Is a finite covering space of a unimodular Lie group unimodular?

Suppose we have two Lie groups $$G$$ and $$H$$ and a covering map $$f:G\rightarrow H$$ with a finite number of sheets which is a Lie group homomorphism.

I want to know whether $$G$$ is unimodular whenever $$H$$ is unimodular.

I'd like to know this because I am trying to prove that $$SL_n(\mathbb C)$$ is unimodular. I have managed to come up with a map

$$SL_n(\mathbb C)\times\mathbb C^*\xrightarrow \phi GL_n(\mathbb C),$$

given by $$(A,z)\mapsto zA$$. Then $$\phi$$ turns out to be a Lie group homomorphism which is also a covering map of $$n$$ sheets. I have also showed that $$GL_n(\mathbb C)$$ is unimodular by defining a measure which is an integral of a complex form(of type $$(n^2,n^2)$$) of $$\mathbb C^{n^2}$$. This sort of imitates an argument found in Serge Lang's book $$SL_2(\mathbb R)$$.

So if this argument using covering maps works, then $$SL_n(\mathbb C)\times\mathbb C^*$$ being unimodular implies $$SL_n(\mathbb C)$$ is unimodular.

Is there another way to show $$SL_n(\mathbb C)$$ is unimodular using $$\phi$$?

I'd like to have a sort of elementary answer, please, as I don't know that much about Lie groups. I just know what a Lie group is, what a left invariant measure is, and the definition of the Lie algebra.

Let $$G$$ be a Lie group and $${\cal G}$$ its Lie algebra, $$G$$ is unimodular if and only if $$tr(ad(g))=0$$ for every $$g\in{\cal G}$$ where $$tr(ad(g))$$ is the trace of $$ad(g):{\cal G}\rightarrow {\cal G}$$ defined by $$ad(g)(x)=[g,x]$$. Let $$G\rightarrow H$$ be a finite covering, the Lie algebras of $${\cal G}$$ and $${\cal H}$$ are isomorphic, we deduce that $$G$$ is unimodular if and only if $$H$$ is unimodular.
• nice ${}{}{}{}{}$ Oct 2, 2018 at 0:35