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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?

Thanks.

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

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