 Solution

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The renormalization-group beta function impliesIntegrating the one-loop running coupling equation givesHere the sign of ${\beta }_i$ is the sign convention in the question: a negative coefficient makes the inverse coupling increase towards high energy.

For gauge coupling unification, let $L=\log (M_{\mathrm{GUT}}/M_Z)/(2\pi )$ and let $a_G$ be the common inverse coupling at the unification scale. The normalized hypercharge coupling is $(5/3){\alpha }_1$, soSubtract the second relation from the first to determine $L$, then subtract it from the third. Provided $3{\beta }_1/5-{\beta }_2\ne 0$, this provesThe factors $3/5$ belong to the inverse normalized coupling; dropping them changes the prediction.

Use the Standard Model gauge group $SU(3{)}_C\times SU(2{)}_L\times U(1{)}_Y$ and convention $Q=T_3+Y$. One fermion family and one complex Higgs doublet have the following Standard Model representations. The left-handed quark and lepton doublets contain two Weyl spinors each; the right-handed fields are single Weyl species.The row $u_R$ denotes the right-handed up quark. Equivalently an all-left-handed table replaces the three right-handed fields by $u_R^c:(\overline{3},1{)}_{-2/3}$, $d_R^c:(\overline{3},1{)}_{1/3}$, and $e_R^c:(1,1{)}_1$; these are alternative descriptions, not extra fermions. The minimal model has no right-handed neutrino. For completeness, the family-independent gauge fields are the spin-one gluons $(8,1{)}_0$, weak gauge bosons $(1,3{)}_0$, and hypercharge gauge boson $(1,1{)}_0$; they are not counted once per family.

For the Standard Model one-loop gauge coefficients, include three fermion families and one Higgs doublet. With $T(\mathbf{N})=1/2$, the colour Dynkin index sum in one family isThe first factor two counts the two members of $Q_L$; the colour trace is already included in the Dynkin index. There is no coloured scalar. HenceFor $SU(2{)}_L$, the three colours of $Q_L$ and the lepton doublet give $3(1/2)+1/2=2$ per family. The Higgs contributes $T_2(\varphi )=1/2$. ThusFor $U(1{)}_Y$, sum squared hypercharges over every colour and doublet component. One family contributesThe two complex Higgs components contribute $2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction term, soThis ${\beta }_1$ is for the unnormalized coupling $g_Y$ used in the question. For $g_1^{\mathrm{GUT}}=\sqrt{5/3}{g}_Y$, the corresponding coefficient is $(3/5){\beta }_1=41/10$. This distinction is required for a consistent unification calculation.