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Discrete Symmetries from Broken SU(N)

arXiv:hep-ph/9207231v1 10 Jul 1992SHEP-91/92-24Discrete Symmetries from Broken SU(N)and the MSSMP. L. WhitePhysics Department, University of Southampton,Southampton SO9 5NH, UK.AbstractIn order that discrete symmetries should not be violated by gravitational effects, it isnecessary to gauge them.

In this paper we discuss the gauging of ZZN from the breaking ofa high energy SU(N) gauge symmetry, and derive consistency conditions for the resultingdiscrete symmetry from the requirement of anomaly cancellation in the parent symmetry.These results are then applied to a detailed analysis of the possible discrete symmetriesforbidding proton decay in the minimal supersymmetric standard model.1

1 IntroductionIt has now been known for some time that discrete symmetries are likely to be stronglyviolated by gravitational effects [1]. This is a problem for the minimal supersymmetricstandard model (“MSSM”), where the possible renormalisable interaction terms in theLagrangian are constrained not merely by Lorentz and gauge invariance, but also by therequirement that a discrete symmetry, R parity, is not broken by the Lagrangian.

Thissymmetry is introduced purely in order to avoid the presence of Yukawa couplings whichwould lead to phenomenologically unacceptable proton decay, and a number of other possi-ble discrete symmetries have also been studied [2]. In order to ensure that such symmetriesare respected by the gravitational interactions it is necessary to gauge them [3], [4].

Gaug-ing a discrete symmetry means obtaining it as the residual symmetry after the spontaneousbreaking of some high energy gauge symmetry.Since this high energy symmetry must originally have been anomaly free, it is possibleto derive consistency conditions for the discrete symmetry, and this has been done for thecase of a U(1) gauge symmetry breaking to a ZZN discrete symmetry [5]. These resultshave been applied to the possible symmetries of the MSSM [6], with the result that onlya very restricted choice of such symmetries is possible.In this paper we shall discuss another possibility for the production of discrete sym-metries, namely that of the breaking of SU(N) to its ZZN centre, derive the resultingconsistency conditions, and apply our results to the MSSM.

The structure of the paper isas follows. Following this introduction, section 2 briefly discusses how this breaking mayoccur in a non-trivial way.

In section 3, we derive the consistency conditions which must beobeyed by the ZZN charges of the theory. These are then applied to the MSSM in section4.

Section 5 is the conclusion.2 Symmetry Breaking of SU(N) to its CentreIn this section we shall begin by briefly discussing how an SU(N) symmetry can bebroken without breaking its ZZN centre (consisting of those matrices in SU(N) of formexp( 2nπiN )×1 where 1 is the N ×N unit matrix), an idea which was mentioned in reference[4]. The simplest such mechanism is that this breaking could occur through a field in theadjoint representation acquiring a vacuum expectation value (hereafter “vev”), but it isalso possible to break the SU(N) by giving a vev to a field in any representation whichhas zero charge under the ZZN.The ZZN charge of a field which is in an irreduciblerepresentation of SU(N) with nu upper and nl lower vector indices is given by nl −nu,and is defined modulo N. Although this gives many possible breaking mechanisms, weshall only describe breaking through the adjoint and N index symmetric representations.To break SU(N) completely down to ZZN is trivial, and can be done either by repeatedbreaking or by choosing a representation of high enough dimension that when it gains a vevthe group breaks completely.

However, for reasons which will become apparent when wecome to discuss consistency conditions which must be satisfied by the discrete symmetry,this leads to extremely tight constraints on the resulting ZZN and so is not the mostinteresting case.Another possibility is that breaking occurs to give part or all of thestandard model gauge group SU(3)C × SU(2)L × U(1)Y , either directly from SU(N) orwith some non-trivial mixing with other symmetry groups. This can occur in many ways,2

given enough effort in constructing the potentials. As an example we construct a modelin which N = 3 and SU(3) × U(1) breaks to SU(2) × U(1) × ZZ3.

From this it is obvioushow one can go about constructing more elaborate theories.We begin with a gauge group SU(3) × U(1)X, which we break by giving a vev to ascalar A in the adjoint representation of zero X charge. The vev will be taken to be ofform< A >=√3aT8 = a10001000−2(2.1)where a is a constant of mass dimension 1.

