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How do you find all the generators $(\Bbb Z_{19})^\times$?

I know the generators for $\Bbb Z_n$ is all elements, $a$, where $\gcd(a,n)=1$, so for $\Bbb Z_{19}=\{1,2,\dots, 18\}$ since $19$ is prime.

I am not sure how to do it for $(\Bbb Z_n)^\times.$

Shaun
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user520403
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  • $(\Bbb Z/p)^{\times}=U(p)\cong C_{p-1}$ is a cyclic group for $p$ prime. Take $p=19$. For $n$ not prime, it is more complicated, but for example $U(pq)\cong U(p)\times U)q)$ for different odd primes $p,q$. – Dietrich Burde Dec 01 '19 at 20:08
  • $2$ is a generator of $\mathbb Z_{19}^\times$ because $2^{18}\equiv1$ but $2^6,2^9\not\equiv1\pmod{19}$; so is $2^a$ where $a\in{5,7,9,11,13,17}$ is relatively prime to $18$ – J. W. Tanner Dec 01 '19 at 20:09

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