Why is the "actual" gauge group of the standard model $SU(3) times SU(2) times U(1) /N$?

In this paper John Baez says the actual gauge group of the standard model is $SU(3) \times SU(2) \times U(1) /N$. Can someone explain the logic behind this line of thought?

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  Firstly, does this group $N$ has some name?

  
  


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  How can we see that $N$ acts "trivially on all the particles in the Standard Model" ?

  
  


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  Does this factoring change anything or why isn't this "real" gauge group mentioned anywhere else?

  
  

Answer

Baez actually has another paper (with Huerta) that goes into more detail about this. In particular, Sec. 3.1 is where it's explained, along with some nice examples. The upshot is that the hypercharges of known particles work out just right so that the action of that generator is trivial. Specifically, we have

Left-handed quark     Y = even integer + 1/3
Left-handed lepton    Y = odd integer
Right-handed quark    Y = odd integer + 1/3
Right-handed lepton   Y = even integer

Because these are the only values for known fermions in the Standard Model, that generator does nothing. So basically, you can just take the full group modulo the subgroup generated by $(\alpha, \alpha^{-3}, \alpha^2)$ -- where $\alpha$ is a sixth root of unity.

There's also this paper by Saller, which goes into greater detail about the "central correlations" of the Standard Model's gauge group, but in a more technical presentation. Saller also goes into some detail in chapter 6.5.3 of his book.