Why is the order of a multiplicative modulo group divisible by the order of an element in its group?

Question

In this question:

Find natural numbers $n,m$ such that the residue class $[m]_n \in \mathbb{Z}_n$ has order $5$.

My solution was $m = 2$ and $n = 31$ as $2^5 = 32 \equiv 1 \pmod{31}$.

However, the solutions stated that:

If $(\mathbb{Z}_n, \times)$ has an element of order $5$, then we know that its order is divisible by $5$.

Is my answer not a counter-example to this statement?

Answer 1

The Lagrange Theorem for groups helps here: the element in question generates a subgroup. The order of the element is the order of the subgroup. The order of the subgroup divides that of the group, by Lagrange's theorem.

Answer 2

Here's an easier explanation. Note, $1^b=1$ for any integer b. This means once you get to 1, you can power your result to any power. Modular arithmetic groups, obey normal arithmetic, otherwise they wouldn't be useful in disproving numerical results. Your order is an exponent, by the exponent rule $(x^y)^{z}=x^{y\cdot z}$, we get the same result as multiplying your order by anything. This makes them congruent to each other, and to 1. Therefore any multiple of an order of 1 mod something, is also an exponent that produces 1 for the base used. The order of the group, is the highest exponent required to make any base element get to 1. After getting to one the congruences repeat, so the group order, must be a multiple of every element's individual orders.