Prove that the order of every element of $Z_{100}$ is a divisor of 100:
$i)$ Recall that the order of an element is the least positive integer (let's say $n$) that satisfies $a^{n} = e$
Additionally, recall that a divisor is a number that divides a integer exactly (i.e. no remainder)
Therefore, taking the order of an element (by definition of order) implies that the element can be applied to itself (via the group's operation) '$n$' times to get back to the identity exactly
$ii)$ Thus, it must be true that $n\mid 100$ (i.e. for some $a \in Z_{100}$, $|a| = n$ and that |a| divides $100$) because if $n$ isn't a divisor of 100 then $n$ couldn't be an order of an element; since this would then imply that upon dividing 100 by the order $n$ that a remainder was produced which contradicts the definition of order
Consider the following:
$n = |a|q + r$ and $r \in\{0, 1, …. |a|-1\}$ : by the Division Algorithm
$e = a^{n}$ : Definition of $|a| = n$
$= a^{|a| q}a^{r}$ : Substitution
$= (a^{|a|})^{q}a^{r}$ : Rules of Exponents
$= e^{q}e^{r}$ : Definition of Order
$= a^r = e$ : Definition of Identity
This now states that $a^r = e$; but if $a^r = e$ and $0 \le r \lt |a|$ then this infers that $r = 0$ since by the definition of order: |a| is the least positive integer that returns the identity, and by the fact that $r \in \{0, 1, …, |a|-1\}$, the only element $r$ can be to satisfy $a^r = e$ without breaking the definition of order is $r = 0$
Therefore proving that for any given $a \in Z_{100}$ and $|a| = n$, that $n\mid 100$
Best Answer
Your proof is fine. It could be shorter though.
Suppose $g\in \Bbb Z_{100}$ has order $n\nmid 100$. Then $\langle g\rangle$ has order $n$, so, by Lagrange's Theorem, $n\mid 100$, a contradiction. The result follows.