5 - Generators, Cyclic Groups

A contrived definition, followed by a discussion on generators:

Containment of a subset of a group

Let $G$ be a group and $S$ be a subset of $G$ . Then a containment group of $S$ , denoted by $\bar{S}$ is the collection of subgroups $H$ of $G$ such that $S \subset H$ . That is $\bar{S} = {H ⩽ G : S \subset H}$ .

Now, Notice that any $H \in \bar{S}$ has the following properties:

Since $H$ contains $S$ , for each $s \in S$ , $s^{- 1} \in H$ .
For any $m \in N$ , and $s_{1}, s_{2}, \dots s_{m} \in S$ , the product $s_{1} s_{2} . . . s_{m} \in H$ .
Putting it all together, for any $m \in N$ , $H$ contains the element $s_{1} s_{2} \dots s_{m}$ where $s_{i} \in S$ or $s_{i}^{- 1} \in S$ .

Now, define $⟨ S ⟩ = {s_{1} s_{2} \dots s_{m} : m \in N, s_{i} \in S o r s_{i}^{- 1} \in S}$ . From the above discussion, it is clear that $⟨ S ⟩ \subset H$ for each $H \in \bar{S}$ .
Moreover, of course $S \subset ⟨ S ⟩$ . We show that $⟨ S ⟩$ is a group (precisely it is a subgroup of $G$ ).

We will repeatedly abuse the definition of $⟨ S ⟩$ (note that a definition is not a person so this is okay) for the following proof.

$e \in ⟨ S ⟩$ : pick any $s \in S$ . Then, by defenition, it is clear that $s s^{- 1} = e \in ⟨ S ⟩$ .
Suppose $x, y \in ⟨ S ⟩$ Then $x = s_{1} s_{2} \dots s_{m}$ and $y = t_{1} t_{2} \dots t_{n}$ for some $m, n \in N$ , such that $t_{i} o r t_{i}^{- 1} \in S$ and $s_{i} o r s_{i}^{- 1} \in S$ . Hence, $x y = s_{1} s_{2} \dots s_{m} t_{1} t_{2} \dots t_{n} \in ⟨ S ⟩$ .
Suppose $x \in ⟨ S ⟩$ Then $x = s_{1} s_{2} \dots s_{m}$ , for some $m \in N$ such that $s_{i} o r s_{i}^{- 1} \in S$ . Then $x^{- 1} = s_{m}^{- 1} s_{m - 1}^{- 1} \dots s_{1}^{- 1} \in ⟨ S ⟩$ .

Hence, each $H \in \bar{S}$ contains the subgroup $⟨ S ⟩$ . Therefore, $⟨ S ⟩$ is the smallest group that contains $S$ , known as the subgroup generated by $S$ . In particular, for each $H \in \bar{S}$ , $S \subset ⟨ S ⟩ ⩽ H$ .

Group generated by a set

let $G$ be a group, and $S$ a subset of $G$ , then inheriting the operator from $G$ , the subgroup of $G$ generated by $S$ , is the set $⟨ S ⟩ = {s_{1} s_{2} \dots s_{m} : m \in N, s_{i} \in S o r s_{i}^{- 1} \in S}$ .

Cyclic group

For a group $G$ , the group generated by any element $g \in G$ is denoted by $⟨ g ⟩$ . This group is called the cyclic subgroup of $G$ generated by $g$ .

Notice that $⟨ g ⟩$ just contains the products of $g$ with itself : $g^{i}$ and the products of $g^{- 1}$ with itself: $g^{- i}$ , where $g^{0} := e$ . The smallest $n \in N$ for which $g^{n} = e$ is (if it exists) known as the order of $⟨ g ⟩$ .

In such a case, it is clear that $⟨ g ⟩ = {g^{0} = e, g^{1}, \dots g^{n - 1}}$ . where ${g^{i}}^{- 1} = g^{n - i}$ .

Otherwise, if no such $n$ exists, the order of $⟨ g ⟩$ is infinite.

The order of an element $g$ of a group $G$ is the order of its cyclic subgroup $⟨ g ⟩$ .

Two cyclic groups of the same order $⟨ g ⟩$ , $⟨ h ⟩$ are isomorphic to each other, where $g^{i} \mapsto h^{i}$ is the desired isomorphism.

Moreover, a cyclic group of finite order, say $n$ is isomorphic to the rotations of a regular $n$ -gon.

necessary and sufficient condition for a cyclic group

Let $G$ be a group. Then $G$ is cyclic if and only if, there exists $g \in G$ , $G = ⟨ g ⟩$ .

\begin{proof} Suppose $G$ is cyclic, then by definition there exists an element $g \in G$ such that $G = {g^{i} : i \in Z}$ .
Now suppose that there exists $g \in G$ such that $G = ⟨ g ⟩$ , then by definition, $G$ is cyclic.

\end{proof}

Although the above proposition seems to bash on the obvious, it's extremely useful: for example in showing that the multiplicative group of integers, modulo a prime is cyclic.