Proving a multiplicative group module n is generated in a handy way

Question

Prove that $(\mathbb{Z}/23\mathbb{Z})^*$ is generated by $5\pmod{23}$

Is there a tactic or theorem that could help me finding the power of $5\pmod{23}$ for each element?

I was thinking about using Euler's theorem or Fermat's Little theorem, but I don't see how they could help me get for example $4\pmod{23}$ as generated by $5\pmod{23}$.

Thanks in advance

@DonThousand Wouldn't that be more suited to show that $5$ generates the additive group $\Bbb Z/23\Bbb Z$? — Arthur, Apr 02 '20 at 20:12
@Arthur Yes, but I think it can be used for both. Clearly, Euler's Theorem is the most useful, but OP is confused by them so. — Rushabh Mehta, Apr 02 '20 at 20:13
Would it help you to know that you are trying to show 5 is a primitive root modulo 23? — Eric Towers, Apr 02 '20 at 20:14

score 2 · Answer 1 · edited Apr 02 '20 at 20:15

2

You can check that $5^{11} \ne 1 \pmod{23}$ and $5^{2}\ne 1 \pmod{23}$ using exponentiation by squaring.

edited Apr 02 '20 at 20:15

gt6989b

54,422

answered Apr 02 '20 at 20:10

vujazzman

2,028

1

also, if $5^{11} $ were $\equiv1\bmod23$ then $5$ would be a quadratic residue mod $23$, but it's not, as can be shown by quadratic reciprocity – J. W. Tanner Apr 02 '20 at 20:18

Proving a multiplicative group module n is generated in a handy way

1 Answers1