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How to show SU(n) is a normal subgroup of U(n)

  1. Jan 10, 2010 #1

    I'd like to show that SU(n) is a normal subgroup of U(n).

    Here are my thoughts:

    1)The kernel of of homomorphism is a normal subgroup.

    2)So if we consider a mapping F: G-> G'=det(G)

    3)Then all elements of G which are SU(n), map to the the identity of G', therefore SU(n) is a normal subgroup of GL(n,C).

    Firstly, is this correct?

    Also, a very stupid question, but can I ask whether my musings constitute a formal 'proof'? I mean, would I need to show prove statements (1) and (2) (ie. show it obeys the homom. property) or can I just say that I've assumed that they are for the purposes of the proof at hand?

  2. jcsd
  3. Jan 11, 2010 #2
    you are correct - formal proofs can assume other known theorems - it is a matter of how much your reader needs to see and how much he already knows.
  4. Jan 11, 2010 #3
    thanks wofsy... I should have given a bit of context.

    I'm going to sit an exam on introductory group theory, lie algebras and representations very soon... I feel a bit unsure when I see "prove that" questions (my excuse, if you can call it that, is that my background isn't in mathematics but I am doing a masters in one and I don't know many of the things I am expected to know...)
  5. Jan 11, 2010 #4
    I would say that theorems in your book or theorems proved in class you can refer to without proof. But the best way to know is to ask you teacher.
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