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(self answered question, thanks for the hints Derek Holt provided:-))

problem 18,section 4 chapter 2 in Herstein's abstract algebra: Using the results of Problem 15 and 16,prove that if p is an positive odd prime number, then $(p-1)!\equiv -1 \pmod p$. (self answered question, thanks for the hints Derek Holt provided:-))


Problem 15: If $x^2\equiv 1 \pmod p$, then $x\equiv 1\pmod p$ and/or $x\equiv -1\pmod p$
Problem 16: If $G=${$a_k|1\leq k\leq n$} is a finite abelian group, then $(\prod^n_1 a_k)^2=e.$

pxc3110
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  • This has been asked on MSE before, but Wikipedia has a few proofs. – fuglede Apr 25 '14 at 10:49
  • @fuglede I'll study wikipedia's proof later, but first I want to know how to solve it in the way required by the author of the book. – pxc3110 Apr 25 '14 at 11:01
  • Hint: to see that it is not $1$, note that each element apart from $1$ and $-1$ has an inverse mod $p$ which is not equal to itself. – Tobias Kildetoft Apr 25 '14 at 11:04
  • The basic idea is that you can pair off every element of $U_p$ iwth its inverse in $U_p$, but this leaves self-paired elements $x$, which are those satisfying $x^2 \equiv 1 \bmod p$. Prove that the only self-paired elements are $1$ and $p-1$, and deduce the result. – Derek Holt Apr 25 '14 at 11:05
  • You may also be interested in the generalization http://math.stackexchange.com/questions/474214/product-of-elements-of-a-finite-abelian-group/ – Tobias Kildetoft Apr 25 '14 at 11:08
  • No, $-1$ (which equals $p-1$) is its own inverse. – Tobias Kildetoft Apr 25 '14 at 11:29
  • In what argument? – Tobias Kildetoft Apr 25 '14 at 11:32
  • Your argument starts with a false statement, as I just mentioned. – Tobias Kildetoft Apr 25 '14 at 11:33
  • If $(p-1)!=1$ then $ 2 = (p-1)!+1$ (mod p) making $n=2$ (mod p) a prime, since 1 is always a rest after division with t, 0<t<p. This should hold for all primes, but it is not true since e.g. $2= 7+2 =9$ mod (7) $=3*3$ mod(7) (with $0<3<7$). – Mikael Jensen Apr 25 '14 at 12:05
  • @MikaelJensen What do you mean by a prime (mod $p$)? – Tobias Kildetoft Apr 25 '14 at 12:08
  • I am thinking of a number a where no two numbers b and c can exist < p so that b*c=a (mod p) – Mikael Jensen Apr 25 '14 at 12:13
  • @MikaelJensen so you seem to be concluding that if $(p-1)!$ does not depend on $p$ (mod $p$) then it cannot be $1$? – Tobias Kildetoft Apr 25 '14 at 12:17
  • I am not sure about what "depends on p (mod p) means" (nor am I sure about my own proof but I came to the same problem in the end of my private proof as pxc3110, and that is how I reasoned). – Mikael Jensen Apr 25 '14 at 12:21
  • @MikaelJensen I mean that you have "this should hold for all primes" as part of the argument. But it is not a priori obvious that it does not vary depending on $p$. – Tobias Kildetoft Apr 25 '14 at 12:22
  • You need to point out where in my chain of reasoning I go wrong, e.g I start with 2=1+(n-1)! which always must be 1 after division with t<p. Then I find a counterexample. – Mikael Jensen Apr 25 '14 at 12:24
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    @MikaelJensen I is hard to point to something specific, since the conclusion is correct. But the argument hinges on the fact that $(p-1)!$ has the same remainder mod $p$ for any prime $p$, and this is not at all obvious. – Tobias Kildetoft Apr 25 '14 at 12:26
  • It is (p-1)!+1 that has the same remainder (1) after division. (p-1)! also has the same remainder of course (0). – Mikael Jensen Apr 25 '14 at 12:27
  • @MikaelJensen Wy? – Tobias Kildetoft Apr 25 '14 at 12:28
  • (p-1)! is a product of all numbers 1-(p-1) and herefore can be divided by any number betwen 1-(p-1). – Mikael Jensen Apr 25 '14 at 12:31
  • @MikaelJensen Why does that say anything about the remainder mod $p$? – Tobias Kildetoft Apr 25 '14 at 12:31
  • Perhaps it doesn't but give me a counterexample, that is really what I would need. – Mikael Jensen Apr 25 '14 at 12:34
  • This is my first part by the way (which may also be flawed): When k goes from 1 to p-1, $ (p-1)!/k$ will form a permutation of the numbers 1 to p-1, since $ (p-1)!/k1 =(p-1)! /k2$ implies that $k1=k2$ (for p prime). For this permutation, multiplication of all the left hand terms = multiplication of all the right hand terms gives the equality $ [(p-1)!]^{p-1}/ (p-1)!= (p-1)! $, so that $[(p-1)!]^(p-1) = (p-1)!^2$, and $[(p-1)!]^(p-1) = 1$ (Fermat’s little theorem ) and therefore (p-1)! is either 1 or -1. – Mikael Jensen Apr 25 '14 at 12:47
  • I tried to get the exponents right:This is my first part by the way (which may also be flawed): When k goes from 1 to p-1, $ (p-1)!/k$ will form a permutation of the numbers 1 to p-1, since $ (p-1)!/k1 =(p-1)! /k2$ implies that $k1=k2$ (for p prime). For this permutation, multiplication of all the left hand terms = multiplication of all the right hand terms gives the equality $ [(p-1)!]^{p-1}/ (p-1)!= (p-1)! $, so that $[(p-1)!]^{p-1} = (p-1)!^2$, and $[(p-1)!]^{p-1)} = 1$ (Fermat’s little theorem ) and therefore (p-1)! is either 1 or -1. – Mikael Jensen Apr 25 '14 at 12:53
  • @Kildetoft Assume that $(p-1)! = 1$ (mod p). Then $(p-1)! +1 = 2 $ (mod p) and $2 =1$ (mod p) (since $(p-1)!=0$) for all numbers between 1 and (p-1) – for any prime p. Therefore $(p-1)!=-1$ Surely this is obvious. – Mikael Jensen May 02 '14 at 13:38
  • See also: http://math.stackexchange.com/questions/1368459/proof-of-wilsons-theorem-using-concept-of-group – Martin Sleziak Jan 30 '16 at 21:56
  • @Martin Sleziak That's later than mine – pxc3110 Feb 09 '16 at 07:23
  • @pxc3110 But it is still a useful link for readers of your question. Which is why I posted it. – Martin Sleziak Feb 09 '16 at 07:27

