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Every Lie group $G$ is naturally contained in its holomorph Hol($G$) = $G \rtimes $ Aut($G$)

Is Hol$(G)$ always a Lie group?

If the answer is yes our main questions:

1.For a left invariant metric $g$ of Hol($G$), are there always two left invariant metrics $g_{1}$ and $g_{2}$ of $G$ and Aut($G$), respectively such that $g$ is conformal to the product of $g_{1}$ and $g_{2}$? (motivated by $G=\mathbb{R}$, as an example of this situation)

  1. For what type of Lie groups, $G$ is a totally geodesic submanifold of its holomorph? As we see, $G=\mathbb{R}$ does not satisfies this property, but to what extent those Lie groups with this propery are studied?

3.Motivated by Poincare upper halph plane $\mathbb{H}^{2}\simeq \mathbb{R} \rtimes \mathbb{R}^{+}$, is it true to say that Aut($G$) is always a totally geodesic submanifold of Hol($G$)?

And finally: can one write the lie algebra of Hol($G$) in term of Lie algebras of $G$ and $Aut(G)$ and also the natural action of $Aut(G)$ on the Lie algebra of $G$?

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    $\begingroup$ $Hol(G)$ is always a Lie group, since $G$ and $Aut(G)$ are Lie groups and the action is smooth. The Lie algebra is $\mathfrak{g}\rtimes Der(\mathfrak{g})$. $\endgroup$
    – Dietrich Burde
    Commented Jan 19, 2015 at 21:51
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    $\begingroup$ @DietrichBurde thank you very much for your comment. Could you please give a reference for your last part of your comment. Is it easy to proof? $\endgroup$
    – Ali Taghavi
    Commented Jan 20, 2015 at 9:05
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    $\begingroup$ A reference is here; $G$ should be connected and simply connected to conclude that $Aut(\mathfrak{g})\simeq Aut(G)$. $\endgroup$
    – Dietrich Burde
    Commented Jan 20, 2015 at 9:53
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    $\begingroup$ Some assumptions are also needed to ensure that $\mathrm{Aut}(G)$ is a Lie group ($G$ virtually connected is enough), see the comments in math.stackexchange.com/questions/1589303/… $\endgroup$
    – YCor
    Commented Dec 28, 2015 at 15:08

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