Lemma 3

Let $a\in G$ and let $\mathfrak{m}$ be a complement of $\mathfrak{g}^1(a)$ in $\mathfrak{g}$. Let U be a neighbourhood of 0 in $\mathfrak{g}$ and exp an exponential map from U to G. The map

$f: (x, y) \rightarrow$ (exp $y$)$a$(exp $x$)(exp $y$)$^{-1}$

from $(\mathfrak{g}^1(a)\times \mathfrak{m})\cap U$ to G is étale at $(0,0)$.

The tangent linear maps at 0 of the maps $x \rightarrow a$(exp $x$) and $y \rightarrow$ (exp $y$)$a$(exp $y$)$^{-1}$ are the maps $x \rightarrow ax$ and $y \rightarrow ya-ay=a(a^{-1}ya-y)$ from $\mathfrak{g}$ to $T_aG =a\mathfrak{g}$ (Chap. III, §3, no. 12, Prop. 46). Consequently, the tangent map of $f$ at $(0,0)$ is the map $(x, y) \rightarrow ax+a(a^{-1}ya-y) =a(x+a^{-1}ya-y)$ from $\mathfrak{g}^1(a)\times \mathfrak{m}$ to $a\mathfrak{g}$. This map is injective. Indeed, if $x\in \mathfrak{g}^1(a), y\in \mathfrak{m}$ and if $x+a^{-1}ya-y$ = 0, then (Ad($a$)$-1$)$y$ = Ad($a$)$x\in \mathfrak{g}^1(a)$ since Ad($a$)$\mathfrak{g}^1(a)\subset \mathfrak{g}^1(a)$. This implies that $y\in \mathfrak{g}^1(a)$ and consequently that $y$ = 0. Since Ad($a$) is injective on $\mathfrak{g}^1(a)$, it follows that $x$ = 0. Since dim $\mathfrak{g}=$ dim$\mathfrak{g}^1(a) +$ dim $\mathfrak{m}$, this shows that $f$ is étale at $(0,0)$.