# Centerless groups

1. Feb 20, 2014

### gentsagree

I imagine a matrix group, with multiplication as the composition rule, to always possess the quality of having centre (I,-I), as I can't see when both elements wouldn't commute with all others. On the other hand, though, a centerless group is defined as having trivial centre, i.e. Z=I (which means, Z doesn't include -I).

I imagine non-matrix groups could show this property, but I can't think of any.

Could somebody give a couple of examples of centreless groups, and what "constraints" must be relaxed (from my matrix group example above) in order to achieve them?

2. Feb 20, 2014

### pasmith

A matrix group must contain the identity $I$, but need not contain $-I$.

The group $D_3$, which is the symmetry group of an equilateral triangle, has trivial center. The group has a 2-dimensional representation generated by a reflection in the x-axis
$$M_m = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$
and a rotation through $2\pi/3$ about the origin,
$$M_\rho = \begin{pmatrix} \cos(2\pi/3) & -\sin(2\pi/3) \\ \sin(2\pi/3) & \cos(2\pi/3) \end{pmatrix}$$

3. Feb 20, 2014

### gentsagree

Thanks, I see how it would work for D3. However, I was referring to the center of SU(2), which is (I,-I).

If the composition rule is multiplication, how is it possible to find an element of the group which doesn't commute with -I?

4. Feb 20, 2014

### pasmith

SU(2) is a matrix group. There are other groups of 2x2 matrices, and the fact that the center of SU(2) is {I, -I} has no bearing on the center of any other matrix group.

It isn't, but that doesn't matter if the group in question doesn't contain -I in the first place!

The center of a group $G$ consists exactly of those $g \in G$ such that for all $h \in G$, $gh = hg$.

Thus, if $G$ is a matrix group and $-I \notin G$, we don't care that $(-I)M = M(-I)$ for all $M \in G$; $-I$ fails to be in the center of $G$ by virtue of not being in $G$ in the first place.