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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).

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 $\prod U({n_i})$. 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.
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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.
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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).

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 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 $U(n)$.

Hope this helps.
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