Every connected semisimple subgroup of $GL(n,\mathbb{R})$ is in $SL(n,\mathbb{R})$?

Let $G$ be a connected, semisimple subgroup of $GL(n,\mathbb{R})$. Here, semisimple means that $G$ has no normal, connected abelian subgroup other than the trivial group.

In Mostows book Strong rigidity of locally symmetric spaces, on p. 11, he writes

Let $G$ be an analytic subgroup of $GL(n,\mathbf{R})$. [...] Assume moreover that $G$ is semi-simple [...]. Since $G$ is its own commutator subgroup, each element of $G$ has determinant one.

I'm trying to understand this: Does the following proof work?

• $G/[G,G]$ is semisimple as a quotient of the semisimple group $G$ (is this correct?)
• $G/[G,G]$ is connected since $G$ is connected (quotients of connected spaces are connected)
• Since $G/[G,G]$ is abelian, it must be trivial, i.e. $[G,G]=G$.

Its also not clear to me why $\det(g) = 1$ for each $g \in G$. Why does this follow?

Note that the determinant of a commutator is always $1$. The result follows. $$\det [g,h]=\det (g^{-1}h^{-1}gh)=1$$