Let $k$ be a field. It might as well be algebraically closed, but I do not want to assume that it has characteristic $0$. I will write "group" for "affine group scheme over $k$", not assuming smoothness.

Two groups can have the same Lie algebras without being equal. For example, if $k$ has characteristic $2$, then every maximal torus in $\operatorname{SL}_2$ has the same Lie algebra as the centre $\mu_2$. Even two smooth groups can have the same Lie algebras without being equal: for example, all maximal tori in $\operatorname{SL}_2$ have the same Lie algebra. At least it is true that, if a smooth group $H$ is contained in a connected group $G$, and their Lie algebras are equal, then $H$ equals $G$; and so, if two connected subgroups $H_1$ and $H_2$ of $G$ have equal Lie algebras *and* smooth intersection, then they are equal.

I'm looking more for a result in line with Borel - Linear algebraic groups, Theorem 13.18(4)(d): given a maximal torus $T$ in a smooth, *reductive* group $G$, and a root $\alpha$ of $T$ in $G$, there is a unique smooth, connected subgroup of $G$ that is normalised by $T$ and whose Lie algebra is the $\alpha$-weight space of $T$ on $\operatorname{Lie}(G)$. The key ingredients here are reductivity and the torus action.

So I'm interested in any more general results of this sort that allow one to deduce equality of groups from equality of their Lie algebras. If that's too broad, I'll focus a bit: suppose that $G$ is a smooth, reductive group; $H_1$ and $H_2$ are smooth, connected, reductive subgroups; and $T$ is a torus in $H_1 \cap H_2$ that is not necessarily maximal in $G$, but is maximal in both $H_1$ and $H_2$. In this setting, if the Lie algebras of $H_1$ and $H_2$ are equal, then can we conclude that the groups are equal?

EDIT: I forgot to add, in case it helps, that, in my situation, $\operatorname C_G(T)^\circ$ (the connectedness automatic if $G$ itself is connected) is a torus.

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