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$\DeclareMathOperator\SU{SU}\DeclareMathOperator\rank{rank}$Take a simply connected Lie group $G$ such as $\SU(N)$ and a maximal rank subgroup $H$, i.e. $\rank(G) = \rank(H)$. Assume that $H$ takes the following form $H = H_1 \times H_2 \times \dotsb \times H_n$. Is it true that the $H_i$ are not simply connected?

For instance, it seems to be true given the classification tables in Antoneli, Forger, and Gaviria - Maximal subgroups of compact Lie groups, but I am not able to find a proof.

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    $\begingroup$ Are you assuming that $n>1$? After all, $\mathrm{Spin}(7)$ contains $\mathrm{Spin}(6)$ as a maximal rank subgroup and $\mathrm{G}_2$ contains $\mathrm{SU}(3)$ as a maximal rank subgroup, and all of these groups are simply-connected. Also, $\mathrm{Spin}(5)$ contains $\mathrm{SU}(2)\times\mathrm{SU}(2)$. $\endgroup$
    – Robert Bryant
    Commented Jun 26 at 22:28
  • $\begingroup$ Oh, also, note that $\mathrm{Sp}(n)$ contains $\mathrm{Sp}(1)\times\cdots\times\mathrm{Sp}(1)$ ($n$ times) as a maximal rank subgroup (of rank $n$), and $\mathrm{Sp}(m)$ is simply-connected for all $m$. $\endgroup$
    – Robert Bryant
    Commented Jun 27 at 3:36
  • $\begingroup$ If $G$ is allowed to be non-semisimple, you should say what you mean by "rank". $\endgroup$
    – YCor
    Commented Jun 27 at 7:28

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