Generated Cyclic Subgroups

# Generated Cyclic Subgroups

 Definition: Let $(G, \cdot)$ be a group. Then a Generated Cyclic Subgroup of $(G, \cdot)$ from the elements $a_1, a_2, ..., a_n \in G$ denoted $\langle a_1, a_2, ..., a_n \rangle$ is a subgroup of all elements formed from all possible products, $\cdot$ of $a_1, a_2, ..., a_n$ and $a_1^{-1}, a_2^{-1}, ..., a_n^{-1}$. If $G = \langle a_1, a_2, ..., a_n \rangle$ then we say that $G$ is Generated by $a_1, a_2, ..., a_n$.

We first note that $(\langle a_1, a_2, ..., a_n \rangle, \cdot)$ is necessarily a subgroup of $(G, \cdot)$. Clearly $\langle a_1, a_2, ..., a_n \rangle \subseteq G$ and so from the Subgroups and Group Extensions page, we only need to check whether $\langle a_1, a_2, ..., a_n \rangle$ is closed under $\cdot$ and that for each element in $\langle a_1, a_2, ..., a_n \rangle$ we have that the inverse of this element with respect to $\cdot$ is also contained in $\langle a_1, a_2, ..., a_n \rangle$.

We see that $\langle a_1, a_2, ..., a_n \rangle$ is definitely closed under $\cdot$ since any possible product of $a_1, a_2, ..., a_n$ and $a_1^{-1}, a_2^{-1}, ..., a_n^{-1}$ is in $\langle a_1, a_2, ..., a_n \rangle$ by definition. Also, for each element in $\langle a_1, a_2, ..., a_n \rangle$ we must have that its inverse with respect to $\cdot$ is also contained in $\langle a_1, a_2, ..., a_n \rangle$.

Let's now look at an example of a generated subgroup. Consider the group of integers module $8$ denoted $\mathbb{Z}_8 = \{ 0, 1, 2, ..., 7 \}$, under the operation $\cdot$ defined for $x, y \in \mathbb{Z}_8$ by:

(1)
\begin{align} \quad x \cdot y = (x + y) \mod 8 \end{align}

Consider the generated subgroup $\langle 3, 5 \rangle$. Notice that $3 \cdot (3 \cdot 3) = 3 \cdot 6 = 1$ and that $3 \cdot 1 = 4$, $(3 \cdot 1) \cdot 1 = 5$, $((3 \cdot 1) \cdot 1) \cdot 1 = 6$, etc… will generate all elements in $\mathbb{Z}_8$. In other words, $\mathbb{Z}_8 = \langle 3, 5 \rangle$.

Now consider the generated subgroup $\langle 4 \rangle$. We have that $4 \cdot 4 = 0$ and $0 \cdot 0 = 0$, so $\langle 4 \rangle = \{ 0, 4 \}$ is a nontrivial subset of $\mathbb{Z}_8$, and so $(\langle 4 \rangle, \cdot)$ is a nontrivial generated subgroup of $\mathbb{Z}_8$.

For a different example, consider the group of real-valued functions that are continuous on the interval $[a, b]$ with respect to function addition, $(C[a, b], +)$. Clearly $f(x) = x$ is a continuous real-valued function on any interval $[a, b]$, and so we can consider the generated subgroup $\langle x \rangle$:

(2)
\begin{align} \quad \langle x \rangle = \{ 0, \pm x, \pm 2x, ... \} = \{ kx : k \in \mathbb{Z} \} \end{align}

In this particular case we see that $\langle x \rangle \neq C[a, b]$.