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Are there any non-trivial group extensions of $SU(N)$?

If not, can one show/prove there are no non-trivial group extensions of $SU(N)$?

It is possibly partial related to the homotopy group property. Or one can try to argue from the exact sequence.

Proof/Show: Let us call $Q=SU(N)$. If the above claim is true, namely, we cannot find some larger $G$ with its normal subgroup $N$, such that $G/N=Q$. Here what we mean by "non-trivial" means that a non-identity normal $N \neq 1$, and there the $G$ is not a product group (i.e. $G \neq Q \times N)$, where $Q=SU(N)$.

p.s. The statement has been corrected for clarity, please upvoting it for the sake of getting better discussions/attentions.

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wonderich
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  • What do you mean by nontrivial? The direct product of $SU(n)$ with any other group will be a nontrivial extension. If by nontrivial you mean that aren't direct or semidirect products? – Chickenmancer May 04 '17 at 19:27
  • Yes, by nontrivial I mean that aren't direct products, but may be semidirect products, or act projectively. – wonderich May 04 '17 at 19:28
  • Are you familiar with group cohomology? – Chickenmancer May 04 '17 at 19:31
  • Thanks, I know for example https://math.stackexchange.com/questions/2260699/short-exact-sequence-and-cohomology-group – wonderich May 04 '17 at 19:39
  • p.s. Justin made a comment earlier that $U(N)/SU(N)=U(1)$, but it was not quite correct to my case because I am looking for $G/N=SU(N)$ for some non-product group $G \neq N \times SU(N)$. – wonderich May 04 '17 at 20:36
  • I believe that my questions is very motivated and well-defined. – wonderich May 04 '17 at 22:49
  • Could you edit the question? Non-trivial doesn't mean what you want it to mean in this context. I would of just said "Group extensions that are not direct products" or whatever. This way you can clear up some of the confusion caused. – Ali Caglayan May 05 '17 at 00:04
  • $U(n)$ is not, as a group, a product $S^1 \times SU(n)$. If you allow the extension to have higher dimension then that does give an example. I don't know the answer for finite groups. –  May 05 '17 at 00:04
  • Thanks Mike, but I am looking for $1 \to N \to G \to SU(N) \to 1$, not the other way you mentioned. – wonderich May 05 '17 at 00:23
  • @wonderich did you figure this out? – QGravity Apr 06 '24 at 16:42

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