[This is a sequel to the previous question [sub-coroot systems](https://mathoverflow.net/questions/441161/sub-coroot-systems), that has been answered! :-) ]


Let $T$ be a maximal torus of a compact Lie group $K$,
and let $\Lambda \subset {\mathfrak t}$ be the coroot lattice for $(K,T)$, where $\mathfrak t$ is the Lie algebra of $T$.


Assume now that $K'$ is another compact Lie group that admits a homomorphic embedding $i:K\hookrightarrow K'$ verifying the condition that $i(Z\_K)\subset Z\_{K'}$, where $Z\_K, Z\_{K'}$ are the centers of $K$ and $K'$, respectively. Let $T'$ be a maximal torus of $K'$ such that $i(T)\subset T'$. This gives a vector space inclusion ${\mathfrak t} \subset {\mathfrak t}'$, into the Lie algebra of $T'$.


Question: Is it true that
$$ \Lambda = {\mathfrak t} \cap \Lambda'$$


where $\Lambda' \subset {\mathfrak t}'$ is the coroot lattice of $(K',T')$?


If this is not always true, are there some simple conditions under which it becomes true?



Sub-coroot lattices
As for your [other question](https://mathoverflow.net/questions/441161/sub-coroot-systems), I will work with complex Lie groups, but you can pass to maximal compact subgroups if you prefer.


Consider the natural embedding $\operatorname{SO}\_4(\mathbb C) \to \operatorname{SL}\_4(\mathbb C)$, where $\operatorname{SO}\_4$ is taken with respect to the quadratic form $(x\_1, x\_2, x\_3, x\_4) \mapsto x\_1 x\_4 + x\_2 x\_3$. If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}\_4(\mathbb C)$ by $\alpha\_1, \alpha\_2, \alpha\_3$, in the obvious fashion, then the common restriction of $\alpha\_1$ and $\alpha\_3$ to the diagonal torus in $\operatorname{SO}\_4(\mathbb C)$ is a simple (with respect to the upper-triangular Borel) root $a$, and the common restriction of $\alpha\_1 + \alpha\_2$ and $\alpha\_2 + \alpha\_3$ is a simple root $b$. Then $a^\vee$ equals $\alpha\_1^\vee + \alpha\_3^\vee$, $b^\vee$ equals $\alpha\_1^\vee + 2\alpha\_2^\vee + \alpha\_3^\vee$, and $\Lambda$ equals $\mathbb Z a^\vee + \mathbb Z b^\vee$, but $(\Lambda' \cap \mathfrak t)/\Lambda$ has order $2$, generated by the image of $\alpha\_2^\vee$.


