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Just today, I realised that I had been mis-interpreting the FTFGAG. One can speak of the $2^\infty$-torsion subgroup or the $2'$-torsion subgroup, the $3^\infty$-torsion or the $3'$-torsion subgroup, or even just the torsion subgroup of a FGAG or the maximal torsion-free quotient … so surely one can speak of the torsion-free subgroup, right? In fact, when I was corrected on this, my first thought was to reply: "just take the subgroup consisting of all infinite-order elements", and it only occured to me as I was saying it to wonder how the identity element would squeeze its way into this so-called 'subgroup'.