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I know that $2^{100} \equiv 1 \pmod {125}$ because $\phi(125)=100$. $125=5^3$ is also the perfect power of an odd prime, so it has at least one primitive root. So, it is reasonable to check if $2$ is a primitive root mod $125$.

To check this, it would suffice to find every divisor of $100$ as a power of $2$, but that would take a longer time than I believe to be necessary, because I read once that we only need to check the divisors $2^2\cdot 5$ and $2\cdot 5^2$ as powers of $2$. Indeed, neither of them are the orders of $2$ mod $125$, and I am also told that $2$ is a primitive root mod $125$. However, I don't quite understand why we only need to check $2^2\cdot 5$ and $2\cdot 5^2$.

On top of that question, how can we then generalize this way of checking to other mods?

mpnm
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1 Answers1

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Let us say that $|G|=n$ and you want to check if $g \in G$ is a primitive root. Then what you should check is the following: $$\text{is } \quad g^{n/q} \not\equiv 1 \pmod{n} \, \quad \forall \quad \text{prime divisors } q \text{ of } n?$$ If this holds then $g$ is a primitive root otherwise not.

Proof: We will prove this by contradiction. Suppose $g^{n/q} \neq 1$ for all prime divisors $q$ of $n$ but $\text{ord}(g)=m <n$. Note that order of $g$ is $m$ tells us that $g^m=1$. Moreover since the order of an element must divide the order of the group, therefore $m | n$. In other words, there exists an integer $t$ such that $n=mt$. Since $m<n$, therefore $t>1$. By prime factorization theorem, $t$ must have at least one prime factor, let us call it $p$. Observe that since $p | t$ and $t | n$, therefore $p$ is also a prime factor of $n$. Now consider $$g^{n/p} = g^{mt/p}=\left(g^{m}\right)^{t/p}=1^{t/p}=1.$$ But this violates the condition that $g^{n/q} \neq 1$ for all prime divisors $q$ of $n$. Hence our assumption that $m<n$ is false. Therefore $m=n$ and hence $g$ is a generator of $G$.

So in your problem: all you need to do is check the following:

Is $$2^{100/2}\equiv 2^{50} \equiv 1 \pmod{125} ?$$ Is $$2^{100/5}\equiv 2^{20} \equiv 1 \pmod{125} ?$$ If the answer to both these is NO, then $2$ is primitive root.

Anurag A
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