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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 $\tilde k\in K, \tilde n\in N$?

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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.

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