Thus we have breaking of formSU(3) × U(1)X →SU(2) × U(1)8 × ZZ3 × U(1)X(2.2)where U(1)8 is the U(1) symmetry generated by the SU(3) generator T8. We now introducea field Bijk in the three index symmetric representation with non-zero X charge, and giveit a vev in the 333 component.

This then leaves the full symmetry breakingSU(3) × U(1)X →SU(2) × U(1)8 × ZZ3 × U(1)X→SU(2) × U(1)Y × ZZ3(2.3)Here U(1)Y is a linear combination of U(1)8 and U(1)X. The potential which will givethis breaking isV (A, B) =λ1tr(A2) −6a22 + λ2tr|(A + 2a′)imBmjk + cyclic|2+ λ3tr|B|2 −b22(2.4)Here λi > 0∀i, and at tree level we impose a = a′.It is simple to check that thisbreaking is stable under (small) radiative corrections, even if the equality a = a′ is broken.Similarly, choosing another form of the potential will allow us to break SU(2) × U(1)X toU(1), as in the standard model.3 ZZN Consistency ConditionsIn this section we derive consistency conditions for the residual ZZN symmetry.

These comefrom two sources: firstly the requirement that in the original SU(N) gauge theory therewere no anomalies; and secondly the observation that for each irreducible representation thetotal ZZN charge must be zero mod N. The latter condition will be much more restrictivein the case where the breaking of SU(N) is to ZZN only. Since this discussion is ratherelaborate, the reader who is uninterested in the derivation is recommended to skip to thesummary of results at the end of the section.We shall use Dynkin indices [7] to describe representations of SU(N) (for reviews see[8]).

For SU(N) the Dynkin indices take the form of a set of N −1 non-negative integerswhich we shall call an. Each distinct set of Dynkin indices corresponds to an irreduciblerepresentation of SU(N), and such properties as the dimension and Casimir invariants ofthe representation can be represented as functions of these indices.

The Dynkin indices of3

the most common representations of SU(N) are then given in Table 1 below. It is clearthat for all the representations listed here the ZZN charge of a particle in an irreduciblerepresentation with Dynkin indices an is given by the relationQ =N−1Xp=1pap(3.1)and in fact it is easy to use the rules for finding direct product of representations to checkthat this is true for all irreducible representations of SU(N).Table 1RepresentationDynkin indices (a1 .

. .

aN−1)singlet(0 . .

.0)fundamental(10 . .

.0)2 index symmetric(20 . .

.0)2 index anti-symmetric(010 . .

.0)N −1 index anti-symmetric(0 . .

.01)adjoint(10 . .

.01)We now wish to use the anomaly cancellation constraints in the form of Dynkin indices.We firstly have the constraint that SU(N) × SU(N) × U(1) anomalies must cancel. Thisis equivalent to the statement thatXall reps ΛqΛI2(Λ) = 0(3.2)where qΛ is the U(1) charge of the fermions in the representation Λ, and I2(Λ) is the secondorder index of Λ, whose relation to the second order Casimir invariant C2(Λ) is given byI2(Λ) = C2(Λ)D(Λ)(3.3)where D(Λ) is the dimension of the representation.

From this we may use the standardresults (using the notation of Slansky [8]) that, in root space with δ equal to half the sumof the positive roots,C2(Λ) = (Λ + 2δ, Λ)(3.4)Here the symbol Λ is used to represent the vector in weight space corresponding to thehighest weight of the representation Λ. This may be expanded in terms of the Dynkinindices an (which are the components of Λ in the Dynkin basis) to give for SU(N)C2(Λ) =N−1Xm=1N(N −m)mam + m(N −m)a2m +m−1Xn=02n(N −m)anam(3.5)4

and similarlyD(Λ) =Ypositive roots α(Λ + δ, α)(δ, α)=N−1Yp=1h 1p!N−1Yq=ppXr=q−p+1(1 + ar)i=(1 + a1)(1 + a2)(1 + a3) . .

. (1 + aN−1)×2 + a1 + a22 2 + a2 + a32 2 + a3 + a42.

. .2 + aN−2 + aN−12×3 + a1 + a2 + a33.

. .3 + aN−3 + aN−2 + aN−13×.

. .

×N −1 + a1 + . .

. + aN−1)N −1(3.6)The corresponding equation for the cancellation of the SU(N) × SU(N) × SU(N)anomaly is that the object I3(Λ) should vanish, where I3(Λ) is given by C3(Λ)D(Λ).

