Is there a formula for finding all generators of a cyclic group.

Question

Queston;

Given that 2 is a generator of cyclic group U(25), find all generators.

I am only conversant with the finding the mod which is very long with this question. Pls can someone enlighten me on how to get it done faster.

I am new to the cyclic group.

solution U(25) = {1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24}

2^20 = 1 (mod 25)

Thanks

Answer 1

In a cyclic group of order $n$ generated by $g$, the order of $g^k$ is $\dfrac{n}{\gcd(n,k)}$.

In particular, the generators are $g^k$ with $\gcd(n,k)=1$.

In your case, $g=2$ and $n=\phi(25)=20$.

Therefore, the generators of $U(25)$ are $2^k$ for $k$ coprime with $20$, that is, $k$ odd not a multiple of $5$.

Answer 2

The formula you need is the one that tells you that if $g$ is a generator of the cyclic group $G$ of order $n$, then $g^{k}$ is a generator if and only if $\gcd(n, k) = 1$.