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Nov 30, 2024 at 0:41 history edited Arturo Magidin CC BY-SA 4.0
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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
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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