Prove that $(\mathbb{Z}/2^n\mathbb{Z})^\times$ is generated by $−1$ and $5$.

Asked Oct 18 '24 at 01:54

Active Oct 18 '24 at 10:52

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Prove that $(\mathbb{Z}/2^n\mathbb{Z})^\times$ is generated by $−1$ and $5$.

So, I am trying to understand this problem. What makes sense to me is that $\langle -1 \rangle$ will always generate $\{1, 2^n-1 \}$ and $\langle 5 \rangle$ should generate the rest, but I am struggling to see how it would work out in a couple of examples, specifically $(\mathbb{Z}/8\mathbb{Z})^\times = \{1,3,5,7 \}$, but neither of my generators get to $3$.

edited Oct 18 '24 at 06:47

Bill Dubuque

282,220

asked Oct 18 '24 at 01:54

Bernardo Falcao

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 18 '24 at 01:54
4

$3\equiv (-1)(5)\pmod 8$. You also need to take products of $-1$ and powers of $5$. – Arturo Magidin Oct 18 '24 at 02:09
You can find this in Vinogradov, or other number theory texts. It's well-known and I think due to Gauß. – suckling pig Oct 18 '24 at 02:14
It's not cyclic for $n\gt2,$ which is interesting. – suckling pig Oct 18 '24 at 02:17
1

Duplicate almost surely. – Bill Dubuque Oct 18 '24 at 06:47
Yes, with the site search I found 3 duplicates within $\approx$ 5 minutes. – Martin Brandenburg Oct 18 '24 at 12:04

Prove that $(\mathbb{Z}/2^n\mathbb{Z})^\times$ is generated by $−1$ and $5$.

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