Proposition 5

Let $a\in G$ and let W be a neighbourhood of $e$ in G. There exists a neighbourhood V of $a$ with the following property: for all $a'\in V$, there exists an element $g\in W$ such that $\mathfrak{g}^1(a')\subset$ Ad($g$)$\mathfrak{g}^1(a)$.

Put $\mathfrak{g}^1=\mathfrak{g}^1(a)$ and let $\mathfrak{g}=\mathfrak{g}^1+\mathfrak{g}^+$ be the Fitting decomposition of Ad($a$)$-1 ($§1, no. 1). Let H be a Lie subgroup germ of G with Lie algebra $\mathfrak{g}^1$. For all $h\in H$, Ad($h$)$\mathfrak{g}^1\subset \mathfrak{g}^1$. Since $[\mathfrak{g}^1,\mathfrak{g}^+]\subset \mathfrak{g}^+$, there exists a neighbourhood U of $e$ in H such that Ad($h$)$\mathfrak{g}^+\subset \mathfrak{g}^+$ for all $h\in H$. Since the restriction of Ad($a$)$-1$ to $\mathfrak{g}^+$ is bijective, U can be chosen so that the restriction of Ad($ah$)$-1$ to $\mathfrak{g}^+$ is bijective for all $h\in U$. Then $\mathfrak{g}^1(ah)\subset \mathfrak{g}^1(a) =\mathfrak{g}^1$ for all $h\in U$. By Proposition 4, Int(W)($aU$) is a neighbourhood of $a$ in G. If $a'\in$ Int(W)($aU$), then $a'=g(ah)g^{-1}$ with $g\in W$ and $h\in U$; it follows that $\mathfrak{g}^1(a') =$ Ad($g$)$\mathfrak{g}^1(ah)\subset$ Ad($g$)$\mathfrak{g}^1(a)$.