Symmetric group and the empty set

Question

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.

Answer 1

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.

Answer 2

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 $X=\emptyset$.