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Consider the simple group $A_\lambda$, the alternating group on the set $\lambda$, which I will assume has regular cardinality. Recall that this is the smallest subgroup of all permutations of $\lambda$ containing the finite even permutations. In particular, only finitely many elements of $\lambda$ are moved by elements of $A_\lambda$.

I would like to find a subgroup $L \lt A_\lambda$ such that $A_\lambda/L$ is countable. I considered using the subgroup that fixes a specific countable subset $\omega \hookrightarrow \lambda$, but from my back-of-envelope scribblings this didn't seem to work out - or I just couldn't see it!

Since $|A_\lambda| = \lambda$, we need $|L| = \lambda$ also, which puts constraints on what $L$ can be.

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If $G$ is a simple group with a subgroup $H$ of countably infinite index, then the left-multiplication action of $G$ on the left-cosets of $H$ defines an embedding of $G$ into the group of all permutations of $\mathbb{N}$. Thus $\lambda$ is at most $2^{\aleph_0}$. – Chris Eagle Nov 23 '11 at 23:57
Ah, that's good to know. I originally just had $\lambda$ uncountable, but needed stronger. Thanks for disabusing me of this notion, it will save me a lot of time :) Feel free to add this as an answer. – David Roberts Nov 24 '11 at 0:01

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