Lie Group Automorphism which are diffeomorphism

Question

Is every smooth automorphism of a Lie Group $G$ a diffeomorphism?

Answer 1

Let $F: G\to G$ be an automorphism. Let $dF_e: T_eG \to T_eG$ be the induced map on the Lie algebra $T_eG$.

As $F(gh) = F(g)F(h)$, Putting $h = e^{tX}$ for all $X \in LG$, we have

$$(*)\ \ \ \ dF_g \circ d(\ell_g)= d(\ell_{F(g)}) \circ dF_e$$

Now if $dF_e(X) =0$ for some nonzero $X\in T_eG$, then consider $f(t)= e^{tX}$. We have

$$\frac{d}{dt} F\circ f(t) = dF_{f(t)} \circ f'(t) = dF_{e^{tX}}\circ d(\ell_g) (X) = 0$$

by $(*)$. Thus $F(e^{tX}) = F(e) = e$ for all $t$.But that is impossible. So $dF_e$ is invertiable. By $(*)$, $dF_g$ is invertible for all $g$ and so $F$ is a diffeomorphism.

Answer 2

An automorphism between compact Lie groups is a bijective continuous function between compact manifolds, hence it is closed. So such an automorphism needs to be omeomorphism and its inverse is continuous and smooth.