Let $S$ be an infinite set of cardinality $\alpha$ and $G$ be a subgroup of $Sym(S)$. Let $\sigma(g)=\{s\in S \mid sg\neq s\}$ for each $g\in G$ and define $$Sym(S,\, \alpha)=\{g\in Sym(S)\mid |\sigma(g)|<\alpha\}.$$ I was trying to establish the cardinality of $Sym(S,\, \alpha)$.

What is the cardinality of $Sym(S,\, \alpha)$, and how to prove it?

Old Question

I argued pretty much as follows:

  • Every element $g\in Sym(S,\, \alpha)$ may be constructed in the following way: take a subset $T$ of cardinality strictly smaller than $\alpha$ and consider a permutation of $S$ which fixes all the elements of $S\backslash T$.

  • How many subsets of cardinality strictly less than $\alpha$ are there? For each cardinality $\beta <\alpha$ I find $\alpha^\beta=\alpha$ (this is not true) subsets of cardinality $\beta$. Since there are $\alpha$ cardinalities before $\alpha$ the total number is $\alpha$.

  • For each $L$ of this subsets I need to consider a permutation. Since $|L|<\alpha$ the set $|Sym(L)|=2^{|L|}\leq \alpha$.

So, at the end, it follows that $|Sym(S,\, \alpha)|=\alpha$.

Are my arguments correct?

  • $\begingroup$ I would avoid the notation $o(\sigma(g))$ since $\sigma(g)$ is not a group (I would rather use $|\sigma(g)|$). Also, in your third step $2^{|L|}\leq \alpha$ follows from the extended continuum hypothesis, doesn't it? $\endgroup$ Mar 2, 2016 at 9:22
  • $\begingroup$ $\aleph_\omega^{\aleph_0}\gt\aleph_\omega.$ $\endgroup$
    – bof
    Mar 2, 2016 at 9:24
  • $\begingroup$ @ClémentGuérin Yes the third step use GCH. I do not know how to avoid it. $\endgroup$
    – W4cc0
    Mar 2, 2016 at 11:12
  • $\begingroup$ @bof You're right, I'll change the question asking also what is the actual cardinality of $Sym(S,\, \alpha)$, and how to prove it. $\endgroup$
    – W4cc0
    Mar 2, 2016 at 11:14

1 Answer 1


The cardinality of $Sym(S,\alpha)$ is $\alpha^{<\alpha}$: that is, it is the supremum of the cardinals $\alpha^\beta$ where $\beta$ ranges over all cardinals less than $\alpha$.

First, I claim that $|Sym(S,\alpha)|\geq\alpha^{<\alpha}$. Indeed, fix a cardinal $\beta<\alpha$ and identify $S$ with the set $\beta\times(\alpha+1)$. For any function $f:\beta\to\alpha$, consider the permutation of $S=\beta\times(\alpha+1)$ which swaps $(x,\alpha)$ and $(x,f(x))$ for each $x\in \beta$ and is otherwise the identity. The support of this permutation has cardinality $\beta$, and so this defines an injection from the set of functions $\beta\to\alpha$ to $Sym(S,\alpha)$. Thus $|Sym(S,\alpha)|\geq\alpha^\beta$. Since $\beta<\alpha$ was arbitrary, this means $|Sym(S,\alpha)|\geq\alpha^{<\alpha}$.

Conversely, I claim $|Sym(S,\alpha)|\leq\alpha^{<\alpha}$. Indeed, if $g\in Sym(S,\alpha)$, then $g$ is uniquely determined by a pair of maps $\beta\to S$ for some $\beta<\alpha$: one map which is a bijection from $\beta$ to the support of $g$, and one map which then says where each element of the support is sent by $g$. There are $\alpha^\beta\cdot\alpha^\beta=\alpha^\beta$ pairs of functions $\beta\to S$. This shows that $$|Sym(S,\alpha)|\leq\sum_{\beta<\alpha}\alpha^\beta.$$ Now $\alpha^\beta\leq \alpha^{<\alpha}$ for each $\beta<\alpha$, and there are at most $\alpha$ cardinals less than $\alpha$. So the sum above is a sum of at most $\alpha$ terms, each of which is at most $\alpha^{<\alpha}$. Thus $$|Sym(S,\alpha)|\leq\alpha^{<\alpha}\cdot\alpha=\alpha^{<\alpha}.$$


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