A closed finite subset of a group is subgroup I am reading book of abstract algebra. The proof of a lemma are as below. I am just wondering why is H  of the special form: $a, a^2,... , a^{n +1}$?
 A: Here is a simpler proof.
Let $a \in H$ and consider the map $\mu: x \mapsto ax$ on $G$.
$\mu$ sends $H \to H$ because $H$ is closed.
$\mu$ is injective because $G$ is a group. Hence, $\mu$ defines a bijection $H \to H$ because $H$ is finite.
Therefore, $au=a$ for some $u \in H$ and so $u=e$ and $e \in H$.
Also, $av=e$ for some $v \in H$ and so $v=a^{-1}$ and $a^{-1} \in H$.
A: $H$ is not necessarily on the form $\{a,a^2,\ldots,a^{n+1}\}$.
You just pick an element $a\in H$ (such that $a\neq e$), and then you consider all the powers $a,a^2,\ldots,a^{n+1}$. These powers need not cycle through all elements of $H$.
It may very well be that they repeat themselves with a period of $2$, i.e. $a=a^3=a^5=\ldots$, and $e=a^2=a^4=\ldots$. There can be a lot of other stuff in $H$ besides the powers of $a$.
The only thing important to the proof is that you know that the powers repeat themselves at some point, which they will do, because there are $n+1$ of them, and $H$ only has $n$ elements.
A: The book does not say that $H$ is of the special form $\{a, a^2,... , a^{n +1}\}$.
All it says is that each $a^j$ is an $H$ and hence $\{a, a^2,... , a^{n +1}\}$ is a subset of $H$,
that is $\{a, a^2,... , a^{n +1}\}\subseteq H$.
It does not say that $\{a, a^2,... , a^{n +1}\}= H$. 
Since $H$ has $n$ many elements, but $n+1$ many powers of $a$ are listed, then some of these powers of $a$ must coincide (the same element of the group is listed once as $a^i$ and listed later again, this time as $a^j$), that is $a^i=a^j$ for some $i,j$. 
