Can non-geometrically reduced reduced subschemes happen for reductive groups?

Question

The title is meant to be punchy, but also a tongue-in-cheek acknowledgement of the prevalence of ‘reduce’-derived words in this area. (Unfortunately, I overlooked the fact that the question in the title is the opposite of the one in the body. Fortunately, @LaurentMoret-Bailly's answer is quite clear about this: ‘yes’ to the question in the title, ‘no’ to the question below.)

Let $k$ be a separably closed field, let $\overline k/k$ be an algebraic closure, and suppose that $G$ is a connected $k$-group scheme of finite type.

I know that $G_\text{red}$ need not be geometrically reduced, but the standard examples of this have $G_\overline k$ unipotent. Is that for convenience, or is it necessary?

To put it more precisely, suppose that $(G_\overline k)_\text{red}$ is a reductive $\overline k$-group scheme. Does it follow that $G_\text{red}$ is geometrically reduced?

Answer 1

No, it does not follow that $G_\mathrm{red}$ is geometrically reduced (thus the answer to the title question is yes).

Let $p=\mathrm{char}(k)>0$ and let $H=\mathbb{G}_a \rtimes \mathbb{G}_m$ be the group of affine transformations of $\mathbb{A}^1_k$.

Let $(x,y)\in H$ act on $\mathbb{A}^1_k$ by the "twisted" action $z\mapsto x^p+y^p z$. If $z\in k$ is not a $p$-th power, its stabilizer $G$ is easily seen to be reduced, but $G_{\overline{k}}$ is isomorphic to the stabilizer of the origin which is $\alpha_p \rtimes \mathbb{G}_m$.

In particular, $(G_{\overline{k}})_\mathrm{red}$ is isomorphic to $\mathbb{G}_m$.