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I read this post, Is the symmetry group of two spin 1/2 particles $SU(2) \times SU(2)$ or $SU(4)$?

If the picked answer is correct, can I believe that an $N$-degenerate system respects $SU(N)$ symmetry? But if so, I am confused in this way: can we understand that a spin $j$ particle contains $2j+1$ internal degenerate states? Then the symmetry of this particle is $SU(2j+1)$, right? If $j>2$, then it is larger than $SU(2)$!

I must be wrong somehow since I know little about group theory. So, basically, what kind of system respects $SU(N)$?

Update 1

Since I have not enough reputation, I could not add comment on the linked post, which I am still confused, so I may also post here(related to this issue): There are $M\geq 2$ particles, if each particle respects $SU(N)$ symmetry, and if

  • A. there is no interaction between them, then what is the symmetry group, if
    • A1.they are non-identical particles?
    • A2. they are identical particles?
  • B. there is pairwise interaction between them, and the interaction is $SU(N)$ invariant,
    • B0. what do I mean by $SU(N)$ invariant?
    • B1. symmetry group for non-identical case?
    • B2. for identical case?
share|improve this question
Yang-Mills has an $SU(N)$ symmetry. –  JamalS Apr 13 at 21:32
The title question (v1) seems to be a list question. –  Qmechanic Apr 13 at 21:33
@Qmechanic i am not sure which question you think is a list question. If the first one, "a N-degenerate system respects $SU(N)$ symmetry?", just a Yes or or No question. –  Mathieu Apr 15 at 4:55

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