# $\mathbb{Z}_6$ symmetry in the Standard Model

It's a well-known result that the spontaneous symmetry breaking of $SU(5)$ would lead not to the usual $G_{SM}=SU(3)\times SU(2)\times U(1)$, but to $G_{SM}/\mathbb{Z}_6$. However, it's also often argued that $G_{SM}/\mathbb{Z}_6$ (actually, also $G_{SM}/\mathbb{Z}_2$ and $G_{SM}/\mathbb{Z}_3$) would be indistinguishable from $G_{SM}$, since its effect is to forbid left-handed fermions with zero hypercharge (which were not seen experimentally and so wouldn't be a problem for the phenomenologist building his $G_{SM}$ from the ground up).

How can we demonstrate the statement in italic, and does the $\mod\mathbb{Z}_6$ symmetry play another role once integrated in the unified $SU(5)$? (in particular in connection with homotopy groups and possible magnetic monopoles)

• I dont think the statement in italic is correct. The /Z6 acts on the global property of the SM, and this is not equivalent to GSM/Z2 or GSM/Z3. See Huck's "Global structure of the standard model, anomalies, and charge quantization" and refs therein for more details.
– xi45
Dec 21 '17 at 23:55
• Possible duplicates: physics.stackexchange.com/q/176518/2451 & physics.stackexchange.com/q/149039/2451 Dec 22 '17 at 16:59