Left invariant Riemannian metric induce bi-invariant volume element. This is problem 3-11 from Jack Lee's Riemannian Manifold.
Suppose $G$ is a compact connect Lie group with a left invariant metric $g$, let $dV$ denote the Riemannian volume element of $g$. Show that $dV$ is bi-invariant.
The problem came with a hint, which goes as follows: Show that $R_p^* dV$ is left invariant and positively oriented, and hence equals $\varphi(p)dV$ for some $\varphi:G\to R^+$. Then show that $\varphi: G\to R^{+}$ is Lie group homomorphism, and therefore the image is compact.
I think I'm fine with the latter half of the hint, but I can't see how $R_p^* dV$ should be left invariant and positively oriented. Another fact(from a previous problem) that I think would be helpful is that $g_{ij}:=g(X_i,X_j)$ is a constant if $g$ is left invariant, and $\{X_i\}$ is a left invariant frame. Thank you in advance for any answers.
 A: Left-invariance: $L_h^*R_g^*({\rm d}V) = (R_g\circ L_h)^*({\rm d}V) = (L_h\circ R_g)^*({\rm d}V) = R_g^*L_h^*({\rm d}V) = R_g^*{\rm d}V$, since left translations commutes with right translations and ${\rm d}V$ is already left-invariant.
Positivity: by linear algebra we know that $(R_g)^*({\rm d}V) = (\det {\rm d}(R_g))\,{\rm d}V$. But $R_g$ is a diffeomorphism and $G$ is connected, so $\det {\rm d}(R_g)$ is always positive or always negative. But the dependence on the parameter $g$ is also continuous, and $R_e = {\rm Id}_G$ is orientation-preserving.
Homomorphism: $R_g^*R_h^*{\rm d}V = R_g^*(\varphi(h){\rm d}V) = \varphi(h)R_g^*({\rm d}V) = \varphi(h)\varphi(g){\rm d}V$. But we also have that $$R_g^*R_h^*({\rm d}V) = (R_h\circ R_g)^*({\rm d}V) = R_{gh}^*({\rm d}V) = \varphi(gh){\rm d}V,$$ so $\varphi(gh) = \varphi(g)\varphi(h)$ (as $\mathbb{R}^+$) is abelian. It's a Lie group homomorphism (i.e., also smooth) due to the determinant expression from the previous item.
Constancy of $\varphi$: every subgroup of $\mathbb{R}^+$ is either discrete or dense. In particular, the image of $\varphi$. By compactness of $G$, it must be discrete as opposed to dense. By connectedness of $G$, it must be a single point. This single point must be $1$ due to $\varphi$ being a homomorphism.
So $R_g^*({\rm d}V) = {\rm d}V$ for every $g\in G$ as well.
