# Covering space of an abelian topological group is abelian if the covering map is a homomorphism

I'm trying to show that if $(E, \cdot)$ and $(G, \cdot)$ are both topological groups, $G$ is abelian, and $(E, p)$ is a covering of $G$ such that $p:E\to G$ is a homomorphism with respect to $\cdot$, then $E$ must also be abelian. I think it should suffice to show this under the usual "niceness" conditions for covering spaces, where $E$ and $G$ are path connected and locally path connected.

Of course, $p$ being a homomorphism and $E$ being abelian means that $p(x \cdot y) = p(y \cdot x)$. So I know that $x \cdot y$ and $y \cdot x$ are in the preimages of all the same sets of $G$. My assumption is that I now need to call upon the fact that $p$ is a covering map somehow, but I'm drawing a blank as to how. Am I on the right track here? If so, what property of $p$ as a covering map am I missing? If not, what might a better approach be?

Consider $m:E\times E\rightarrow E$ defined by $m(x,y)=xyx^{-1}y^{-1}$.
$p(m(x,y))=1_G$; this implies that the image of $m$ is contained in $p^{-1}(1_G)$. Since $p$ is covering, $p^{-1}(1_G)$ is discrete, and since $E$ is connected, the image of $m$ is connected, and thus is $1_E$.