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Looking only at the electric charge, three generations of sleptons can be arranged in a 24 of SU(5), as we learnt ten years ago.
[tex]\begin{pmatrix}
\nu & \nu & \nu & e & e \\
\nu & \nu & \nu & e & e \\
\nu & \nu & & e & e \\
\bar e & \bar e & \bar e & \nu & \nu \\
\bar e & \bar e & \bar e & \nu & \nu \\
\end{pmatrix}
[/tex]

SU(5) iself can be decomposed as SU(3) times SU(2), amounting to two kinds of flavors. Five preons if you wish. They have electric charges -1/3 and +2/3 respectively,
[tex]
5 = 3 +2 = (-1/3, -1/3, -1/3) + (2/3, 2/3)
[/tex]

An inconveniet of this representation is that we (meaning, I) do not know how to fold it into spinors. It is interesting because it looks unique in some sense I described ten years ago, and because it gives a lot of extensions. For starters, quark charges.

The 24 above, got out from [itex]5 \times \bar 5[/itex], gave us three generations of leptons. The 15 out from [itex]5 \times 5[/itex] will give three generations of (still-uncolored) anti-squarks

[tex]\begin{pmatrix}
-2/3 & -2/3 & -2/3 & 1/3 & 1/3 \\
& -2/3 & -2/3 & 1/3 & 1/3 \\
&& -2/3 & 1/3 & 1/3 \\
&& & 4/3 & 4/3 \\
&&&& 4/3 \\
\end{pmatrix}
[/tex]

and well, three "half-generations" of a peculiarly charged squark, which adds some puzzlement to the folding into fermions.

All of them, [itex] 15 + \bar {15} + 24 = 54 [/itex] should be easy to arrange into a symmetry of a bigger group. The 54 of SO(10) is in mind.

Note also that colouring with SU(3) seems compatible, given that [itex]3 \times 3[/itex] contains a [itex]\bar 3[/itex]

This was the status ten years ago. Then early this year I noticed a paper of Dubois-Violette using some arrangement in the way of Jordan Algebras and I wonder now if this 54 can be folded into a 27 with the 4/3 things in the diagonal. Something like
[tex]\begin{pmatrix}
2 & 8 & 8 \\
\bar 8 & 2 & 8 \\
\bar 8 & \bar 8 & 2 \\
\end{pmatrix}
[/tex]

The other big point of the summer has been to examine the coloration. We want the 24 to combine with a colour singlet (which we can get out of 8+1), and we want the 15 to combine with a colour triplet (which we can get out of 3+6). So the issue at hand escalates quickly: (15+15+24) * 9 = 486. Or with U(5) instead of SU(5): (15+15+25) * 9 = 495

Interesting hint.