Theexpression for C3(Λ) in terms of the Dynkin indices is [9]C3(Λ) =N−1Xp,q,r=1dpqr(ap + 1)(aq + 1)(ar + 1)(3.7)with the totally symmetric tensor dpqr defined bydpqr := 12p(N −2q)(N −r)where p ≤q ≤r(3.8)Note that we have changed both the notation and the normalisation of reference [9], sincethe normalisation is irrelevant for our purposes, and the factor half in (3.8) will simplifylater algebra.We now wish to find how these conditions constrain the possible ZZN symmetries ofthe theory. In order to do this, we shall need to derive a number of relations involving theDynkin indices.

We first note that D(Λ) = D(a1 . .

. aN−1) is obviously an integer (since itis the dimension of a representation).

Similarly, D(a1 . .

. aN) is defined as the dimensionof a representation of SU(N + 1) and so is also an integer, where am is taken to be thesame in both cases for m < N, and aN is arbitrary.

We may now use equation (3.6) tofind thatD(a1 . .

. aN) =(1 + aN)2 + aN−1 + aN2× .

. .

×N + a1 + . .

. + aNND(a1 .

. .

aN−1)= 1N!" NYm=1(m + aN−m+1 + .

. .

+ aN)#D(a1 . .

. aN−1)(3.9)5

A useful definition is then f0(a1 . .

.aN) fromD(a1 . .

.aN) =: 1N f0(a1 . .

. aN)D(a1 .

. .aN−1)(3.10)Now, since D(a1 .

. .aN) is an integer, it is clear thatf0(a1 .

. .

aN)D(a1 . .

.aN−1) = 0mod N(3.11)This equation is true for all aN (so long as aN is a non-negative integer) and so it is alsotrue if we replace aN by (aN + 1). Thus we definefi+1(aN) =fi(aN + 1) −fi(aN)=∂fi(aN)∂aN+ 12!∂2fi(aN)∂a2N+ 13!∂3fi(aN)∂a3N+ .

. .

(3.12)from which it is clear thatfi(a1 . .

. aN)D(a1 .

. .aN−1) = 0mod N∀i(3.13)In particular we have, suppressing all the am dependence except that on aN,fN(aN) =∂Nf0(aN)∂aNN(3.14)fN−1(aN) =∂N−1f0(aN)∂aN−1N+ (N −1)2∂Nf0(aN)∂aNN(3.15)fN−2(aN) =∂N−2f0(aN)∂aN−2N+ (N −2)2∂N−1f0(aN)∂aN−1N+(N −2)3!+ (N −2)(N −3)8 ∂Nf0(aN)∂aNN(3.16)These equations may be derived using (3.12) and remembering that∂N+1f0(aN)∂aN+1N= 0(3.17)We may now expand out the explicit form of these three equations, and we discoverthat after some simplificationfN(a1 .

. .aN) =N!

(3.18)fN−1(a1 . .

. aN−10) =Xppap(3.19)fN−2(a1 .

. .

aN−10) =12(N −1)Xp pa2p −p2ap + 2Xq

We must now use (3.5) and (3.7), together with (3.19) and (3.20) to derive constraintson the ZZN charges of the theory.We begin with the requirement that the SU(N) ×SU(N) × U(1) anomaly cancels. The contribution to this anomaly of a representation Λis qΛC2(Λ)D(Λ) which can be expanded out with (3.5) to giveqΛC2(Λ)D(Λ) = −qΛ(Xppap)2D(Λ) + qΛN 2D(Λ)(Xpap)+ qΛND(Λ)N−1Xp=1(−p2ap + pa2p +Xqp3N2 apaq(p2q + q2p −Np2 −2Npq + N 2p)+Xpa3p(−32Np2 + 12N 2p) +Xq>papa2q(−92Npq + 32N 2p)+Xq>pa2paq(−3Np2 −32Npq + 32N 2p) +Xp

Extensive use of (3.19), (3.20), and (3.13) reduces this toC3(Λ)D(Λ) = (Xppap)3D(Λ)mod N 2(3.26)from which we conclude that the anomaly cancellation condition is thatXiQ3i = 0mod N 2(3.27)It should be noted that we are assuming that the ZZN charge Q of a particle in therepresentation labelled by ap is given by Pp pap, although this is in fact only true modN and some of our equations require definitions to be valid mod N 2. However, it is easyto check that it does not matter if we select another value for Q (equal to the first modN) so long as we pick the same value for all fermions in each representation.

