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By the way, $A_\omega ^{fin}$ (group of even permutation of countable set) can't be embedded into compact group (hence $S_\omega^{fin}$ also can't).

It is well-known (proof uses theory of representations) that every compact group is isomorphic to a subgroup of cartesian product of unitary groups, so WLOG we may assume that $A_\omega^{fin}$ is embedded in $\prod U({n_i})$. Now let $f_i$ be a homomorphism $f:A_\omega^{fin} \rightarrow U({n_i})$ such that $\prod f_i$ is an embedding in $\prod U({n_i})$. Obviously Ker $f_i$ can't be trivial for all i, so for some i $Ker(f_i) \neq A_\omega^{fin}$ , but since $A_\omega^{fin}$ is simple $Ker (f_i) =(0)$ and $f_i$ is injective.

So, we've just proved that if simple group can be embedded in compact group it can be embedded in $U(n)$ for some n.

Now, it's easy to finish the proof (there are lot's of methods, actually). For example note that there are infinitely many pairwise commuting different elements in $A_\infty^{fin}$ A_\omega^{fin}$such that$x^2 = e $but there are only finitely many such elements in$U(n)$. Hope this helps. 3 added 2 characters in body; deleted 5 characters in body; added 1 characters in body By the way,$A_\omega ^{fin}$(group of even permutation of countable set) also can't be embedded into compact group (hence$S_\omega^{fin}$also can't). Proof(by a contradiction): It is well-known (proof uses theory of representations) that every compact group is isomorphic to a subgroup of cartesian product of unitary groups, so WLOG we may assume that$A_\omega^{fin}$is embedded in$\prod U({n_i})$and . Now let$f_i$be a homomorphism$f:A_\omega^{fin} \rightarrow U({n_i})$such that \prod f_i$\prod f_i$is an embedding in$\prod U({n_i})$. Obviously Ker$f_i$can't be trivial for all i, so for some i$Ker(f_i) \neq A_\omega^{fin}$, but since$A_\omega^{fin}$is simple$Ker (f_i) =(0)$and$f_i$is injective. So, we've just proved that if simple group can be embedded in compact group it can be embedded in$U(n)$for some n. Now, it's easy to finish the proof (there are lot's of methods, actually). For example note that there are infinitely many pairwise commuting different elements in$A_\infty^{fin}$such that$x^2 = e $but there are only finitely many such elements in$U(n)$. Hope this helps. 2 added 6 characters in body By the way,$A_\omega ^{fin}$(group of even permutation of countable set) also can't be embedded into compact group (hence$S_\omega$S_\omega^{fin}$ also can't).

It is well-known (proof uses theory of representations) that every compact group is isomorphic to a subgroup of cartesian product of unitary groups, so WLOG we may assume that $A_\omega^{fin}$ is embedded in $\prod U({n_i})$ and let $f_i$ be a homomorphism $f:A_\omega^{fin} \rightarrow U({n_i})$ such that \prod f_i is an embedding in $\prod U({n_i})$. Obviously Ker $f_i$ can't be trivial for all i, so for some i $Ker(f_i) \neq A_\omega^{fin}$ , but since $A_\omega^{fin}$ is simple $Ker (f_i) =(0)$ and $f_i$ is injective.
So, we've just proved that if simple group can be embedded in compact group it can be embedded in $U(n)$ for some n.
Now, it's easy to finish the proof (there are lot's of methods, actually). For example note that there are infinitely many pairwise commuting different elements in $A_\infty^{fin}$ such that $x^2 = e$ but there are only finitely many such elements in $U(n)$.