Is the inverse of an element in a group different than the element (except for $e$)? Also, are all the subgroup of a cyclic group cyclic? Suppose the cyclic group $G=\{e,a,a^2,a^3,a^4,a^5,a^6,a^7,a^8,a^9,a^{10},a^{11}\}$ under some operation, say *.
$$G=\langle a\rangle$$
Now, each element must have an inverse, s.t. $b*b^{-1}=e$, and the inverse $b^{-1}$ is unique and different than $b$ except for $b=e$.
Here, the inverse of $a$ is $a^{11}$ and so on. However, the inverse of $a^6$ is $a^6$ (since $a^6*a^6=a^{12}=e$), which is my first issue here since $a^6\neq e$.
The second issue is that there is a theorem that says:
"Every subgroup of a cyclic group is cyclic."
Source: Contemporary Abstract Algebra - CH.4, Theorem 4.3: Fundamental Theorem of Cyclic Groups.
We now define $H=\{e,a^3,a^9\}$, which is a subset of G and also a subgroup (under *) [closed, associative, has an identity and all elements have inverse]. But H is not cyclic [needs to have $a^6$ to be so].
I would like to know why my example is incorrect, or that I did not quite understood the subject.
 A: It is not required that $b^{-1}\neq b$. In fact, $(a^6)^{-1}=a^6$. Your subset is not a subgroup because $(a^3)^2=a^6$, which is not in the set.
A: You don't need to have $g^{-1}\neq g$ in a group. Elements of order two coincide with their inverses, as $a^6$ does in your example.
$H$ is not a subgroup, for example because it is not closed under multiplication, indeed $a^3$ is in the group but $a^3\cdot a^3$ is not.
A: What's the problem with the fact that for some element $g$ of a group $g$ you have $g^2=e$ and $g\ne e$? That happens often you you have provided an example. In fact, wheneve an element $g$ has order $2n$, for some $n\in\Bbb N$, $(g^n)^2=e$, but $g^n\ne e$.
And $H$ is not a subgroup since, as you wrote, $a^6(=a^3*a^3)$ is missing.
A: Suppose $G$ is a finite group, say $|G|=n$. 
A basic theorem in the theory of finite groups (known as Cauchy's theorem) says that for every prime divisor $p$ of $|G|$ there exists a subgroup $H<G$ with $p$ elements.
If $n$ is even $2$ is a prime divisor of $n$, thus the existence of a subgroup
$$
H=\{e,g\}<G
$$
is guaranteed. But since $H$ is a group in its own right you must have necessarily have $g=g^{-1}$ although $g\neq e$.
Thus the existence of elements equal to their inverses is actually rather common in groups.
