I don't understand your question. Please clarify $G,R_G$, what is supposed to be a $\Bbb{Z}_p$-module, $r$, and so on.

@reuns, here $G$ is the Galois group of the finite galois extension and $R_G$ is its representation over $\mathbb{Z}_p$

What do you mean by "the Galois representation of $G$"?

@EricWofsey, I mean the group representation of the Galois group $G$ over $\mathbb{Z}_p$

Still unclear. Please define $R_G$ and what is supposed to be a $\Bbb{Z}_p$-module.

@reuns, I asked the question for general representation map. My purpose was to know whether we can say that a finite extension is abelian by studying the properties of $p$-adic representation of the corresponding Galois group

Take any faithful representation ($K$ or $\Bbb{Z}_p[G]$ or whatever) and check if the elements of the group commute. $G$ acts on $K$ or $\Bbb{Z}_p[G]$ in the natural way.

$Gal(\mathbb{Q}_p(C[p^n])/\Bbb{Q}_p)$ doesn't have to be abelian. It is naturally a subgroup of $GL_2(\Bbb{Z}/p^n\Bbb{Z})$.

@reuns, For your comment for any faithful representation of $G$ on $\mathbb{Z}_p$, take two different elements $g_1,g_2 \in G$ such that $\phi:G \to \text{Sym}(\mathbb{Z}_p)$ with $g \mapsto \phi_g$ defined by $\phi_g(x)=g \cdot x$. Then $\phi_{g_1}\phi_{g_2}(x)=\phi_{g_1}(g_2 \cdot x)=g_1g_2 \cdot x$. So $G$ is abelian if and only if the maps $\phi_{g_1}$ and $\phi_{g_2}$ commutes

@reuns, thanks for your last comment. $Gal(\mathbb{Q}_p(C[p^n])/\mathbb{Q}_p)$ is a subgroup $GL_n(\mathbb{Z}_p/p^n \mathbb{Z}_p)$, where $n$ is the number of varaibles of the algebraic curve $C$.

@reuns, my question is- how to show $Gal(\mathbb{Q}_p(C[p^n])/\mathbb{Q}_p)$ is abelian ? Does the freeness property of the corresponding Tate module $T_pC$ helps anything ?

$G$ doesn't act on $\Bbb{Z}_p$ in a natural way (except if you have an homomorphism from $G$ to the $p-1$-th roots of unity) why do you think it does. $T_p$ is an abelian group and a free $\Bbb{Z}_p$-module by definition. No idea what you mean for the number of variables. For an elliptic curve it is $GL_2$, for a dimension 2 abelian variety it is $GL_4$.

thanks. I mean the Tate module $T_pC$ is a $\mathbb{Z}_p$-module and equipped with linear continuous action of the Galois group $G$. In other words, $T_pC$ is $\mathbb{Z}_p[G]$-module. Now I want to show $G$ (in my question) is abelian. the question-Is there any way to conclude $G$ is abelian by looking on $T_pC$ ?

$G$ doesn't act on $T_p$, it acts on $T_p/p^n T_p$ or $C[p^n]$. You are confusing with $Gal(\Bbb{Q}_p(C[p^\infty])/\Bbb{Q}_p)$. In general $G$ is not abelian, that's the answer, period.

@reuns, For Lubin-Tate group $F$, then $Gal(\mathbb{Q}_p(F[p^n])/\mathbb{Q}_p) \subset GL_1(\mathbb{Z}_p/p^n \mathbb{Z}_p)=(\mathbb{Z}_p/p^n \mathbb{Z}_p)^*$ and the corresponding $G$ is abelian

@reuns, what do you mean by $period$ here?

There are a bunch of problems here. Firstly, your question only makes sense if $C$ is an elliptic curve. Arbitrary algebraic curves don't have a group structure, so don't have torsion, or a Tate module. Or do you mean the Jacobian of a smooth projective curve? Secondly, it's not true in general that $G$ is abelian. For example, if $E=X_0(11)\colon y^2 + y = x^3 - x^2 - 10x -20$ viewed over $\mathbb Z_2$ has $\mathrm{Gal}(\mathbb Q_2(E[2])/\mathbb Q_2) = S_3$.

@Mathmo123, $C$ can be formal group as well. I have just a question. $C[p^n]$ is a free $\mathbb{Z}/p^n \mathbb{Z}$-module of rank $2$ when $C=E$=elliptic curve. Now, $C[p^n]$ is also a $\mathbb{Z}_p$-module. Is it free $\mathbb{Z}_p$-module ?

$E[p^n]$ is a free $\mathbb Z_p$-module: every point is annihilated by $p^n$. The Tate module $\varprojlim_nE[p^n]$ is a free $\mathbb Z_p$ module. But... none of this has anything to do with $E$ being defined over $\mathbb Z_p$.

@Mathmo123, It is not clear to me how $E[p^n]$ is free $\mathbb{Z}_p$-module. For, the p-adic integers $\mathbb{Z}_p$ is a PID and $E[p^n]$ is finitely generated as $\mathbb{Z}_p$-module. Then by structure theorem, $E[p^n]=F+T$, where $F$ is the free part of $E[p^n]$ and $T$ is the torsion part. However since every element of $E[p^n]$ is annihilated by $p^n$, we have $F=0$. This shows $E[p^n]$ is not free $\mathbb{Z}_p$-module instead a torsion-module. Am I wrong ?

Sorry - that was a typo. It is not a free $\mathbb Z_p$ module, because every point is annihilated by $p^n$. As you say, it's a torsion module. The Tate module is a free $\mathbb Z_p$-module

@Mathmo123, thanks for confirmation. A final question. Can $E[p^n]$ be a cyclic $\mathbb{Z}_p$-module ? or can $E[p^n]$ be principally generated as $\mathbb{Z}_p$-module ? That is I want to write $E[p^n]$ as $E[p^n]=\mathbb{Z}_p \cdot \alpha$ for some $\alpha \in E[p^n]$

It depends on what you mean by $E[p^n]$. Usually, if $E$ is an elliptic curve over a field $K$, then $E[p^n]$ means the $\overline K$-points of $E[p^n]$ - that's the set that has a Galois action. So in that case, no, when $K$ has characteristic $0$, we always have $E[p^n]\simeq (\mathbb{Z}/p\mathbb Z)^2$. But it's quite possible over $K$ that $E[p^n](K)$ - i.e. the points defined over $K$ - is cyclic or even empty. Note here that $K$ has nothing to do with $p$. In your case, you just so happen to have $K = \mathbb Q_p$.

@Mathmo123, thanks for your nice comment. In your 3rd line, you wrote $E[p^n] \simeq (\mathbb{Z}_p/p \mathbb{Z}_p)^2$. Is it typo or you really mean $E[p^n] \simeq (\mathbb{Z}_p/p^n \mathbb{Z}_p)^2$ ? All I know is that $E[p^n]$ is a free $ \mathbb{Z}_p/p^n \mathbb{Z}_p$-module. Is $E[p^n] $ a free $\mathbb{Z}_p/p \mathbb{Z}_p$-module as well ?

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