Generated Subgroups
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Generated Subgroups

Definition: Let $(G, *)$ be a group. Then a Generated Subgroup of $(G, *)$ from the elements $a_1, a_2, ..., a_n \in G$ denoted $<a_1, a_2, ..., a_n>$ is a subgroup of all elements formed from all possible products, $*$ of $a_1, a_2, ..., a_n$ and $a_1^{-1}, a_2^{-1}, ..., a_n^{-1}$.

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

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

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 $*$ defined for $x, y \in \mathbb{Z}_8$ by:

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

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

Now consider the generated subgroup $<4>$. We have that $4 * 4 = 0$ and $0 * 0 = 0$, so $<4> = \{ 0, 4 \}$ is a nontrivial subset of $\mathbb{Z}_8$, and so $(<4>, *)$ 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 $<x>$:

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

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

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