Induced module is abelian

Question

Let $G$ a profinite group and $H\subset G$ a normal open subgroup such that $G/H$ is abelian. Let $M$ be a discrete $H$-module. We say that the $H$-module is abelian if the image of the action map $\rho:H\rightarrow \text{Aut}(M)$ has abelian image. Show that the following are equivalent:

1. The induced $G$-module Ind$^G_H(M)$ is abelian.
2. ker$(\rho)$ is a normal subgroup of $G$ and $G/$ker$(\rho)$ is abelian.

I am a bit stuck on this problem and any help would be appreciated. I got confused abou the notation but I realized we can reformulate the statements taking $\rho:H\rightarrow \text{Aut}(M)$ and $\phi:G\rightarrow \text{Aut}(\text{Ind}^G_H(M))$ and prove that the following are equivalent

1. $G/\text{ker}(\phi)$ abelian.
2. ker$(\rho)$ normal in $G$ and $G/$ker$(\rho)$ abelian.

Thank you in advance to everyone who will try help me.