Let $G$ be a connected topological group and $N$ a discrete normal subgroup of $G$.
Is it true that $Z(G)/N = Z(G/N)$, where $Z(G)$ denotes the center of $G$?
I know that every discrete normal subgroup of $G$ is contained in its center.
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Ah, I see, every. – Igor Rivin Dec 07 '13 at 04:28
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This is probably true for Lie groups if you care about this case. – Moishe Kohan Dec 07 '13 at 15:16
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Yes it's true and this is just the same argument as showing that $N$ is central.
Define $Z_{G/N}(G)$ as the inverse image of $Z(G/N)$ in $G$. The equality is equivalent to asking whether $Z_{G/N}(G)=Z(G)N$. The inclusion $\supset$ is trivial.
For $z\in Z_{G/N}(G)$, we have, for all $g\in G$, $gzg^{-1}z^{-1}\in N$. Since $N$ is discrete and $G$ is connected, this implies that for all $g\in G$, we have $gzg^{-1}z^{-1}=1$. So $z\in Z(G)$. (And hence $Z(G)N=Z(G)$, which retrieves that $N$ is central.)
YCor
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