The special unitary group is a Lie Group

I am trying to find a proof of the fact that the special unitary group is a Lie Group, but I can't come up with any good ideas and I couldn't find anything by searching at google.

• You could show that it is a sub manifold of the collection of all $n \times n$ matrices by showing that 0 is the only critical value of the determinant. – Wintermute Apr 18 '17 at 18:00
• @Wintermute: That argument would work for $\operatorname{SL}(n,\mathbb C)$, but the OP asked about $\operatorname{SU}(n)$. – Jack Lee Apr 18 '17 at 18:54
• Any ideas, please? – perlman Apr 20 '17 at 21:52