order of a equals 2 but o(e)= 1 why? let G be any group.and let 'a' belongs to G .a is not equal to e.
and let 'a' is a self inverse element .
since identity(e) is also a self inverse element ,
but o(a)=2 ,& o(e)=1 ,why?
 A: $o(x)$ is the mimimum number of times we need to multiply $x$ by itself to get to $e$. For example $o(x) = n$ means $\underbrace{x\cdot x \dots x}_{n\text{ times}}=e$. For your $a$, assuming $a\neq e$, $n$ is two since $\underbrace{a\cdot a}_{2\text{ times}} = e$ and one for $e$ since $\underbrace{e}_{1\text{ time}} = e$.
A: Note that $a^k=e$ only implies that $o(a)|k\in\Bbb N$. Thus you have $o(a)|2$, meaning $o(a)$ could be $1$ or $2$. In case $a=e,o(a)=1$. Otherwise $o(a)=2$.
A: Order is the smallest positive power to get $a^k = e$.  And $e^1 = e$.  So the order is $1$.  
$o(a) = k$ does not just mean that $a^k= e$.  It also means that for all $i: 1\le i < k$ that $a^k\ne e$.  After all if $a^k = e$ then $a^{mk} =(a^k)^m =e^m = e$ as well.  But only one value can be the order.  And that is the smallest.
If $a \ne e$ and $a^2 = e$ then $o(a) = 2$.  But if $a^2 = e$ and $a^1=e$ then $o(a) = 1$.  But the only time we have $a^1=e$ is when $a^1=a = e$.  $e$ is the only element with order $1$.
A: One way to define the order of an element is as the order of the cyclic subgroup generated by that element, and since $\langle e \rangle = \{e\}$, $o(e) = 1$.
