**Fact/motivation:**
$\DeclareMathOperator{\inv}{inv}\DeclareMathOperator{\GL}{GL}$
If $G$ is a smooth manifold of dim.$n$ and a group, s.t. the multiplication
$$
m \colon G \times G \to G
$$
is smooth, then it implies that the inversion
$$
\inv \colon G \to G
$$
is smooth too. This is e.g. due to the '*Implicit Function Theorem*' and can be found for example in https://services.math.duke.edu/~bryant/ParkCityLectures.pdf (The author does not require for a Lie group to have smooth inversion).

But I don't know of any similar statement for regular maps in algebraic geometry. So my question is:

**Question:**

Does a similar statement (not) hold for algebraic (affine) groups? That is, is there any group, for instance of the form $G = V(I) \subset \mathbb{A}^n(k)$ over some *algebraically closed* field $k$ ($I \subset k[X_1,\ldots, X_n]$ ideal), such that the multiplication is regular but the inversion is not?

**Non-example:**

All I could think of was for example $\mathbb{A}^1(k)^{\times}$, but this is in general not closed inside $\mathbb{A}^1(k)$ (unless $k$ finite but then every map is polynomial), and if one 'closifies' it, i.e. understands it rather as $V(XY-1) \subset \mathbb{A}^2(k)$, this forces the inversion automatically to be polynomial (so this is just $\mathbb{G}_m(k)$). Same of course applies for $\GL_n$.