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
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.