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I have a set $X=\left \{1,...,n\right \}$ and the symmetric group on a set of n elements has order n!.

When $n=0$,why do we have $S_{\left \{1,...,0\right \}}=S_{\varnothing }$ ?

I know that n=0 means there are no elements in the set and $0!=1$ but why do we have ${\left \{1,...,0\right \}}={\varnothing }$? Is it not simply ${\varnothing }=\left \{\right \}$?

Thank you.

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2 Answers 2

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Your $X$ makes no sense when $n=0$ so it must be defined separately (if you really wanted to, but I don’t know anyone who cares about $S_0$). The natural thing is for $S_0$ to stand for the group of all bijections on the set with no object, the empty set. There is exactly one such function (the empty function) so $S_0$ is trivial, just like $S_1$. And, $0!=1!=1$ so everything is coherent.

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  • $\begingroup$ I was wondering what the teacher meant when they wrote $S_{\left \{1,...,0\right \}}=S_{\varnothing }$, it is only a shortened notation then I guess.I thought it was strange.Thanks again. $\endgroup$
    – user159729
    Jul 16, 2018 at 1:11
  • $\begingroup$ I wouldn’t have written that myself because it looks either wrong or weird or both. Don’t tell your teacher I said that. $\endgroup$
    – Randall
    Jul 16, 2018 at 1:12
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Think of $X = \{1,...,n\}$ as being shorthand for $X = \{i \in \mathbb{Z} \mid 1 \le i \le n\}$.

So if $n=0$ then $X=\emptyset$.

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  • $\begingroup$ I understand that but the notation ${\left \{1,...,0\right \}}={\varnothing }$ is what I have a problem with.The set ${\left \{1,...,0\right \}}$ is not empty but I'm guesssing it 's not literal. $\endgroup$
    – user159729
    Jul 16, 2018 at 1:33
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    $\begingroup$ The notation $\{1,...,n\}$ is kind of a bad notation, so the way to understand it in the "exceptional" cases is to rewrite it so that it becomes good notation. That's what I was trying to say. $\endgroup$
    – Lee Mosher
    Jul 16, 2018 at 1:41

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