Why does the equality $\langle x\rangle=\{x^{-1},1,x\}$ not generally hold?

Question

So if $X$ is a not empty subset of a group $(G,\cdot,1)$ then the collection $$ \mathcal X:=\{H\in\mathcal P(X):H\,\text{subgroup containing X}\} $$ is not empty since it contains $G$ and moreover its intersection $\langle X\rangle$ is the most little subgroup containing $X$. Now let's we observe that the collection $$ H_x:=\{x^{-1},1,x\} $$ is a subgroup containing $x$ since it contains the unit $1$ and as well the inverse of any its element and since it is closed with respect the operation $\cdot$ defined in $G$. So I observe that if $x$ is in $\langle x\rangle$ then surely its inverse $x^{-1}$ is there contained as obviously the unit so that the inclusion $$ H_x\subseteq\langle x\rangle $$ holds but if $H_x$ is a subgroup containing $x$ then even the inclusion $$ \langle x\rangle\subseteq H_x $$ must holds and thus finally the equality $$ \tag{1}\label{1}\langle x\rangle =H_x $$ holds. Now really many (all actually) authors says that $G$ is cyclic if there exists $x\in G$ such that $$ G=\langle x\rangle $$ so that if what we above observe was true then any cyclic group must have at least three elements and this seems to me really stranger because consulting the Web I understood that this cannot be: so I thought to put a specific question where I ask clarification why \eqref{1} does not holds. So could someone help me, please?

Answer 1

To show that the equality does not hold in general, it suffices to provide a single counterexample.

Consider $G=(\Bbb Z, +)$. The identity of $G$ is $0$. Also, $1\in G$ with inverse $-1\in G$. Therefore, we have

$$H_1=\{-1,0,1\}.$$

But $1+1=2\notin H_1$, so $H_1$ cannot be a group as it is not closed. (All subgroups are themselves groups.)

In fact, $\langle 1\rangle=G$.

Answer 2

Generally, as Milten observed into the comments above, $H_x$ is not a subgroup since $x^2$ is not in $H_x$ which so is not closed: e.g $H_3$ is not a subgroup of $(\Bbb R,\cdot)$. Moreover, we can observe that if there exsists a prime number $p\in\Bbb N_+\setminus\{1\}$ such that $$ |G|=p $$ then for any $x\in G\setminus\{1\}$ the equality $$ G=\langle x\rangle $$ holds: indeed, we know (see here for details) that $$ |G|=\big|\langle x\rangle\big|\cdot\big|G:\langle x\rangle\big| $$ so that if $x$ is not $1$ and so $\langle x\rangle$ is not the trivial subgroup $\{1\}$ then necessarly $$ \big|\langle x\rangle\big|=p\quad\text{and}\quad\big|G:\langle x\rangle\big|=1 $$ so that $G$ is just $\langle x\rangle$ and this proves the existence of a class of not trivial cyclic subgroups, that is the class of finite groups with prime order.