Def. of $H(S) :$ $H(S) = \{g\in G \mid \exists n\in \mathbb N , \exists \{g_1, g_2,... g_n\}\subseteq S\cup S^{-1} , g = g_1 ...g_n\}$.
Note : it can be shown that $H(S)$ is simply the same set as $\langle S \rangle$.
Context : group theory.
The reference I am using is Reversat & Bigonnet , Algèbre pour la licence ( Undergraduate Abstract Algebra).
The authors propose a formal definition of $\langle x \rangle$ as the result of applying to a singleton $\{x\}$ the general definition of a set H(S) such that :
$H(S) :$ $H(S) = \{g\in G \mid \exists n\in \mathbb N , \exists \{g_1, g_2,... g_n\}\subseteq S\cup S^{-1} , g = g_1 ...g_n\}$.
In words ( as understand it), $H(S)$ is the set of all $g_{\in G}$ such that , for som $n$, $g$ is the product of $n$ factors coming from an $n$-elements subset of $S\cup S^{-1}$ , with $S^{-1}=$ the set of the inverses of the elements of $S$.
I know that if $x$ is a member of some group $G$ , $\langle x \rangle$ can have more than $3$ elements.
But, when I try to apply the formal definition, I get ( apparently) a contrary result.
If $S = \{x\}$ , then $S\cup S^{-1} = \{x, x^{-1}\}$
A finite ( necessarily finite) subset of $S\cup S^{-1}$ is
$\emptyset$
$\{x\}$
$\{x^{-1}\}$
$\{x, x^{-1}\}$
The products ( with a finite number of factors) that can be made out of these subsets are : $x$ , $x^{-1}$ , $xx^{-1}$.
So it seems that I get $\langle x \rangle = \{e, x,x^{-1} \}$.
How to escape this conclusion? What mistake do I make in the application to $S= \{x\}$ of the general definition of $H(S)$.
If the factors that " compose" $g\in H(\{x\})$ must come ( in this particular case) from a subset of $\{x, x^{-1}\}$ , how can $g$ be the product of more that $2$ factors?