Does $SO(7)$ have a universal covering by $SU(6)$?

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A question which I'm sure is inane to an expert on Lie groups, but since $SU(2)$ has a universal covering to $SO(3)$, and $SU(2)$ is isomorphic to the unit quaternions, this question might be entertained for the following reason:

The Quaternions enjoy a complexification into the Octinions in which they remain a division algebra (though fail to be associative). Although I'm sure the non-associativity of $\mathbb{O}$ kills the prospect of finding an isomorphism from its unit ball to $SU(6)$, it might be the case that the geometry extends between the two higher dimensional Lie groups, even if the algebraic structure probably doesn't.

The question first arose due to comments on Can octonions be used to rotate 7-dimensional vectors?

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The universal cover of $SO(n)$ is the group $Spin(n)$. For small values of $n$, there are isomorphisms between various compact simply connected Lie groups that you might not expect, corresponding to accidental isomorphisms between small Dynkin diagrams. We have the one you mentioned; $Spin(3)=SU(2)$. There is also $Spin(4)=SU(2) \times SU(2)$, and $Spin(5)=Sp(2)$, and $Spin(6)=SU(4)$.

However, we know that $Spin(7)$ isn't $SU(6)$ since $SU(6)$ is $35$ dimensional and $SO(7)$ is $21$ dimensional.