I am currently working through the book "Lie Groups, Lie algebras and Rep." of Brian C. Hall and I came across the following problem (Chapter 1, Problem 11 in the 2nd. ed.)
A subset $E$ of a matrix Lie group $G$ is called discrete if for each $A$ in $E$ there is a neighborhood $U$ of $A$ in $G$ such that $U$ contains no point in $E$ except for $A$. Suppose that $G$ is a connected matrix Lie group and $N$ is a discrete normal subgroup of $G$. Show that $N$ is contained in the center of $G$.
My attempt for a proof is the following,
The center of $G$ consists of all the elements $A$ in $G$ such that $AB = BA$ for all $B \in G$. Since $N$ is a normal subgroup, $BAB^{-1} \in N$ for all $A \in N$ and $B \in G$. Because $G$ is connected, for all $A$ and $B$ in $G$ there exists a continuous path $A(t)$, $a \leq t \leq b$, lying in G with $A(a) = A$ and $A(b)$ = B. Therefore, suppose we have the continuous path $B(t)$ going from $B(a) = I$ to $B(b) = B$ in some interval $a \leq t \leq b$ and take the product $B(t)AB(t)^{-1}$ for some $A \in N$. Given that $N$ is a normal subgroup, the product $B(t)AB(t)^{-1}$ must be in $N$. In particular, at $t = a$, $A = B(a)AB(a)^{-1}$, but since $N$ is discrete, at a neighborhood of $t = a$ we must have $A = B(a + \varepsilon)AB(a + \varepsilon)^{-1}$, therefore, $AB = BA$ in a neighbourhood of $t = a$. Then, if at some point $B(t)AB(t)^{-1}$ were equal to an element of $N$ other than $A$ we would have a "jump" on path as $t$ enters another neighborhood, but this is not possible since, by assumption, $B(t)AB(t)^{-1}$ is continuous. Therefore, $AB = BA$ independently of $t$ and we conclude that any member of $N$ must be in the center of $G$.
However, I am unsure whether this is a "formal" proof or if there is some handwaving arguments I made. Is this proof is valid? How could I improve it in terms of language?
*Note: In the book, the author defines a matrix Lie group to be connected as path connected. Hence my definition of connectedness used above.
Best Answer
Let $n\in N$, consider the map defined on $G$ by $f(g)=gng^{-1}$ the image of $f$ is connected since $f$ is continuous and $G$ is connected and it is contained in $N$ since $N$ is normal, since $N$ is discrete this image contains only one element which is $f(e)=n=gng^{-1}$, we deduce that $N$ is in the center.