Cyclic Group

We define $G = \{a^n:n\in\Z\} = \langle a \rangle$ is a Cyclic Group.

• $\forall g\in G, \langle g \rangle$ is a subgroup of $G$ (cyclic subgroup). $g$ is the genrator here.

• cyclic group is abelian

• $H\le G$ is a subgroup of $G$, $g\in G$, $g\in H \iff \langle g\rangle \subseteq H$

• $|a| = |\langle a\rangle|$

• subgroups of cyclic groups are cyclic, $|\langle a\rangle| = n \implies$ the order of any subgroup of $\langle a \rangle$ is a divisor of $n$; $\forall k|n, k > 0$, the group $\langle a\rangle$ has exactly one subgroup of order $k$

• Let $H \le G$ be a subgroup, and let $g\in G$, then $g\in H$ is equivalent to $\langle g \rangle \sube H$

• let $k\in \Z^+, k|n \implies \phi(d) =$ number of elements (with order $d$) in a cyclic group of order $n$

  • $|\langle a\rangle| \ne \infty,$ the number of elements of order $d$ is a multiple of $\phi(d)$

• order:

  • The smallest positive integer $n, g^n = e \implies |g| = n$ or infinite order (order of element of group)
  • $|g| = n \implies g^k\cdot g^l = g^{k+l(mod \ n)}$
  • $a^k = e = a^n\implies a^{gcd(k,n)}= e$ or $|a| = gcd\{n\in\N : a^n = e\}$

Cyclic Group Theorem

Given precondition: $G$ be some group and $a\in G$.

Theorem 1

$|\langle a\rangle| = \infty \implies (a^i = a^j \iff i = j)$

$|\langle a\rangle| \ne \infty \implies (a^i = a^j \iff n | (i-j))$

From above, we have corollary that:

• the order of element $a$ is the same as the order of group $\langle a\rangle$, i.e. $|a| = |\langle a\rangle|$
• $|a| = n \land a^k = e \implies n|k$
• $|G| \ne \infty$, $a,b \in G, ab=ba \implies |ab|||a||b|$

Theorem 2

$|a| = n \implies \langle a^k \rangle = \langle a^{gcd(n,k)}\rangle$ where $k \in \Z^+$

$|a| = n \implies |a^k| = n/gcd(n,k)$ where $k \in \Z^+$

From above, we have corollary that:

• $\forall b \in \langle a\rangle, |a| \ne \infty \implies |b| | |a|$
• $|a| = n$. $\langle a^i\rangle = \langle a^j\rangle \iff gcd(n,i) = gcd(n,j) \iff |a^i| = |a^j|$
• $|a| = n$.

Theorem 3 Fundamental Theorem of Cyclic Group

$\forall \langle b\rangle \le \langle a\rangle, |a| = n \implies |b| | n$, that is, $\forall k\in \Z^+, k|n \implies \langle a^{n/k}\rangle \le \langle a\rangle$ and $|a^{n/k}| = k$

Theorem 4 Number of Elements of Each Order in a Cyclic Group

$\forall d\in \Z^+, d|n \implies \phi(d) =$ number of elements (with order $d$) in a cyclic group of order $n$

From above, we have corollary that:

• In a finite group, the number of elements of order $d$ is a multiple of $\phi(d)$