Timeline for Prove that if $g \in \mathbb{Z}_{p_1}^{*}$ has the property $g^{p_2} \equiv 1 \mod p_1$ then $S$ is a proper subgroup of $\mathbb{Z}_{p_1}^{*}$
Current License: CC BY-SA 4.0
9 events
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| Nov 30, 2024 at 0:41 | history | edited | Arturo Magidin | CC BY-SA 4.0 | edited body |
| Nov 28, 2024 at 19:56 | comment | added | Bill Dubuque | Re: your wrong path: generally in any group of order $\,n,\,$ if $\,g^k = 1\,$ and $\,k\,$ is coprime to $\,n\,$ then $\,g = 1,\,$ since - by Lagrange - the order of $\,g\,$ divides the coprimes $\,k,n\,$ so it must be $1$, hence $\,S = \langle g\rangle = \{1\}\,$ (which is trivially closed under multiplication). This is what happens in your case when the prime $\,p_2\nmid p_1-1.\,$ So nothing is "wrong" in this case. But none of this is relevant to your original problem. | |
| Nov 28, 2024 at 19:41 | comment | added | Bill Dubuque | Above is a simply a special case of ubiquitous mod order reduction. Note also $\,(g^i)^{-1} = g^{k-i}\,$ by $\,g^{k-i}g^i = g^k = 1.\ \ $ | |
| Nov 28, 2024 at 19:40 | comment | added | Bill Dubuque | You are going down the wrong path. $p_1$ plays no role here (other than the fact that $\Bbb Z_{p_1}^*$ is a monoid, i.e. closed under multiplication). Similarly primality is not relevant. For $\,g\,$ in any monoid, if $\,g^k = 1\,$ for $\,k\ge 1\,$ then $\,S = \{1,g,g^2,\ldots,g^{k-1}\}\,$ is closed under multiplication simply because a product has form $\,g^n,\,$ and $\,\color{#c00}{g^k=1}\Rightarrow g^n = g^{n\bmod k},\,$ since by division $\,n = kq+r,\ 0\le r\le k-1,\,$ so $\, g^n = (\color{#c00}{g^k})^q g^r = \color{#c00}{1}^kg^r = g^r.\ \ $ | |
| S Nov 28, 2024 at 13:32 | history | suggested | Long-Ping Li | CC BY-SA 4.0 | Formatting can be improved. |
| Nov 28, 2024 at 13:31 | review | Suggested edits | |||
| S Nov 28, 2024 at 13:32 | |||||
| Nov 27, 2024 at 12:03 | vote | accept | KSI | ||
| Nov 27, 2024 at 11:46 | answer | added | EDX | timeline score: 2 | |
| Nov 27, 2024 at 10:58 | history | asked | KSI | CC BY-SA 4.0 |