need help: r is the smallest positive integer such that $a^r = e$, the identity, and $$ $=$ {$e$, $a$, $a^2$ , ... ,$a^{r-1}$} I need help understanding this theorem:
Let G be a group and a be an element in G with order r.Then r is the smallest positive integer such that $a^r = e$, the identity, and $<a>$ $=$ {$e$, $a$, $a^2$ , ... ,$a^{r-1}$}.
If someone could explain this theorem in the simplest possible way that would be great.
 A: First note that $\langle a \rangle = \{ a^n \mid n \in \mathbb{Z} \}$, where we define $a^0=e$ and $a^{-n} = (a^{-1})^n$.
If $\langle a \rangle$ is finite, then we must have $a^i=a^j$ for some $j>i$, and then $a^{j-i} = e$. But then writing $m=j-i$, we have
$$a^{pm+q} = (a^m)^p a^q = e^p a^q = a^q$$
for all $p,q$, and so $a^q = a^{q'}$ whenever $q \equiv q' \bmod m$.
In particular, if $a^m=e$ for some $m>0$, then the set $\langle a \rangle$ has $\le m$ elements, one for each remainder modulo $m$.
So assume that $\langle a \rangle$ has exactly $r$ elements. Then:


*

*$e,a,a^2,\dots,a^{r-1}$ must all be distinct, since otherwise (by the above) we have $a^i=a^j$ for some $0 \le i < j < r$, and then $\langle a \rangle$ has $\le m <r$ elements, where $m=j-i>0$ (contradiction);

*$a^r = a^k$ for some $0 \le k < r$, since otherwise $a^r$ is an $(r+1)^{\text{st}}$ distinct element of $\langle a \rangle$ (contradiction); and

*$k=0$ in the previous point, since if $k>0$ then $0 \le r-k < r$ and $a^{r-k}=e$, contradicting the fact that $e,a,\dots,a^{r-1}$ are distinct.


So indeed $a^r=e$ and $a^s \ne e$ for any $0<s<r$.
