Serre tensor construction on finite flat group schemes 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.
 A: Thinking about this more, it might be easier than you think. You don't seem to even care about the actions of $\mathcal{O}_E$ or $\mathcal{O}_{E'}$ in your question so you should consider what happens if you just forget them. Maybe $G'$ is complicated as an $\mathcal{O}_{E'}$-group scheme, but if you just consider $G'$ as a group scheme with $\mathcal{O}_E$-action then you can construct it as $G\otimes_{\mathcal{O}_E}\mathcal{O}_{E'}$ where $\mathcal{O}_{E'}$ is just considered as a $\mathcal{O}_E$-module. As a $\mathcal{O}_E$-module $\mathcal{O}_{E'}$ is simply free of rank $r$, and so it should just follow that $G'$ is isomorphic to $G^r$. Now your questions become easy because if we fix such an isomorphism (e.g. by picking an $\mathcal{O}_E$-basis of $\mathcal{O}_{E'}$) then we can set $H'=H^r$ and this gives you the construction you want. 
NB I am assuming that by $\#\omega_G=\frac{1}{r}\#\omega_{G'}$ you mean actually that the length of $\omega$ gets divided by $r$; the orders themselves are powers of $p$ and so of course usually not multiples of $r$.
What do you think?
