Here is a question I get from sitting in my Lie algebra class: Fix a Lie algebra $\mathfrak{h}$, we know there is a unique simply connected Lie group $H$ which serves as the universal cover of other connected Lie groups with the same Lie algebra. Now assuming $H$ is a $n_1$ sheeted cover of $H_1$, and a $n_2$ sheeted cover of $H_2$, (we are not restricting the Deck groups yet). we know $\phi: \mathfrak{h_1}\cong \mathfrak{h_2} (\cong \mathfrak{h})$ but we only cares about the isomorphism between the first two factors. Now as we travel along all $a\in \mathfrak{h_1}$, $(a, \phi (a))$ will form a Lie subalgebra in $\mathfrak{h_1}\bigoplus\mathfrak{h_2}$, denoted by $\mathfrak{h'}$. By the existence theorem of Lie subgroups we have a uniqueLie subgroup $G\subset H_1\times H_2$ corresponding to $\mathfrak{h'}$. Note $\mathfrak{h_1},\mathfrak{h_2}, \mathfrak{h'}$ are isomorphic Lie subalgebras in $\mathfrak{h_1}\bigoplus\mathfrak{h_2}$, however since they sort of "sit in different directions" in the ambient Lie algebra when we form the exponential map we are producing non isomorphic Lie subgroups. Anyhow, Project down from $G$ to $H_1$ and $H_2$ induce an iso on the Lie algebra therefore we know $G$ actually covers $H_i$. That's the set up of the question.

Now if I know $n_1$ and $n_2$ are coprime to each other, then automatically $G$ has to be the universal cover $H$. However, if say the maximal common divisor of $n_1$ and $n_2$ is 2, and that there are two non isomorphic Lie groups that could cover both $H_1$ and $H_2$, namely H and the one doubly covered by H donoted $\tilde{H}$, (here I need some compatible condition on the Deck group, but let assume $\tilde{H}$ covers $H_i$). So my question is which one is isomorphic to $G$ when I draw the graph? I am worried a little bit that the there may not a canonical way of choosing this isomorphism $\phi$ to make this question make sense, (or is there)? My guess is that if there is a "canonical choice", then the $G$ we get from the graph should be the smaller $\tilde{H}$, while my colleage thinks that by choosing different $\phi$, you can get both. When $H_1$ and $H_2$ have a lot of non isomorphic common covers, then by choosing different $\phi$ you can produce all of them as subgroups of $H_1\times H_2$? (fixing $H_1$ and $H_2$.) I really hope someone ccan shed lights on this, thanks very much!