Cyclic Groups

Thm

Let $G$ be a group and $\{H_i{\}}_{i\in I}$ a nonempty family of subgroups. Then ${\bigcap }_{i\in I}{H}_i$ is a subgroup of $G$.

Let $a,b\in {\bigcap }_{i\in I}{H}_i$. Then it follows that $a,b\in H_i$ for all $i\in I$. Since each $H_i$ is a subgroup, $a{b}^{-1}\in H_i$ for all $i\in I$. Therefore, $a{b}^{-1}\in {\bigcap }_{i\in I}$.

Subgroup generated by X

Let $G$ be a group and $X\subset G$. The subgroup generated by the set X (denoted $⟨X⟩$) is the smallest subgroup containing $X$. That is, $⟨X⟩={\bigcap }_{i\in I}{H}_i$ where $\{H_i{\}}_{i\in I}$ is the family of subgroups of $G$ containing $X$.

The elements of $X$ are called generators of the subgroup $⟨X⟩$. The generators are not uniquely determined. It may be possible that $⟨X⟩=⟨Y⟩$ but $X\ne Y$.

If $X=\{a_1,\dots ,a_n\}$, we write $⟨a_1,\dots ,a_n⟩$ instead and call the subgroup finitely generated.

Cyclic Group

If $G=⟨g⟩$, then $G$ is called the cyclic group generated by $g$.

We seek to characterize all cyclic groups upto isomorphism.

Thm

Every subgroup $H<\mathbb{Z}$ is cyclic, i.e., $H=⟨m⟩$ for some minimal $m\in {\mathbb{Z}}_{\ge 0}$.

Clearly $⟨0⟩$ is the trivial subgroup; assume that $m>0$. Notice that $⟨m⟩=\{km\mid k\in \mathbb{Z}\}\subset H$. Let $h\in H$. Then by the division algorithm, $h=qm+r$ for some $q,r\in \mathbb{Z}$ where $0\le r<m$. Notice that $qm\in H$ so $r=h-qm\in H$. Since $m$ is minimal in $H$ and $0\le r<m$, it follows that $r=0$ or $r=m$. Therefore, $H\subset ⟨m⟩$.

Classification of Cyclic Groups

Every infinite cyclic group is isomorphic to $\mathbb{Z}$ and every finite cyclic group of order $m$ is isomorphic to $Z_m$.

If $G=⟨a⟩$, then the map $\alpha :\mathbb{Z}\to G$ given by $k\mapsto a^k$ is an epimorphism. If $\text{Ker}\alpha =0$ then $G\cong \mathbb{Z}$. Otherwise, $\text{Ker}\alpha$ is a nontrivial subgroup of $\mathbb{Z}$. By theorem 4, $\text{Ker}\alpha =⟨m⟩$ for some minimal $m\ge 0$ satisfying $a^m=e$. Notice that

$$a^r=a^s\iff a^{r-s}=e\iff r-s\in \text{Ker}\alpha =⟨m⟩\iff m\mid (r-s),$$

i.e., $r=s$ in ${\mathbb{Z}}_m$. Therefore, $G\cong {\mathbb{Z}}_m$.

Let $G$ be a group of and $a\in G$ have infinite order ($|⟨a⟩|=\infty$). Then

1. $a^k=e$ if and only if $k=0$
2. the elements $a^k$ for all $k\in \mathbb{Z}$ are distinct

If $a$ has finite order $m>0$, then

3. $m$ is the least positive integer such that $a^m=e$
4. $a^k=e$ if and only if $m\mid k$
5. $a^r=a^s$ if and only if $r\equiv s\mathrm{mod}m$
6. $⟨a⟩$ is the distinct elements $a,a^2,\dots ,a^{m-1},a^m=e$
7. if $k\mid m$ then the order $|a^k|=m/k$