Order of an element in a finite Group

No, it's not a typo: the order of an element need not be equal to the order of the group (think about e.g. the unit element $e$), it's enough that it divides it in order to obtain $a^{|G|}=e$:

Say, order of $a\in G$ is $k$, and $k$ divides $n=|G|$, i.e. $n=kq$ for some $q\in\Bbb Z$. Then we have $$a^n=a^{kq}=(a^k)^q=e^q=e$$

The text is not saying that $n$, the order of the group, is the order of $a$. The text is saying that $n$ is a number that could be different than the order of $a$ that is also such that $a^n =e$.

By Legrange theorem, you know that $|a|$ divides $n$ so $n = k*|a|$ for some integer $k$.

So $a^n = a^{k|a|} = (a^{|a|})^k= e^k =e$.

That shouldn't be surprising. Although it is easy to see how the statement could be misinterpreted.

(Note: It's very easy to prove $a^j =e$ if and only if $j$ is multiple of the order of $a$. So this is just pointing out that $n$ is a multiple of the order of $a$.)

It's not a typo: $ a^n = e $ does not mean that the order of $a$ is $n$, only that it is at most $n$ (actually, it must be a divisor of $n$).