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Exercise 1369Topology › Topology of Mn(ℝ) › Exponential, logarithm and group topologyL3Added 6 Nov 2026

Centres of GL_n and SL_n, and discrete normal subgroups

#centre#discrete subgroup#connected group

Statement

  1. Compute the centres of GLn(R)GL_n(\mathbb{R}) and SLn(R)SL_n(\mathbb{R}).
  2. Show that a discrete normal subgroup of a connected topological group is central.
Hint

For part 1, test commutation with the matrix units Eij+IE_{ij}+I or with I+EijI+E_{ij}. For part 2, look at the map g↦gng−1g\mapsto gng^{-1} for fixed n∈Nn\in N.

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Solution

1. Centre of GLnGL_n. Let ZZ commute with all invertible matrices. Then for i≠ji\ne j it commutes with I+EijI+E_{ij}, hence with EijE_{ij}. Now ZEijZE_{ij} has ZeiZe_i as its jj-th column and zeros elsewhere, while EijZE_{ij}Z has the jj-th row of ZZ as its ii-th row and zeros elsewhere. Comparing entries: the (k,j)(k,j) entry for k≠ik\ne i gives zki=0z_{ki}=0, the (i,l)(i,l) entry for l≠jl\ne j gives zjl=0z_{jl}=0, and the (i,j)(i,j) entry gives zii=zjjz_{ii}=z_{jj}. As i≠ji\ne j are arbitrary, ZZ is diagonal with equal diagonal entries. Thus ZZ is a scalar matrix: Z(GLn)={λI:λ∈R×}Z(GL_n)=\{\lambda I:\lambda\in\mathbb{R}^\times\} (conversely scalar matrices are central).

Centre of SLnSL_n. The same argument applies since I+Eij∈SLnI+E_{ij}\in SL_n for i≠ji\ne j. So the centre consists of the scalar matrices of determinant 11: λn=1\lambda^n=1 with λ∈R\lambda\in\mathbb{R}. If nn is odd, λ=1\lambda=1 and Z(SLn)={I}Z(SL_n)=\{I\}. If nn is even, λ=±1\lambda=\pm1 and Z(SLn)={±I}Z(SL_n)=\{\pm I\}.

2. Discrete normal subgroups. Let GG be connected, N⊴GN\trianglelefteq G discrete (every point of NN is isolated in NN, with the subspace topology). Fix m∈Nm\in N and define c:G→Nc:G\to N, c(g)=gmg−1c(g)=gmg^{-1}. It is well defined since NN is normal and continuous (into GG, with values in NN). Its image is a connected subset of GG contained in the discrete space NN (in the subspace topology, a connected subset of a discrete space is a point). So cc is constant, equal to c(e)=mc(e)=m: gmg−1=mgmg^{-1}=m for all gg. Hence mm is central.

Application. In SL2(R)SL_2(\mathbb{R}) (connected) the subgroup {±I}\{\pm I\} is discrete, normal and central, as predicted. Connectedness is needed: in the non-connected group O(2)O(2) the cyclic group generated by R2π/3R_{2\pi/3} is finite (so discrete) and normal (conjugating by a reflection gives R−2π/3R_{-2\pi/3} and conjugating by a rotation changes nothing), but it is not central, since a reflection does not commute with R2π/3R_{2\pi/3}.

Remark. This is the first step in showing that a covering group's kernel is central: for a connected covering G~→G\tilde G\to G of topological groups, the kernel is discrete normal, hence central, hence abelian.