Two affine group schemes over $k$ with isomorphic group for each $k$-algebra $S$ but not isomorphic as affine group schemes

Question

I'm looking for two affine group schemes $G_1$ and $G_2$ (i.e. functors from (affine group schemes over $k$) to (Group) ) such that for every $k$-algebra $S$, $G_1(S)\cong G_2(S)$, but $G_1$ and $G_2$ are not isomorphic as affine group schemes (i.e. as functors). The following is the guideline of the example given to me. In the following $k=\mathbb{F}_p$.

Consider $\varphi:\operatorname{Spec}k[x^{\pm 1}] = \mathbb G_m\to \operatorname{Spec}k[x^{\pm 1}] = \mathbb G_m$ corresponding to the map $x\mapsto x^p$ and define $\mu_p = \operatorname{Spec}\left(k[x^{\pm 1}]/(x^p-1)\right) = \ker\varphi$.

Consider $F_p:\operatorname{Spec}k[x] = \mathbb G_a\to \operatorname{Spec}k[x] = \mathbb G_a$ corresponding to the map $x\mapsto x^p$ and define $\alpha_p = \operatorname{Spec}\left(k[x]/(x^p)\right) = \ker F_p$.

I know how to proof that $\mu_p$ is not isomorphic to $\alpha_p$ as affine group schemes.

Now for all $k$-algebra $S$ we have $\mu_p(S) = \{s\in S \ | \ s^p = 1\}$ and $\alpha_p(S) = \{s\in S \ | \ s^p=0\}$. Then we have a bijection: $$ \mu_p(S) \longleftrightarrow \alpha_p(S), \qquad s\longleftrightarrow s-1 $$ Also $\alpha_p(S)$ and $\mu_p(S)$ are abelian, and each non trivial element of each group has order $p$.

I'm stuck here. I know that the informations stated above are sufficient to imply that $\alpha_p(S)\cong \mu_p(S)$ if the two groups are finitely generated, but I don't know how to conclude (and if it is possible) in the not finitely generated case.

Any hint or ideas? Thank you in advance.

Answer 1

Note that $\mu_p(S)$ and $\alpha_p(S)$ are vector spaces over $\mathbb{F}_p$ thanks to the following actions:

$$ \mathbb F_p \times \mu_p(S) \to \mu_p(S), \qquad (n,s)\mapsto s^n $$

$$ \mathbb F_p \times \alpha_p(S) \to \alpha_p(S), \qquad (n,s)\mapsto n+s $$

In order to prove that $\mu_p(S) \cong \alpha_p(S)$ as abelian groups, it is enough to show that two vector spaces over $\mathbb F_p$ that have the same cardinalty are isomorphic. This is a consequence of this question and the proof is complete.