FIG. 2: Diagram for the process Z ′ → A 0 $_{R}$H 0 R → Δ ++ R Δ −− R + Δ ++ R Δ −− R → 2( µ $^{+}$µ$^{+}$) + 2( µ $^{−}$µ$^{−}$) dominant decay mode of Δ ±± R is into two charged muons. In other parts of the parameter space, the decay A 0 R → Δ ++ R Δ −− R is kinematically forbidden, but the branching fraction for Z ′ → Δ ++ R Δ −− R is still substantial, yielding 4 muons in the final state. In the above , we have assumed that the Δ$_{R}$ scalar triplet couples only to muons. This means that the corresponding scotino n$_{µ}$ is part of the SU (2)$_{R}$ doublet ( n$_{µ}$, µ )$_{R}$ with m$_{n}$$_{µ}$ = 200 GeV. If Δ$_{R}$ couples to electrons, then e $^{+}$e − → e $^{+}$e − scattering through Δ ±± R exchange would be much too big to be consistent with known data. We also assume no flavor mixing, i.e. Δ$_{R}$ does not couple to µe for example, or lepton flavor violating processes such as µ → eee and µ → eγ would be too big. However, Δ$_{R}$ still contributes to the muon anomalous magnetic moment which turns out to have the magnitude of the experimental discrepancy but of the wrong sign. To remedy this situation, one possibility is to add SU (2)$_{L}$ fermion doublets ( N, E )$_{L,R}$ with S = 0 and a neutral scalar singlet χ of S = − 1. The interaction ( ¯ Nν$_{µ}$ + ¯ Eµ ) χ will contribute positively and compensate for Δ$_{R}$. One final complication is that n$_{e}$ should have a mass greater than n$_{µ}$ in order that n$_{µ}$ is dark matter. Since it cannot come from Δ$_{R}$, n$_{e}$ must have a Dirac mass partner, i.e. an n$_{L}$ singlet. Of course, we can avoid all constraints by considering n$_{τ}$ instead as the scotino, in which case Δ ++ R will decay into τ $^{+}$τ$^{+}$. The resulting experimental signature would then be much more difficult to pick out.