Dear MO,

Let $K/\mathbb{Q}_2$ be a finite extension, and let $E/K$ be an elliptic curve with good ordinary reduction, and such that $\mathbb{Q}_2(j(E))=K$. Let $\rho:\operatorname{Gal}(\overline{K}/K)\to \operatorname{GL}(2,\mathbb{Z}_2)$ be the $2$-adic representation associated to $T_2(E)$, the $2$-adic Tate module of $E$, and let $I_K$ be the image of the inertia subgroup via $\rho$. The theory tells us that there is a $\mathbb{Z}_2$-basis $\{P,Q\}$ of $T_2(E)$, such that $I_K$ is a subgroup of $$ \left\{\left( \begin{array}{cc} \ast & \ast \\\ 0 & 1\\\ \end{array}\right) \right\} \subset \operatorname{GL}(2,\mathbb{Z}_2),$$ and, moreover, if $\sigma\in I_K$, then the upper-left corner is $\chi_K(\sigma)$, where $\chi_K$ is the $2$-adic cyclotomic character $\chi_K:\operatorname{Gal}(\overline{K}/K)\to (\mathbb{Z}_2)^\times$. Let $\Phi$ be the image of $\chi_K$.

**My question** is the following: is $I_K$ necessarily a conjugate in $\operatorname{GL}(2,\mathbb{Z}_2)$ of a subgroup of the form
$$ B(\Phi,\Psi)=\left\{\left( \begin{array}{cc} a & b \\\ 0 & 1\\\ \end{array}\right) : a\in \Phi, b \in \Psi \right\} \subset \operatorname{GL}(2,\mathbb{Z}_2),$$
for some fixed additive subgroup $\Psi$ of $\mathbb{Z}_2$? If so, what is the reason?

This would rule out some *weird* subgroups of $\operatorname{GL}(2,\mathbb{Z}_2)$ to appear as inertia subgroups. For example, can the following group $I$ appear as an inertia subgroup $I_K$ in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$?
$$I=\left\langle \left( \begin{array}{cc} 5 & 2 \\\ 0 & 1\\\ \end{array}\right) \right\rangle =\left\{ \left( \begin{array}{cc} 1 & 0 \\\ 0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 5 & 2 \\\ 0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 1 & 4 \\\ 0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 5 & 6 \\\ 0 & 1\\\ \end{array}\right)\right\}\subset \operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z}).$$

**Edit:** The example group $I$ may appear as inertia in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$ if we do not assume that $K$ is the minimal field of definition of $E/K$, i.e., $K=\mathbb{Q}_2(j(E))$, for trivial reasons. For instance, suppose $E / \Bbb Q_2$ has inertia subgroup $B((\mathbb{Z}/8\mathbb{Z})^\times,\mathbb{Z}/8\mathbb{Z})$ in $\mathbb{Q}_2(E[8])/\mathbb{Q}_2$, and let $K$ be the fixed field $\Bbb{Q}_2(E[8])^I$, where $I$ is as in the example above modulo $8$. Then the inertia subgroup for $E/K$ in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$ is as indicated in the example.

Thank you!