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Oct 20, 2014 at 22:34 comment added Avshalom If $\kappa$ is strongly compact, then $\kappa$-freeness implies freeness (like singular compactness), so there is a $L_{\kappa, \kappa}$-sentence saying a group is $\kappa$-free. For the converse, a downward Loewenheim Skolem property is used, as you say, but I am not sure of all names of the people involved.
Oct 20, 2014 at 22:01 comment added Asaf Karagila Interesting. The strongly compact cardinal enters from the Magidor-Vaananen work about Lowenheim-Skolem-Tarski numbers for second-order logic. Right?
Oct 20, 2014 at 21:51 history answered Avshalom CC BY-SA 3.0