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  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$x, y \in \mathbb{Z}_8by: (1) \begin{align} \quad x * y = (x + y) \mod 8 \end{align} Consider the generated subgroup< 3, 5>$. 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]\$.