Let $K/\mathbb{Q}_p$ be a finite field extension and $\mathcal{O}_K\subseteq K$ be its ring of integral elements. Let also $G/\mathcal{O}_K$ be a finite flat $p$-group scheme that is also an $\mathcal{O}_E$-module where $E/\mathbb{Q}_p$ is a finite field extension and $\mathcal{O}_E\subseteq E$ is its ring of integral elements. Let $H \subseteq G$ be any subgroup scheme (it does not have to be $\mathcal{O}_E$-stable). Let $G' = G \otimes_{\mathcal{O}_E} \mathcal{O}_{E'}$ be the Serre tensor construction, where $E'/E$ is a totally ramified extension of degree $r$. Then it is true that $\#\omega_{G} = \frac{1}{r}\#\omega_{G'}$, where $\omega_G = s^*\Omega^1_{G/\mathcal{O}_K}$ and $s:\text{Spec}(\mathcal{O}_K)\to G$ is the identity section. Let $H' \subseteq G'$ be a subgroup such that it pulls back to $H$ under the closed embedding $G \to G'$. Then it should be true that $\#\omega_{H} \ge \frac{1}{r}\#\omega_{H'}$ for every such group $H'$. My question is when equality holds.

When $H$ is an $\mathcal{O}_E$-module (i.e., $\mathcal{O}_E$-stable in $G$), then I can prove when $H'$ is the Serre tensor construction of $H$, which is a subgroup of $G'$, that we have an equality as we do with $G$ (same proof). When it is not $\mathcal{O}_E$-stable, I don't think I can "Serre tensor construct" $H$ to be a subgroup of $G$, but I can, for example, take the image of $H$ in $G'$ under the embedding $G \to G'$ which pulls back to $H$. This is an example of a strict inequality.

Let $K/\mathbb{Q}_p$ be a finite field extension and $\mathcal{O}_K\subseteq K$ be its ring of integral elements. Let also $G/\mathcal{O}_K$ be a finite flat $p$-group scheme that is also an $\mathcal{O}_E$-module where $E/\mathbb{Q}_p$ is a finite field extension and $\mathcal{O}_E\subseteq E$ is its ring of integral elements. Let $H \subseteq G$ be any subgroup scheme (it does not have to be $\mathcal{O}_E$-stable). Let $G' = G \otimes_{\mathcal{O}_E} \mathcal{O}_{E'}$ be the Serre tensor construction, where $E'/E$ is a totally ramified extension of degree $r$. Then it is true that $\text{lg}(\omega_{G}) = \frac{1}{r}\text{lg}(\omega_{G'})$, where $\omega_G = s^*\Omega^1_{G/\mathcal{O}_K}$ and $s:\text{Spec}(\mathcal{O}_K)\to G$ is the identity section. Let $H' \subseteq G'$ be a subgroup such that it pulls back to $H$ under the closed embedding $G \to G'$. Then it should be true that $\text{lg}(\omega_{H}) \ge \frac{1}{r}\text{lg}(\omega_{H'})$ for every such group $H'$. My question is when equality holds.

When $H$ is an $\mathcal{O}_E$-module (i.e., $\mathcal{O}_E$-stable in $G$), then I can prove when $H'$ is the Serre tensor construction of $H$, which is a subgroup of $G'$, that we have an equality as we do with $G$ (same proof). When it is not $\mathcal{O}_E$-stable, I don't think I can "Serre tensor construct" $H$ to be a subgroup of $G$, but I can, for example, take the image of $H$ in $G'$ under the embedding $G \to G'$ which pulls back to $H$. This is an example of a strict inequality.

Let $K/\mathbb{Q}_p$ be a finite field extension and $\mathcal{O}_K\subseteq K$ be its ring of integral elements. Let also $G/\mathcal{O}_K$ be a finite flat $p$-group scheme that is also an $\mathcal{O}_E$-module where $E/\mathbb{Q}_p$ is a finite field extension and $\mathcal{O}_E\subseteq E$ is its ring of integral elements. Let $H \subseteq G$ be any subgroup scheme (it does not have to be $\mathcal{O}_E$-stable). Let $G' = G \otimes_{\mathcal{O}_E} \mathcal{O}_{E'}$ be the Serre tensor construction, where $E'/E$ is a totally ramified extension of degree $r$. Then it is true that $\#\omega_{G} = \frac{1}{r}\#\omega_{G'}$, where $\omega_G = s^*\Omega^1_{G/\mathcal{O}_K}$ and $s:\text{Spec}(\mathcal{O}_K)\to G$ is the identity section. Let $H' \subseteq G'$ be a subgroup such that it pulls back to $H$ under the closed embedding $G \to G'$. Then is it true that $\#\omega_{H} \le \frac{1}{r}\#\omega_{H'}$ for every such group $H'$?

When $H$ is an $\mathcal{O}_E$-module (i.e., $\mathcal{O}_E$-stable in $G$), then I can prove when $H'$ is the Serre tensor construction of $H$, which is a subgroup of $G'$, that we have an equality as we do with $G$ (same proof). When it is not $\mathcal{O}_E$-stable, I don't think I can "Serre tensor construct" $H$ to be a subgroup of $G$, but I can, for example, take the image of $H$ in $G'$ under the embedding $G \to G'$ which pulls back to $H$. I actually think if I use this, I would show the inequality in the other way; if so, the question becomes "Does there exist an $H'$ which pulls back to $H$ such that $\#\omega_{H} \le \frac{1}{r}\#\omega_{H'}$"?

Let $K/\mathbb{Q}_p$ be a finite field extension and $\mathcal{O}_K\subseteq K$ be its ring of integral elements. Let also $G/\mathcal{O}_K$ be a finite flat $p$-group scheme that is also an $\mathcal{O}_E$-module where $E/\mathbb{Q}_p$ is a finite field extension and $\mathcal{O}_E\subseteq E$ is its ring of integral elements. Let $H \subseteq G$ be any subgroup scheme (it does not have to be $\mathcal{O}_E$-stable). Let $G' = G \otimes_{\mathcal{O}_E} \mathcal{O}_{E'}$ be the Serre tensor construction, where $E'/E$ is a totally ramified extension of degree $r$. Then it is true that $\#\omega_{G} = \frac{1}{r}\#\omega_{G'}$, where $\omega_G = s^*\Omega^1_{G/\mathcal{O}_K}$ and $s:\text{Spec}(\mathcal{O}_K)\to G$ is the identity section. Let $H' \subseteq G'$ be a subgroup such that it pulls back to $H$ under the closed embedding $G \to G'$. Then it should be true that $\#\omega_{H} \ge \frac{1}{r}\#\omega_{H'}$ for every such group $H'$. My question is when equality holds.

When $H$ is an $\mathcal{O}_E$-module (i.e., $\mathcal{O}_E$-stable in $G$), then I can prove when $H'$ is the Serre tensor construction of $H$, which is a subgroup of $G'$, that we have an equality as we do with $G$ (same proof). When it is not $\mathcal{O}_E$-stable, I don't think I can "Serre tensor construct" $H$ to be a subgroup of $G$, but I can, for example, take the image of $H$ in $G'$ under the embedding $G \to G'$ which pulls back to $H$. This is an example of a strict inequality.