TITLE: Is there an alternating group on any given infinite set?
QUESTION [10 upvotes]: Given any infinite set $X$ does there exist a surjective group homomorphism $\varepsilon: \mathrm{Perm}(X) \twoheadrightarrow \{+1, -1\}$? 
Here, $\mathrm{Perm}(X)$ denotes the group under composition of all permutations of $X$. 
If such a homomorphism exists, then is it unique?

REPLY [7 votes]: No. Such surjective homomorphism does not exist for any infinite $X$ 
Let's for any cardinality $\alpha$ define $S_\alpha$ as the group of all permutations of a set of cardinality $\alpha$.
The Theorem 6 of "Some remarks on commutators" by Oystein Ore states:

If $\alpha \geq \aleph_0$ then $\forall g \in S_\alpha \exists h, k \in S_\alpha$, such that $g =[h, k]$

From that it follows that $S_\alpha' = S_\alpha$ and thus all non-trivial quotients of $S_\alpha$ are non-abelian. Unfortunately, $C_2$ is a non-trivial abelian group.
However, such homomorphism would be possible if you slightly alter the definition:
Let's for any cardinality $\alpha$ define $FS_\alpha$ as the group of all permutations of the set of cardinality $\alpha$, that have finite support. Then the set $A_\alpha$ of all permutations from $FS_\alpha$ that can be expressed as a product of even number of transpositions will be a normal subgroup of $FS_\alpha$ of index $2$ (and that is exactly the thing you want).