Let $p \in \mathbb{P}$ be a prime and $k$ a field where $\operatorname{char} k\neq p$. equivalently, $p \in k^*$.
Let $G:= \operatorname{Spec}A$ be an affine finite group scheme over $k$ of order $p^d, d \ge 2$. I have to show that $G$ is unramified. I have prepared up to a detail a "proof" and I would like to know if this approach is correct.
Unramified is equivalent to $\Omega_{A/k}=0$. Recall that the order $p^d$ of a an affine finite group scheme $G$ over a field $k$ as the rank of free $k$-algebra $A=k^{p^d}$. What do we know about $\vert G \vert$? Is $\vert G \vert=p^d$? Or does there at least exist a power of $p$ which maps all lements of $G$ to neutral element $e$?
If yes then themultiplication by $p^d$ map on $G$ induces a zero morphism/ everything to identity-map $\overline{p^m}=\bar{e}: G \to G$. As this morphism is functorially this induces as well for every $g \in G$ a map on cotangent space $p^m_g: m_g/m_g^2 \to m_g/m_g^2$ which has also to be a zero map.
The question is how $(\overline{p^d})_g: m_g/m_g^2 \to m_g/m_g^2$ is explicitly described as map between vector spaces $m_g/m_g^2$. Is it given by a multiplication by $p^m$ or at least a power of $p$?
I wish so since if $(\overline{p^d})_g: m_g/m_g^2 \to m_g/m_g^2$ is the multiplication by $p^d$-map then it's an isomorphism of vector spaces since $p \in k^*$. Then we would have an isomorphism between $m_g/m_g^2$ which simultaneously is a zero map. Thus $m_g/m_g^2=0$ and since $g \in G$ was arbitrary $\Omega_{A/k}=0$.
Everything we need is the explicit description of a $(\bar{r})_g: m_g/m_g^2 \to m_g/m_g^2$ map induced by a multiplication by $r$ map of $G$ where $r \in \mathbb{N}$.