# Iwasawa decomposition of inverse

Let $G$ be a semisimple rank one Lie group with finite center. Let $G=KAN$ be the Iwasawa decomposition with $\mathfrak{a}=$Lie($A)=\text{span}\ H$. Then if $G\ni g=kan, a=exp(tH)$ is it true that $$g^{-1}=\tilde k exp(-tH) \tilde n$$ is the decomposition of $g^{-1}$ where $\tilde k\in K, \tilde n\in N$?

The answer is no. Consider $$G=SL(2,\mathbb{R})$$. Let $$g=kan= \left(\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right) \cdot \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) \cdot \left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right) = \left(\begin{matrix} 0 & -1 \\ 1 & 1 \end{matrix}\right).$$ We can write instead $$g=n'a'k'= \left(\begin{matrix} 1 & -\frac{1}{2} \\ 0 & 1 \end{matrix}\right) \cdot \left(\begin{matrix} \frac{1}{\sqrt{2}} & 0 \\ 0& \sqrt{2} \end{matrix}\right) \cdot \left(\begin{matrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix}\right) = \left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right).$$ So we get the Iwasawa decomposition $$g^{-1}=k'^{-1}a'^{-1}n'^{-1}$$ for the inverse, but we see $$\left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) = \exp (-tH) = a^{-1} \neq a'^{-1}= \left(\begin{matrix} \sqrt{2} & 0 \\ 0 & \frac{1}{\sqrt{2}} \end{matrix}\right) .$$ so your suggestion does not hold.
It would be interesting to learn how the $$KAN$$ and $$NAK$$ Iwasawa decompositions (and especially their $$A$$-component) relate to each other.