The Order of the Identity of a Group My lecture notes say that in a group $G$, the identity $e$ is the only element of $G$ which has order $1$. 
I would like to know why the order of $e$ is said to be $1$, when it could clearly be $0$ since the identity $e$ acted on itself zero times is just $e$.
Is it valid to think about an element acting on itself zero times? So for example is it valid to even think about why $e^0 = e$? Cheers.
 A: Don't think of $g^0$ as representing $g$ acting on itself $0$ times but as $g$ acting on something else $0$ times.  Groups frequently represent actions on other objects.  An important class of examples is symmetry groups.  
Consider G as the group of symmetries of a square and $g \in G$ as rotate clockwise by $90^\circ$.  Then $g^2$ is rotate clockwise by $180^\circ$ and $g^3$ is rotate clockwise by $270^\circ$.  So, what is a sensible definition of $g^0$ in this context?  Well, rotate by $0^\circ$.  Similarly we can sensibly define $g^{-1}$ as rotate anticlockwise by $90^\circ$ since this will undo the effect of $g$.  With these definitions, $g^n$ is rotate clockwise by $n \times 90^\circ$ whether $n$ is positive, zero, or negative and $g^m \times g^n = g^{m+n}$ and life is good.  With your convention, things would not be so neat.
So, with this convention, $g^0$ is always $e$ and hence to be useful we must require the order to be $> 0$ and not just $\geq 0$.
A: You can define powers of a group element $g$: $g^0=e$ and $g^{n+1}=g^n\cdot g$ for $n\geq 0$. The order of an element is chosen to be $\geq 1$.
A: The order of an element is a positive integer, by definition.
A: In my book, the order of an element is the order of the cyclic group it generates, so the identity clearly has order 1.
