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It seems like there is a simple field property I'm missing. If we look at the group $Z_n$ for any $ n>=3$ can we say that : It is possible that $(1*2*3*....*(n-1))^2 = 0 $ ?

Can you please advise?

Guy

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    http://mathworld.wolfram.com/WilsonsTheorem.html2012-10-29
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    @lab bhattacharjee Thank you but I know Wilsons Theorem but I can't use it here.2012-10-29
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    Sadly, the abelian group $\Bbb{Z}_n$ does not extend to a field for arbitrary $n$. If we only consider the ones which do extend to fields, note that there are no nilpotent elements, so equivalently, we want to determine whether there exists $p$ such that $(p-1)!$ is divisible by p, a trivial result.2012-10-29
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    @peoplepower finding p such that (p-1)! is divisible by P gives me what?2012-10-29

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Try it: if $n=3$ you get $(1\cdot 2)^2=1$ in $\Bbb Z_3$, not $0$, but if $n=4$ you get $(1\cdot2\cdot3)^2=0$ in $\Bbb Z_4$, so it clearly depends on $n$. Exercise: Prove that $\big((n-1)!\big)^2=0$ in $\Bbb Z_n$ if and only if $n$ is composite.

Of course this means that if $n$ is composite, then $\Bbb Z_n$ is not a field. If you’re interested in fields, you want $n$ to be prime.