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Let $Q$ be a simply-connected compact Lie group. Can one outline the proof (or provide the counter examples if my statement is false) that

there does not exist any group $G$ (with no topology) with a finite normal subgroup $N$, such that $G/N=Q$, where $N$ is not a direct factor of $G$?

Note that this is always false for $Q$ connected but not simply-connected.

(It is easy to come up with counter examples for connected Lie groups, such as $U(1)/(\mathbb{Z}/N\mathbb{Z})=U(1)$, $Spin(n)/(\mathbb{Z}/2\mathbb{Z})=SO(n)$. But those $Q$ are $U(1)$ and $SO(n)$, they are not simply-connected Lie groups.)

I would like to see that such finite group extensions cannot exist for $Q=\mathrm{SU}(N)$ explicitly. Perhaps more generally, one can argue that for $Q$ of type $A_n,B_n,C_n,D_n,E_6,E_7,E_8,F_4,G_2$, the above statement can be proved to be correct(?).

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    $\begingroup$ $G\rightarrow Q$ is a finite covering, use the simple-connectedness of $Q$. The question is not appropriate for this site. $\endgroup$ – abx May 5 '17 at 5:19
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    $\begingroup$ The question doesn't say if it is assumed that $G$ has a structure of Lie group compatible with that of $Q$ (in which case @abx's comment applies) or if this is just abstract algebra (in which case probably the answer is still affirmative, but if so then seems to require input beyond the theory of compact Lie groups). $\endgroup$ – nfdc23 May 5 '17 at 5:25
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    $\begingroup$ I changed the tags since the topological tags such as "general topology" were not appropriate, while (non-topological, non-algebraic) central extensions of simple algebraic groups are naturally considered in K-theory. $\endgroup$ – YCor May 5 '17 at 18:29
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    $\begingroup$ It seems you can get a counterexample for any simple Lie type having a nontrivial Dynkin diagram automorphism: $A_\ell$ for $\ell \geq 2$ or $D_\ell$ or $E_6$: here the finite group of outer automorphisms of your group $Q$ has order 2 or 6. So just form a (non-split) semidirect product. This should apply in particular for special unitary groups of rank $\geq 2$.. $\endgroup$ – Jim Humphreys May 5 '17 at 18:59
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    $\begingroup$ @JimHumphreys you understand the extension in the wrong sense. The finite subgroup should be normal. Btw it follows from my second comment that the answer is positive in classical groups, so $SU(3)$ will not be a counterexample. $\endgroup$ – YCor May 5 '17 at 20:05

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