Let $k$ be an algebraically closed field of characteristic zero. Let $G$ be a group algebraic space over $k$ such that $G\to $ Spec $k$ is locally of finite type.


Suppose that $G(k)$ is finite. 


Does it follow that $G\to$ Spec $k$ is finite (and thus $G$ is a scheme)?


If we assume $G$ is quasi-separated over $k$, then $G$ is in fact a scheme (see [Commutative group algebraic spaces](https://mathoverflow.net/questions/290731/commutative-group-algebraic-spaces)). It follows then that $G$ is finite. 


But what if $G$ is not quasi-separated over $k$? Is $G$ still finite over $k$?



Group algebraic spaces that are locally of finite type and have only finitely many points
The comment of nfdc23 answers the question:


No: for any algebraically closed field $k$, let $C$ be any commutative $k$-group scheme of finite type with positive dimension and $G=C/H$ where $H\to C$ is the $k$-subgroup functor given by the (etale) constant group on $C(k)$. In other words, $G$ is the quotient of $C$ modulo the etale equivalence relation $\delta:H\times C \to C\times C$ defined by $(h,c)\mapsto (c,hc)$ This is a locally finite type algebraic space group with $G(k)=1$ and dimension $\dim(G)=dim(C)>0$. Note that $G$ is not quasi-separated (since $\delta$ is not quasi-compact).


