then $b = \pm a$ so $x^b = x^a$ or $x^{-a}$ and $<x^a> = <x^{-a}>$. (2) $<x^a> = <x^b>$ so $\vert x^a\vert = \vert x^b\vert$ so $\frac{n}{(n,a)} = \frac{n}{(n,b)}$ so (n,a) = (n,b). Conversely assume (n,a) = (n,b) = d we have a = kd, so $x^a \in <x^d>$ which means $<x^d$ and $<x^a> \leq <x^d>$. Since $\vert <x^a>\vert = \frac{n}{d}, \vert <x^d>\vert = \frac{n}{(n,d)} = \frac{n}{d}$ so $<x^a> = <x^d>$. Similarly $<x^b> = <x^d> = <x^a>$.