I've been trying a learn a little more about group schemes by working through a set of exercises on Brian Conrad's website. Exercise 8.3 of http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf reads:

If $k$ is a perfect field and $G$ is a locally finite type $k$-group prove that $G_{red}$ is a closed $k$-subgroup of $G$. Can you find a counterexample if $k$ is not perfect? (There's a third part to the question, but it's irrelevant here)

I'll try to not give away too much for the benefit of others who want to use these exercises, but the first part of the exercise is Lemma 7.10 of http://www.math.columbia.edu/algebraic_geometry/stacks-git/groupoids.pdf in DeJong's stacks project and the hint given for the omitted proof is that $G_{red}\times_k G_{red}$ is reduced if $k$ is perfect.

For the counterexample part, I was able to find a disconnected example over any imperfect field, but it really seemed to depend crucially on being disconnected.

This naturally suggests the question:

If $G_{red}$ is not a subgroup scheme of $G$, must $G$ be disconnected?

edit: After almost a week out an answer in the zero-dimensional case was given and then retracted, but there has not been much action otherwise. Perhaps this is an open problem if $G$ is not finite.

For any finite group scheme $H$, $H_{red}$ is a subgroup of $H$ if and only if $H/H^0$ is a subgroup of $H$ (and in fact if $k$ is perfect, $H \cong H^0 \oplus H_{red}$, see Pink's notes Lecture 6- note this very explicitly depends on $H$ being finite)

For group schemes of locally finite type (as we assume here) there has been much done, for instance in SGA 3 where they prove among other things that if $G$ is connected, it's actually quasi-compact and thus of finite type (Proposition 2.4). Moreso, for a locally finite type group $H$ over a field, $(H^0)_{red}$ is a group *in the category of reduced schemes*. What's unclear is whether they believed it to be a group in the category of schemes but weren't able to prove that or if they knew of a counterexample but didn't include it.