In practicalterms this amounts to selecting one set of numbers to represent the ZZN charges andsticking to it. We shall use the simplest such set, namely {−N−12 , .

. ., N−12 } for N odd,and {−N2 + 1, .

. ., N2 } for N even.A further problem is that the U(1) charges are assumed integer.

If they are not, thenit is necessary to normalise them by multiplying by some overall constant until they are.In addition to (3.23) and (3.27), we can use (3.19) to obtain the result that for eachirreducible representation separatelyXiQi = 0mod N(3.28)Thus if the SU(N) symmetry breaks trivially to ZZN only, and thus has no mixing with thegauge group of the standard model (or their ancestors) at high energies, this constraint canbe applied to each set of fermions with the same quantum numbers (including ZZN charge)separately (since before SU(N) breaking they must have belonged to different irreduciblerepresentations) and so we shall discover that our theory is constrained effectively only togeneration symmetries. This however does not apply in the case where we allow a morecomplicated breaking, and then we may only impose that (3.28) is satisfied if the sum isover all fermions which might originally have been in the same representation, that is thosewith the same ZZN charge.These constraints ignore the possible effects of particles which are not visible at lowenergies because they have acquired large masses [5].

This can occur through a fermiongaining a Majorana mass, which is only possible if it has ZZN charge 0 or N2 (the latteronly for even N when Qi = Qj = N2 ; we are here using our convention for the ZZN chargeswhich gives that −N2 < Qi ≤N2 ) and no U(1) charge. Fermions of ZZN charge 0 do notmatter for the consistency conditions, and so the only effect is for N even when each suchfermion adds N38to the right hand side of equation (3.27), and removes the constraint(3.28) for charge N2 .It is also possible for two fermions, say i and j, to combine to obtain a Dirac mass.This can only occur if Qi + Qj = 0 or Qi + Qj = N , and if qi + qj = 0 (remember that we8

must not violate the U(1) symmetry). Thus we have two effects.

Firstly, the constraint(3.28) is weakened so that the sum now runs not over all fermions of the same ZZN chargeQ but over all fermions of charge Q and −Q. The case where the two fermions both havecharge N2 does not give any further restriction.We have now finished the derivation of the consistency conditions, and will thus sum-marise them below.

For N odd we haveQ(nQ −n−Q) = 0mod N ∀Q(3.29)XiqiQ2i = 0mod N 2(3.30)XiQ3i = 0mod N 2(3.31)while for N even these becomeQ(nQ −n−Q) = 0mod N ∀Q ̸= N2(3.32)XiqiQ2i = 0mod N 2(3.33)XiQ3i = ηN23mod N 2(3.34)In all of these equations the sum over i is the sum over all fermions labelled i whose ZZNcharge is Qi, nQ is the number of fermions of charge Q, and η is an arbitrary integer. Notethat for any given particle content, these are necessary but not sufficient conditions thatthe ZZN symmetry might have originated in an anomaly free high energy SU(N) symmetry.4 Application to the MSSMIn this section we demonstrate the use of our results by applying them to the specificcase of the MSSM.

Here we are severely restricted in which ZZN symmetries are acceptablefrom the requirement that terms in the Lagrangian leading to excessive proton decay beforbidden. The first part of this section will be essentially a summary of the analysis in[6])We shall impose a number of restrictions on the possible symmetries which we consider.These are that the symmetries be generation-blind with three generations; that the onlysinglet be the right-handed neutrino; and that N ≤4.

The second of these is particularlyrestrictive, since the inclusion of particles which interact only at high energy through theSU(N) symmetry, and so only affect the theory below the breaking scale through the ZZN,can seriously weaken the constraints. This effect is more pronounced than the case wherebreaking from U(1) is considered, since then more of the constraints involve mixing withother symmetries, which singlets of course cannot affect.

