I'm trying to prove that for any algebra ${\cal A}$, $Aut({\cal A}) = \{a \in End({\cal A})\ |\ a([x, y]) = [a(x), a(y)], \forall x, y \in {\cal A}\}$ is a closed subgroup of $GL({\cal A})$.
I can prove that it is a subgroup just showing that $a\circ a^{'-1}$ is in $Aut({\cal A})$ which is direct computation. Nevertherless, I don't know how to show that thi group is closed. I've been thinking about proving that it is the isotropy group of $GL({\cal A})$, but I don't get it. I tried to impose that $[x, y]$ is the fixed point for the isotropy subgroup, but I just get:
$$a([x, y]) = a(x)a(y) - a(y)a(x) = xy - yx$$
But this means nothing because $a \in Aut({\cal A})$ is not imposing that $[x, y]$ is fixed.
How do I solve it?