# Group theory

#### RedX

The fundamental representation of SU(N) has a basic form that allows you to deduce that there is a SU(N-1) subgroup. For example, in SU(3), the generators $$T_{1}$$, $$T_{2}$$, $$T_{3}$$ form an SU(2) subgroup.

I'm reading a book right now that goes into the adjoint representation of SU(N) to show that SU(N) has subgroups $$SU(N_1)$$, $$SU(N_2)$$, $$SU(N_3)$$, ... U(1), where $$N_1+N_2+N_3+...=N$$. For example, SU(5) has subgroups SU(3), SU(2), and U(1).

My question is that since SU(N-1) is already subgroup, doesn't $$N_1=N-1$$, forcing $$N_2=1$$, and the rest of the N's zero?

Also, for U(1) to be a subgroup, you have to find a linear combination of generators (call it the generator $$T_0$$) such that the new structure constant has a zero whenever one of its indices is 0? If the structure constant changes, isn't this a new group, and not a subgroup?

And why are the subgroups of SU(5) combined into the direct product group: SU(3)xSU(2)xU(1)?

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#### PRB147

My question is that since SU(N-1) is already subgroup, doesn't $$N_1=N-1$$ , forcing $$N_2=1$$ , and the rest of the N's zero?
Seems that there is no SU(1) group.

#### confinement

Seems that there is no SU(1) group.
Actually SU(1) = SO(1) is the trivial group, i.e. it has a faithful representation consisting of the number 1 under multiplication.

To the OP, maybe it would help if you state the theorem in the book i.e. I cannot tell whether it is a necessary or sufficient condition.

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