order of an element in a modulo group under multiplication

Question

Suppose $G$ is the group $ℤ_{37}^\times$ under multiplication.

Then is there a way that I can prove the order of the element $2$ in $G$ is $36$ without finding all the powers of $2$ until I get unity?

Answer 1

Lagrange's theorem says you only have to check $2^2,2^3,2^4,2^6,2^9,2^{12}$ and $2^{18}$. The checking itself is a lot easier if you exploit relations like $2^6\cdot 2^3=2^9$ and $(2^6)^2=2^{12}$.

Answer 2

Yes, by the Order Test below, $\,2\,$ has order $36\iff 2^{12}\neq 1\neq 2^{ 18}$

Order Test $\,\ \,a\,$ has order $\,n\iff a^n \equiv 1\,$ but $\,a^{n/p} \not\equiv 1\,$ for every prime $\,p\mid n.\,$

Proof $\ (\Leftarrow)\ $ If $\,a\,$ has $\,\rm\color{#c00}{order\ k}\,$ then $\,k\mid n.\,$ If $\,k < n\,$ then $\,k\,$ is proper divisor of $\,n\,$ therefore $\,k\,$ must omit at least one prime $\,p\,$ from the unique prime factorization of $\,n,\,$ hence $\,k\mid n/p,\,$ say $\, kj = n/p,\,$ so $\,a^{n/p} \equiv (\color{#c00}{a^k})^j\equiv \color{#c00}1^j\equiv 1,\,$ contra hypothesis. $\ (\Rightarrow)\ $ By definition of order.