Including such particles is rathercomplicated, since the exact mechanism by which SU(N) breaks will affect how many suchparticles are allowed. For example, if SU(N) breaks to give the SU(2) of the standardmodel, then some of the singlets will have originally come from SU(N) multiplets which9

after breaking give the SU(2) multiplets of the standard model. It is not clear how onemight handle such a situation in general, although it would be possible to analyse it foreach particular case.The MSSM, with the inclusion of right-handed neutrinos but no other singlets, has thegauge group SU(3)×SU(2)×U(1), and particle content (with gauge couplings indicated):q(3, 2, 16)L(1, 2, −12)d(¯3, 1, 13)e(1, 1, 1)u(¯3, 1, −23)ν(1, 1, 0)H (1, 2, −12)¯H (1, 2, 12)(4.1)The gauge symmetry allows a number of Yukawa couplings, including the dimensionfour terms LHe, qHd, q ¯Hu, and µH ¯H all of which must be permitted by the discretesymmetry in order to obtain the correct structure of the standard model, although withthe inclusion of extra singlets it is possible to avoid the necessity of allowing µH ¯H.

Inaddition to these, there is the neutrino mass term L ¯Hν which is an acceptable additionto the standard model (although it is interesting that it is possible to construct modelswhere the discrete symmetry prevents neutrino masses), and terms allowing proton decay.These are lepton number violating termsLLeLQd(4.2)and the baryon number violating termudd. (4.3)We ignore a possible lepton number violating term µL ¯H because with our assumption thatthe term µH ¯H is allowed it can always be removed by a field redefinition if it appears.Although one conventionally requires all of these to be banned by the discrete symmetry,it is in fact sufficient that either (4.2) or (4.3) should be banned together with certainhigher dimension terms.

We shall adopt the view that to be acceptable, a symmetry musteither prevent all of (4.2) and (4.3) or else must ban all lepton or baryon number violatingterms of dimension four or five.In terms of the ZZN charges, the requirement that none of the standard model Yukawainteractions violate the symmetry isQ(H) + Q( ¯H) = 0mod NQ(L) + Q(H) + Q(e) = 0mod NQ(q) + Q(H) + Q(d) = 0mod NQ(q) + Q( ¯H) + Q(u) = 0mod N(4.4)We now begin by considering the specific case N = 2, where our ZZ2 symmetry is arelic of a high energy SU(2) symmetry. It is clear that for this case the constraints (3.32)10

and (3.34) are trivial (of course we would expect the latter, since there are no SU(2)3anomalies). (3.33) gives the constraint6Q(q)2 + 6Q(d)2 −18Q(L)2 + 18Q(e)2 −3Q(H)2 + 3Q( ¯H)2 = 0mod 4(4.5)since we must multiply all the charges by 6 to make them integers, and there are threegenerations.

After some rearrangement, and using (4.4) and the fact that each charge canonly take the values 0 or 1, we find that this constraint is always satisfied. Thus we findthat any ZZ2 symmetry which bans all the unwanted terms in the Lagrangian is acceptablefrom the point of view of anomaly contraints.

Examples of typical theories from whichsuch a symmetry could appear are the wide range of theories involving SU(2)R (wherethe subscript R means “right”), and in fact the usual R-parity could be such a case. Tosee this, note that all discrete symmetries added to the standard model can only reallybe defined modulo the discrete symmetries which are already there and gauged, and theseare the ZZ2 centre of SU(2)L, the ZZ3 centre of SU(3)C, and discrete subgroups of U(1)Y .By this argument, the only inequivalent ZZ2 symmetries are ZZ2L (from SU(2)L), R-parity,and lepton number, although it is not easy to see whether it is possible to construct anSU(2) symmetry which breaks to give the last.The case where N = 3 is less trivial.

The simplest way of solving for all the possiblediscrete symmetries is to write down every possible set of charges satisfying (4.4) and thento test them all against the consistency conditions. This can be done by noticing that given(4.4) we need only assign ZZ3 charges to q, L, ν, and H to define the symmetry completely,and ν will usually be given by the consistency conditions.

We thus only have 33 symmetriesto consider, and many of these are equivalent to one another under reflections (replacingeach charge Q with −Q). The constraints (3.29) to (3.31) are then entirely described byQ(H) + Q( ¯H) = 0Q(q)2 −2Q(u)2 + Q(d)2 = 02Q(q) + Q(d) + Q(u) + 2Q(L) + Q(e) + Q(ν) = 0mod 3(4.6)where we have simplified using the fact that our charges are all in the set {−1, 0, +1}.This gives a fairly straightforward result, and rather than going through all the algebra indetail we shall simply give the possible symmetries, which are listed in Table 2.

Note thatbecause we have 3 generations, the consistency conditions are much weaker than we mighthave expected.Having found all the possible symmetries, we now find whether they give acceptableconstraints on the MSSM Lagrangian. B clearly does not prevent any of the proton decayterms and is of no interest to us.

