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From Cayley's theorem , we know that any group $G$ can be embedded in its permutation group $S(G)$ ; I would like to ask , If $G$ be an infinite group , then does there exist a surjection from $G$ onto the permuttaion group of $G$ ? And even if a surjection exists for some $G$ , can there ever exist a surjective homomorphism of $G$ onto $S(G)$ when $G$ is infinite ?

Please help . Thanks in advance

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    For any infinite set $X$ there are $2^{|X|}$ bijections from $X$ to itself. – Tobias Kildetoft Jul 05 '16 at 09:09
  • @TobiasKildetoft : Umm why is that ? Could you please explain .. –  Jul 05 '16 at 09:42
  • The finitary symmetric group is the subset of $S_{|X|}$ consisting of elements which move only finitely many elements of $X$. Then $|X|=|S_{|X|}|$, so the cardinality argument no longer works. – user1729 Jul 05 '16 at 09:59
  • @user1729 Yes, so if $G$ is the group of all finitary permutations of the infinite set $X$, then $G$ is certainly isomorphic to the finitary symmetric group on the set $G$, but how is that relevant to the question? – Derek Holt Jul 05 '16 at 11:29
  • @Derek I was trying to suggest that the question might be altered and then might admit a positive answer: Let $\operatorname{FSym}(\mathbb{N})$ be the finitary symmetric group on the set $\mathbb{N}$ and let $G$ be a countable group with $G\not\cong \operatorname{FSym}(\mathbb{N})$. Can there exist a surjection $G\rightarrow \operatorname{FSym}(\mathbb{N})$? – user1729 Jul 05 '16 at 11:39
  • (In fact, this does admit a positive answer, e.g. $F_{|\mathbb{N}|}\rightarrow \operatorname{FSym}(\mathbb{N})$.) – user1729 Jul 05 '16 at 11:39
  • @user1729 : But I actually wanted that there be no surjection , much in the same spirit of Cantor's theorem . Still , thanks for your comment –  Jul 05 '16 at 11:51

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No: For any infinite set $X$, the cardinality of $S(X)$ is the same as that of its power set $\mathcal P(X)$ and therefore (by Cantor's theorem) different from the cardinality of $X$ itself.

We need to know that there is a bijection $f: X\times X\to X$. That this exists for every infinite $X$ is a well-known consequence (and, indeed, equivalent) of the axiom of choice.

Now in one direction, every permutation $\sigma\in S(X)$ can be encoded as a subset of $X$, namely $\{ f(x,\sigma(x)) \mid x\in X \}$, so there are at least as many subsets as there are permutations.

In the other direction: Choose two different $x,y\in X$. Then we can encode any subset $A\subseteq X$ as a permutation of $X$, namely the one that swaps $f(x,a)$ and $f(y,a)$ for all $a\in A$, and leaves all other elements of $X$ unchanged. This shows there are at least as many permutations as there are subsets.

The Cantor-Bernstein theorem now tells us that $S(X)$ and $\mathcal P(X)$ are equinumerous.

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