How to understand the grouping of integers from $2$ to $p-2$ in the proof af Wilsons theorem?
I don't understand how you can group the integers $2$ to $p−2$ into $(p-3)/2$ and next how you become the $2*3*...=1 (mod p)$
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The proof seems perfectly understandable to me, so it would help if you could be a bit more specific about what it is that you find confusing. – Ben Grossmann Dec 13 '17 at 17:58
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I don't understand how you can group the integers 2 to p−2 into (p-3)/2 and next how you become the 23...=1 (mod p) – WinstonCherf Dec 13 '17 at 18:02
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1Think about the grouping as a re-arrangement of the terms of the product, so instead of $2 \cdot 3 \cdot 5 \dots$ we have $x \cdot x^{-1} \cdot y \cdot y^{-1} \dots$. – Dan Brumleve Dec 13 '17 at 18:04
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For example, $2 \cdot 4$ and $3 \cdot 5$ are both congruent to $1$ mod $7$. So $2 \cdot 4 \cdot 3 \cdot 5 \cdot 1 \cdot 6 \equiv (7-1)! \equiv -1~ \text{mod}~{7}$ – Dan Brumleve Dec 13 '17 at 18:07
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Does this answer your question? Wilson's theorem intuition – tryst with freedom Mar 12 '22 at 15:09
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I think the best way to illustrate what's happening is to apply the proof to a specific value of $p$. Take, for example, $p = 11$.
We can solve $2x \equiv 1 \pmod {11}$ with $x = 6$. We can solve $3x \equiv 1 \pmod {11}$ with $x = 4$. We can solve $5x = 1 \pmod {11}$ with $x \equiv 9$. We can solve $7x \equiv 1 \pmod{11}$ with $x = 8$.
Note that in the above, we have accounted for all numbers from $2$ to $(11-2) = 9$. We can now rearrange the product $$ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 =\\ (2 \cdot 6) \cdot (3 \cdot 4) \cdot (5 \cdot 9) \cdot (7 \cdot 8) \equiv\\ 1 \cdot 1 \cdot 1 \cdot 1 \equiv 1 \pmod{11} $$ That is, by grouping the integers from $2$ to $11 - 2$ into $\frac{11 - 3}{2} = 4$ pairs of multiplicative inverses, we have shown that $$ 2 \cdot 3 \cdots (11 - 2) \equiv 1 \pmod{11} $$ which allows us to proceed with the proof.
Ben Grossmann
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Thnx, this is more clear now! They also use a non-trivial definition of the inverse with mod p, how does this follow from the normal definition of the inverse that's given by if (a,m)=1 and a as integer then ä is inverse of a modulo m if aä=1(mod m)? – WinstonCherf Dec 13 '17 at 18:20
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For example: if $a = 2$, then we see that $\bar a = 6$ is the inverse to $a$ because we indeed have $a \bar a \equiv 1 \pmod{11}$ – Ben Grossmann Dec 13 '17 at 18:25