Is there a simple group of any (infinite) size?

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.

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 $\kappa$.