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I'm trying to show that for any infinite cardinal $\kappa$ there is a simple group $G$ of size $\kappa$, I tried to use the compactness theorem and then ascending Löwenheim-Skolem, but this is impossible as the following argument shows:

Suppose $F$ is a set of sentences in a first-order language containing the language of groups such that for any group $G$ we have $G\vDash F$ iff $G$ is a simple group, let $\varphi$ be a sentence such that $G\vDash \varphi $ iff $G$ is abelian. Then as for each prime $p$, $\mathbb Z_p$ is simple and abelian, by the compactness theorem there is an infinite abelian simple group $G$, and thus using Löwenheim-Skolem we obtain an uncountable simple abelian group $G$. Pick a non-zero element $a\in G$, then $\langle a \rangle$ is a proper normal subgroup of $G$. Contradiction.

Note that the contradiction is not obtained by the existence of an infinite abelian simple group, something which I don't whether it's true or not, but by the assumption of the axiomatization of the simple groups, and the use of Lowënheim-Skolem theorem.

Now, is there a way to prove this assertion?

Thanks.

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Pardon my ignorance of logic, but is there such a set of sentences $F$? It would appear necessary to quantify over all subgroups of $G$, which is a second-order operation. –  Ryan Reich Jun 2 '13 at 5:58
    
yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic –  Camilo Arosemena Jun 2 '13 at 13:20
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A deleted answer included the following link with nice additional information: mathoverflow.net/questions/32908/… (In particular, Simon's answer shows that simplicity is not first-order.) –  Andres Caicedo Jun 3 '13 at 22:32

1 Answer 1

up vote 14 down vote accepted

For any (finite or infinite) cardinal $\kappa$, if $\kappa \ge 5$ then the finitary alternating group $A(\kappa)$ is simple. This is the group consisting of permutations of $\kappa$ that have finite support and are even. (If $\kappa$ is finite then this is just the ordinary alternating group $A_\kappa$.) If $\kappa$ is infinite, then $A(\kappa)$ has cardinality $\kappa$.

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Hi Trevor. Didn't know about Groupprops, thanks for the link! –  Andres Caicedo Jun 2 '13 at 6:58
    
@Andres Hi Andres, I didn't know about it either; I just heard someone mention this fact in a talk today -- then I happened to see this question and so I googled "finitary alternating group" or something like that and it turned up. –  Trevor Wilson Jun 2 '13 at 7:05

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