It is in fact the ZZ3 centre of the colour group. None ofthese symmetries bans both (4.3) and (4.2), but A,C, and D will protect lepton number,and they are all equivalent modulo ZZ3 colour to the residual ZZ3 from a generationalsymmetry.Thus, there is only one possible anomaly free symmetry for N = 3 which will protectthe proton from excessive decay, although it is noticable that this analysis would not giveany possibilities if we had not included right handed neutrinos, and that there is no reasonhere why such neutrinos should not gain a Dirac mass.11

Table 2: ZZ3 charges consistent with the MSSMSymmetryQ(H)Q( ¯H)Q(q)Q(u)Q(d)Q(L)Q(e)Q(ν)A000001−1−1B001−1−1000C001−1−11−1−1D001−1−1−111Table 3: ZZ4 charges consistent with the MSSMSymmetryQ(H)Q( ¯H)Q(q)Q(u)Q(d)Q(L)Q(e)Q(ν)A000001−1−1B001−1−1000 or 2C001−1−11−1−1D001−1−1220 or 2E001−1−1−111F002221−1−1G1−11020−1−1H1−1−120211I22111111J22111−1−1−1We conclude our analysis by studying the case where N = 4. This again gives fairlysimple results, listed in Table 3.Note that we have ignored cases where all the charges are zero mod 2, since suchsymmetries are merely ZZ2 symmetries.

In general a ZZN symmetry which is a subgroup ofZZM (so that M > N) obeys less restrictive conditions if it is obtained from a breaking ofSU(M) than from SU(N). In this case if we impose the standard model constraints (4.4)and require that the ZZ4 charges are all either 0 or 2 (so as to give a ZZ2 symmetry), thenthe consistency conditions are all trivially satisfied, just as for ZZ2.From the table there are seven possible ZZ4 symmetries which prevent all the unwanteddimension four terms in the lagrangian, and three more which preserve either lepton orbaryon number but not both (although two of those listed are in fact redundant, as they areproducts of others given here).

These symmetries also include some which could prevent12

the neutrino gaining a large Dirac mass.5 ConclusionWe have thus discussed how it is possible to break SU(N) to smaller gauge groups in sucha way as to leave its ZZN centre as a residual symmetry. The requirement that the highenergy SU(N) theory be anomaly free then gives restrictive constraints on the resultingdiscrete symmetry, although these are significantly weakened if we allow the addition tothe model of singlets which interact only through the SU(N) and thus carry non-zerocharge under ZZN.These constraints can then be simply solved in the case of the MSSM for N ≤4 withthe only singlet being the right-handed neutrino, to find that while ZZ2 symmetries are notrestricted at all, only a few possible ZZ3 and ZZ4 symmetries are left.While the constraints given here can be quite restrictive, it may well be possible toderive further ones from a more detailed analysis of the possible symmetry breaking mech-anisms.

For example, if SU(3)C arises from the breaking of SU(N), then each irreduciblerepresentation of SU(N) must contain an SU(3)C multiplet, and so extra singlets can onlybe introduced in conjunction with SU(3)C multiplets.There are a number of interesting possibilities for getting discrete symmetries fromSU(N) which we have not mentioned. Apart from the most obvious cases of breaking theSU(N) to SU(M) (for M < N) or to U(1), and then breaking the latter to a discretesymmetry as usual, we might consider the possibility that after breaking one might havetwo distinct discrete symmetries.

This is essentially covered by the usual analysis, exceptthat if we have both abelian and non-abelian symmetries which break to discrete remnantsZZM and ZZN respectively, then we will have a mixed anomaly cancellation constraint offormXiqiQ2i = 0gcd(N 2,MN)(5.1)where the ZZM and ZZN charges are qi and Qi, and gcd(N 2,MN) is the greatest commondivisor of N 2 and MN. This assumes that we select the ZZM charges to lie in a uniqueset, as for the ZZN ones.Finally, we should mention that there is not any particular reason why discrete sym-metries must not be non-abelian, although it is not easy to see how the breaking mightoccur to give rise to these, and there is no use for such symmetries in the standard model.AcknowledgementsI would like to thank Robert Foot, Ron King, Steven King, and Douglas Ross for varioushelpful suggestions.References[1] S.Hawking; Phys.

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