Let $k$ be a field of characteristic $p>0$, and let $G$ be a finite connected group scheme over $k$. Let $FG$ be the frobenius twist of $G$, i.e. $FG=G \times_{\mathrm{Spec}\ k, Fr} \mathrm{Spec}\ k$, and let $f:G \to FG$ be the usual map that on the level of affine rings is given by $a \otimes \lambda \mapsto \lambda^pa$ [correction: $a^p\lambda$]for $a \in \mathcal{O}(G)$ and $\lambda \in k$.
Let $\Lambda = k[t]/t^2$ be the dual numbers. Why is $f(\Lambda): G(\Lambda) \to FG(\Lambda)$ the zero morphism?
This is claimed in Shatz's book Profinite groups, algebra, and geometry.
Best Answer
Just to turn my comment into an answer (let me know if more clarification is needed).
Since $G$ and $FG$ are connected we know that $G(k)=FG(k)=\{e\}$. Now, for any finite flat group scheme $H$ over $k$ we have that $\mathrm{Lie}(H)=\ker(H(k[\varepsilon])\to H(k))$ where $k[\varepsilon]\to k$ is the unique $k$-algebra map. In particular, if $H$ is connected then $\mathrm{Lie}(H)=H(k[\varepsilon])$. So, if $\mathrm{Frob}:G\to FG$ were injective then, in particular, the induced map $G(k[\varepsilon])\to FG(k[\varepsilon])$ is injective. Or, in other words, we'd have that the map $\mathrm{Lie}(G)\to\mathrm{Lie}(FG)$ is injective. But, it's not hard to see that the Frobenius map induces the zero map on Lie algebras (since $da^p=p da^{p-1}=0$ since the Lie algebra is a $k$-space).