2 Answers2

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Proof:
Let p be an odd prime number.
Consider the group $U_p=${equivalent classes of $a$|$p>a>0$, $gcd(a,p)=1$}
(equivalent relation:$a\equiv b \pmod p$, binary operation:[a][b]=[ab]).
p is a prime,so $U_p=${[a]|$1\leq a\leq p-1$}.
Since $U_p$ is a finite abelian group, $(\prod_{1}^{p-1}[a])^2=\prod_{1}^{p-1}[a]*\prod_{1}^{p-1}[a^{-1}]=[1]$,
so $[(p-1)!]^2\equiv 1 \pmod p$,
therefore, either $(p-1)!\equiv 1\pmod p$$(!)$ or $(p-1)!\equiv -1 \pmod p(!!)$.
Now we'll show that the first statement (!) is incorrect, thus forcing the second statement to be true.
Assume that there exists an element $[a],2\leq a \leq p-1$, such that $[a]^2=[1]$,
therefore $a^2\equiv 1\pmod p$,
or $p|(a-1)(a+1)$,
so $p|(a-1)$ and/or $p|(a+1).$
so $a-1\geq p$ and/or $a+1\geq p$
Therefore, $a=p-1$
So,only $[1]$ and $[p-1]$ are self-paired.
Therefore, consider the product: $x=[1]...[p-1]$,
Apart from $[1]$ and $[p-1]$, all other elements are paired together with their inverses,
so $x=[(p-1)!]=[p-1]$ which doesn't equal to [1] since it requires p|p-2, which is true only when p=1,2.
so $[(p-1)!]\neq [1]$.
So it is false that $(p-1)!\equiv 1\pmod p$.
This forces (!!)to be true,so it must be true that $(p-1)!\equiv -1\pmod p$.
This completes the proof.

pxc3110
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Hints in short:

** The set $\;\Bbb F_p^*:=\{1,2,...,p-1\}\;$ is an abelian (even cyclic) group wrt multiplication modulo $\;p\;$

** Now, pair up the elements of $\;\Bbb F_p^*\;$ in pairs $\;\left(a,a^{-1}\right)\;$ . Note there are exactly two elements in $\;\Bbb F_p^*\;$ that are paired with themselves.

** Finally, in the product $\;(p-1)!=1\cdot 2\cdot\ldots\cdot(p-1)\;$ change the order as to have each element paired with its inverse, as above.

** Well, what did you get in the last step? ...

DonAntonio
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