Something I mentioned early on that I haven't talked about much is commutivity. One of the first things that I defined was the notion of an abelian group - one for which every element of the group commutes with every other element of the group. However, non-abelian groups have certain parts that "act" abelian. The first example of this is called the center of a group.
Definition: Center of a Group
Let G be a group. The center, Z(G), of G is the set of all the elements of G that commute with every element of G. Symbolically, Z(G) = {a∈G : g∙a = a∙g ∀g∈G}.
Let G be a group. The center, Z(G), of G is the set of all the elements of G that commute with every element of G. Symbolically, Z(G) = {a∈G : g∙a = a∙g ∀g∈G}.
In an abelian group, every element commutes with every other element. In a non-abelian group, though, only some of the elements commute with every other element. It is these elements that comprise the center of the group. In order to make the notion of a center a little more clear, I would like to introduce a new group, which requires some knowledge of linear algebra. If you're not familiar with matrices then you can skip it for now. This group, called the general linear group, denoted by GL(n,R) where n is a positive integer and R is the real numbers, is the group of all invertible n by n matrices (matrices with non-zero determinant). It is simple to check that GL(n,R) is a group under matrix multiplication with its identity element the n by n identity matrix (which will be denoted by I). Then if x is a real number, it can be checked that (xI)∙M = M∙(xI) for every matrix, M, in GL(n,R). However, in general, it is not true that N∙M = M∙N for every N and M in GL(n,R). It can be shown that Z(GL(n,R)) = {xI : x∈R}.
The center of a group has a lot of interesting properties and uses. One such property is given below.
Theorem: Center is a Subgroup
The center of a group G is a subgroup of G.
The center of a group G is a subgroup of G.
Proof:
It is quite clear that Z(G)⊆G and that e∈Z(G) so Z(G) is non-empty. To prove that Z(G) is a subgroup, I will use the two-step subgroup test. First choose a,b∈Z(G). In order to show that a∙b∈Z(G) I must show that a∙b commutes with any arbitrary element of G. As such, choose g∈G. Note that (a∙b)∙g = a∙(b∙g) = a∙(g∙b) = (a∙g)∙b = (g∙a)∙b = g∙(a∙b) so then (a∙b)∙g = g∙(a∙b) and a∙b commutes with g. (That was made possible because both a and b commute with g since they are both in the center of G.) Thus a∙b∈G. Second choose a∈Z(G). In order to show that a-1∈Z(G) I must show that a-1 commutes with any arbitrary element of G. As such, choose g∈G. Note that g∙a-1 = e∙(g∙a-1) = (a-1∙a)∙(g∙a-1) = a-1∙(a∙g)∙a-1 = a-1∙(g∙a)∙a-1 = (a-1∙g)∙(a∙a-1) = (a-1∙g)∙e = a-1∙g so then g∙a-1 = a-1∙g and a-1 commutes with g. (This was made possible because a commutes with g since it is in the center of G.) Thus a-1∈G. Finally, by the two-step subgroup test, Z(G) is a subgroup of G.
It is quite clear that Z(G)⊆G and that e∈Z(G) so Z(G) is non-empty. To prove that Z(G) is a subgroup, I will use the two-step subgroup test. First choose a,b∈Z(G). In order to show that a∙b∈Z(G) I must show that a∙b commutes with any arbitrary element of G. As such, choose g∈G. Note that (a∙b)∙g = a∙(b∙g) = a∙(g∙b) = (a∙g)∙b = (g∙a)∙b = g∙(a∙b) so then (a∙b)∙g = g∙(a∙b) and a∙b commutes with g. (That was made possible because both a and b commute with g since they are both in the center of G.) Thus a∙b∈G. Second choose a∈Z(G). In order to show that a-1∈Z(G) I must show that a-1 commutes with any arbitrary element of G. As such, choose g∈G. Note that g∙a-1 = e∙(g∙a-1) = (a-1∙a)∙(g∙a-1) = a-1∙(a∙g)∙a-1 = a-1∙(g∙a)∙a-1 = (a-1∙g)∙(a∙a-1) = (a-1∙g)∙e = a-1∙g so then g∙a-1 = a-1∙g and a-1 commutes with g. (This was made possible because a commutes with g since it is in the center of G.) Thus a-1∈G. Finally, by the two-step subgroup test, Z(G) is a subgroup of G.
I know that's another long and boring proof, but I wrote it out because its a very good example of how to use a subgroup test. As always, feel free to skip the proof if its confusing, but understanding a center and the fact that it is a subgroup is reasonably important. Later, there will be a lot of interesting things that we do with centers and it is crucial that the center is a subgroup (and, in fact, the center turns out to be a normal subgroup, although we haven't gotten there yet).
There is a concept related to the center of a group called a centralizer of an element. The center finds the elements of a group that commute with every single element, whereas the centralizer finds the elements that commute with one single element.
Definition: Centralizer of a in G
Let G be a group and a∈G be fixed. The centralizer of a, C(a) in G is the set of all the elements of G that commute with a. Symbolically, C(a) = {g∈G : a∙g = g∙a}.
Let G be a group and a∈G be fixed. The centralizer of a, C(a) in G is the set of all the elements of G that commute with a. Symbolically, C(a) = {g∈G : a∙g = g∙a}.
The centralizer of a group is not nearly as interesting or as useful as the center, but it is another application of commutivity. I now present the following fact without proof.
Theorem: The Centralizer of an Element is a Subgroup.
Let G be a group. For each a∈G, the centralizer of a in G, C(a), is a subgroup of G.
Let G be a group. For each a∈G, the centralizer of a in G, C(a), is a subgroup of G.
I'm not going to present the proof here because its very similar to the last proof. There are two other facts that become immediately apparent about the relationship between centers and centralizers. First, ∀a∈G, Z(G)⊆C(a). Second, if a∈G then Z(G) = C(a) if and only if a∈Z(G).
This might not seem all that astounding or interesting, and as of right now it shouldn't. The usefullness of the center of a group (and the centralizer of an element) will come later, but for right now it is sufficient just to understand the definitions.
References
Previous Related Post: Cyclic Subgroups
Text Reference: Gallain Chapter 3
Wikipedia: Center of a Group
Wikipedia: Centralizer of a Group
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