Let $S_{\kappa}$ denote the symmetric group on some set of cardinality $\kappa$. Does there exist a generating set $X \subset S_{\kappa}$ such that $|X| < |S_{\kappa}|$ ($\stackrel{?}{=} 2^{\kappa}$)?
More specifically, does there exist a countable set of generators for $S_{\mathbb{N}}$? And if so, can it be constructed without the Axiom of Choice?
It seems clear that the answer to the first and third questions is 'no'. Indeed, if a set of generators $X$ is of infinite cardinality $\alpha$, then the group so generated cannot have cardinality greater than $\alpha$, since it is a quotient of the free group generated by $X$, which in turn is a quotient of the free monoid generated by $X\cup \{x^{-1}:x\in X\}$, and this free monoid has cardinality $\sum_{n\geq0}\alpha^n = \alpha$.
Well, to finish the claim, we need to check that the symmetric group $S_\kappa$ has cardinality $2^\kappa$ (the second question). This is certainly true: suppose given a well-ordering of $\kappa$. Then there are $\kappa^\kappa$ many permutations $f$ where $f(\alpha)$, for "even" $\alpha < \kappa$, is the least ordinal in the set $\kappa-f(\{\beta<\alpha\})$, and for "odd" $\alpha<\kappa$ is any element in $\kappa-f(\{\beta<\alpha\})$. ("Even" means a limit ordinal plus an even finite ordinal, and mutates mutandis for "odd".)
If the set is infinite then $S_{\kappa}$ and $X$ have the same cardinality.
