A discrete normal subgroup is contained in the center

Question

Let $G$ be a connected Lie group and $N$ a discrete normal subgroup of $G$. Then $N$ is contained in the center $Z(G)$.

I've fooled around with this for a little bit and I can't figure out how to use the hypothesis that $N$ is discrete. I think maybe it uses some fact that I do not know.

Re: your initially-missing "connectedness" hypothesis: It's worth remembering that every finite group is a Lie Group (0-dimensional, discrete topology), so when you come to prove a fact like this, and you notice it's not true for finite groups, then surely connectedness will come into play! — John Hughes, Jan 27 '19 at 19:46

Matt Samuel · Accepted Answer · 2019-01-27T19:41:37.297

10

Let $n$ be an element of the subgroup. Then the function $f:G\to N$ sending $g\mapsto gng^{-1}$ is continuous, and since $G$ is connected, so is the image of $f$. Since $N$ is discrete, $f(g) =n$ for all $g$.

edited Jan 27 '19 at 19:41

answered Jan 27 '19 at 18:12

Matt Samuel

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Ah yes, I forgot to say $G$ was connected. That's one of my permanent assumptions. – Tuo Jan 27 '19 at 19:40
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@Tuo I'll edit my answer then. – Matt Samuel Jan 27 '19 at 19:41
Why does $N$ beign discrete impliy $gng^{-1}=n$? – Aug 16 '20 at 13:30
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@HaKuNaMaTaTa The only connected subsets of a discrete space are singletons, and the image of $G$ is connected and contains $n$. – Matt Samuel Aug 16 '20 at 13:31
Sorry but could you explain more on why the function f is continuous? – Hetong Xu May 11 '21 at 02:44
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@HetongXu It's a composition of continuous functions. – Matt Samuel May 12 '21 at 11:07

A discrete normal subgroup is contained in the center

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