I have an old math exam question with the solution included, but there is a certain step of the solution I don't understand.
Task: Determine the order of $7$ in $\mathbb{Z}_{44}^*$
Solution: From the Chinese Remainder Theorem, $\mathbb{Z}_{44}^*$ $\cong$ $\mathbb{Z}_4^*\times \mathbb{Z}_{11}^*$. Therefore, it suffices to determine the orders of 7 in both factors $\mathbb{Z}_4^*$ and $\mathbb{Z}_{11}^*$. In $\mathbb{Z}_4^*$, the order is obviously $2$ (because $7 \not\equiv 1 \ \textrm{mod} \ 4$ but $7^2 = 49 ≡ 1 \ \textrm{mod} \ 4)$, and hence ord($7$) in $\mathbb{Z}_{44}^*$ is an even number. In the second factor, we have $7^2 = 72 = 49 ≡ 5 \not\equiv 1 \ \textrm{mod} \ 11$. So the order is not $2$, but either $5$ or $10$. In both cases, the order (in $\mathbb{Z}_{44}^*$) must then be a multiple of $5$, and therefore can only be $10$.
Question: How do you suddenly come to the conclusion that the order of $7$ must either be $5$ or $10$ in $\mathbb{Z}_{11}^*$ without explicit calculation which would take too long in an exam situation? Why couldn't it be 6 for example.
Some definitions:
$\mathbb{Z}_{n}^*$ $:=$ All numbers that are coprime to n
ord($g$) $=$ The order of an element $g$ in a group $G$, denoted as ord($g$), is the smallest positive integer r such that $g^r = e$ ($e$ being the identity element).
Let say $\operatorname{ord}(7)$ is not a divisor of $10$, for example like $6$ as you said, then we have $$7^{\operatorname{ord}(7)}=7^6\equiv 1 \mod 11\tag{*}$$And by Fermat Little Theorem, $7^{10}\equiv 1\mod 11$, so squaring both sides on $(*)$ give $7^{12}\equiv 7^2\equiv 1^2=1\mod11$, which is contradicts with the order of $7$.
This is also known as Lagrange Theorem in group theory, where the order of an element must divide the order of the group.