I'm working on an exercise in Fulton and Harris that asks you to prove SU$(n)$ is not a complex Lie subgroup of SL${}_n(\mathbb{C})$. If $A \in$ SU$(n)$, then we have that $A^{-1} = {}^t\overline{A}$ (where ${}^t\overline{A}$ is the conjugate transpose).
Since conjugation is not holomorphic, I thought I had it. But then it occurred to me that $A \in \text{GL}{}_n(\mathbb{C})$ (where $A \in \text{SU}(n)$ as above) which is a complex Lie group. What am I missing here?
Perhaps that the easiest way of proving that $SU(n)$ is not a complex Lie group consists in using the fact that its Lie algebra $\mathfrak{su}(n)$ is not a complex Lie algebra. That is easy, since$$\mathfrak{su}(n)=\left\{A\in M_n(\mathbb C)\,\middle|\, A=-\overline{A^T}\right\}$$and clearly, unless $A=0$, if $A\in\mathfrak{su}(n)$, then $iA\notin\mathfrak{su}(n)$.