Centerless Groups: Examples & Constraints

[gentsagree]

I imagine a matrix group, with multiplication as the composition rule, to always possesses 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?

 

[pasmith]

I imagine a matrix group, with multiplication as the composition rule, to always possesses the quality of having centre (I,-I),

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

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?

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,
$$M_\rho = \begin{pmatrix} \cos(2\pi/3) & -\sin(2\pi/3) \\ \sin(2\pi/3) & \cos(2\pi/3) \end{pmatrix}$$

 

[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?

 

[pasmith]

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

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.

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

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.

 

[blue_raver22]

I would like to clarify that a centerless group is a group that does not have a non-trivial center, meaning that the only element that commutes with all other elements is the identity element. This is different from a group that always has a center, as the center can be non-trivial, meaning it contains elements other than the identity element.

Examples of centerless groups include the symmetric group on three or more elements, the alternating group on four or more elements, and the special linear group over a field. These groups do not have a non-trivial center because they have elements that do not commute with all other elements.

In order to achieve a centerless group, the constraint that must be relaxed is the requirement for all elements to commute with each other. This can be achieved by introducing non-commutative elements or operations into the group. For example, in the symmetric group, the elements are permutations, which do not commute with each other. In the special linear group, the elements are matrices with non-commutative operations such as matrix multiplication.

In summary, centerless groups are a special type of group that do not have a non-trivial center, and this can be achieved by relaxing the constraint of commutativity among all elements.