Let $E=E_q$ be the Tate ellipitc curve over a finite extension $K$ of $\mathbb{Q}_p$ for a $q$. Let $T$ be its p-adic Tate module. Let $\mathfrak m$ be the maximal ideal in $K$. I saw in this paper (page 571, second paragraph) the following claim
By the Kummer map, we regard $\hat{E}(\mathfrak m)$ as a subgroup of $H^1(K,\, T)$.
Here $\hat{E}(\mathfrak m)$ sits injectively inside $E(K)$. So I tried to show that $E(K)$ sits as a subgroup in $H^1(K,\, T)$. I don't know what exactly the Kummer map is, but I proceeded as following (produced below) to get $E(K) \otimes_\mathbb{Z}\mathbb{Z}_p$ (not $E(K)$) as a subgroup of $H^1(K,\, T)$. Since $K$ is a finite extension of $\mathbb{Q}_p$, by Tate's uniformization I know how $E(K)$ looks like. And from the structure of $E(K)$, it follows that $E(K)$ doesn't sit inside injectively inside $E(K) \otimes_\mathbb{Z}\mathbb{Z}_p$.
So my question is, is there any other way of embedding $E(K)$ inside $H^1(K,\, T)$ that I fail to see?
The details of my calculations: Start with the Kummer exact sequence: $$ 0 \rightarrow E(\overline{K})[p^n] \rightarrow E(\overline{K}) \xrightarrow{[p^n]} E(\overline{K}) \rightarrow 0 $$ The long exact exact sequence gives the following inclusion $$ E(K)/p^nE(K) \hookrightarrow H^1(K, E(\overline{k})[p^n]) $$ Taking the inverse limit gives $$ \varprojlim_n E(k)/p^nE(K) \hookrightarrow \varprojlim_m H^1(K,E(\overline{K})[p^n]) \cong H^1(k, \varprojlim_{n} E(\overline{K})[p^n]) \cong H^1(K, \varprojlim_n E(\overline{K})[p^n])=H^1(K,\,T) $$ Since $E(K)$ is Noetherian, $\varprojlim_n E(k)/p^nE(K) \cong E(K) \otimes_\mathbb{Z}\mathbb{Z}_p$ so $E(K) \otimes_\mathbb{Z}\mathbb{Z}_p$ sits inside $H^1(K,\,T)$.
On the other hand, to see that $E(K)$ does not sit inside $E(K) \otimes_\mathbb{Z}\mathbb{Z}_p$ we observe that $E(K)$ is isomorphic to $K^\times/q^\mathbb{Z}$ and $K^\times$, by proposition 5.7 of chapter 2 of Neukirch, has a direct summand of $\mathbb{Z}/{(p^f-1)\mathbb{Z}}$ which gets killed under the tensor product by $\mathbb{Z}_p$.