Meanings of centralizer, center, and normalizer

In my text's set of definitions we have

The centralizer of a subset A of a group G: $\displaystyle C_G(A) = \{ g \in G|gag^{-1} = g ~\forall a \in A \}$

The center of a group G: $\displaystyle Z(G) = \{ g \in G | gz = zg ~ \forall z \in G \}$

The normalizer of a subset A of a group G: $\displaystyle N_G (A) = \{g \in G|g A g^{-1} = A \}$

Now the question: *Why* do these groups have these names? Is there some intrinsic meaning to them or are they just terms that have become accepted?

Thanks!

-Dan

Re: Meanings of centralizer, center, and normalizer

If something is in the center, say z, we have:

gz = zg, in other words: multiplication (by g) from "either side" is the same: in fact we have gzg' = z(gg') (we can take z "out of the center" of a triple product).

Another common english meaning to "central" is "essential" or "at the heart of things"...this meaning becomes reinforced when one considers K-algebras for a field K, when we require that the field K lie in the center of the K-algebra A. For the algebra of nxn matrices over K, this means that we need the matrices kI, for k in K, to commute with everything, and it turns out these are the ONLY matrices that do so.

Of course, Z(G) = C_{G}(G). Essentially the centralizer of a set S, C_{G}(S) is the set of all elements of G that "centralize" S, in other words, commute with every element of S. For such elements, we can write gsg' = s(gg'). Working in the centralizer of a set lets us simplify some expressions we get: for example, if g is in the centralizer, then:

gsg^{-1} = s.

Now, a normal subgroup is a subgroup H of G such that: gHg^{-1} = H. Not every subgroup of G is normal, sometimes because G is extremely "non-abelian" (has a small center). One might ask: what is the largest subgroup K of G containing H such that H is normal in K? Clearly, for such a subgroup, we must have:

kHk^{-1} = K, so we see that K has to be contained in N_{G}(H).

On the other hand, H is clearly normal in N_{G}(H), so by the above we see that the normalizer IS the largest such subgroup. In other words, K is an "extension group" of H (in G) that makes H normal. For VERY "non-normal" groups, we will have, unfortunately, N_{G}(H) = H (such subgroups are called "self-normalizing").

When the set being normalized (or centralized) is a single element, say x, we have that: N_{G}(x) = C_{G}(x), because:

gx = xg, for all g in G, is equivalent to:

gxg^{-1} = x, for all x in G.

In any group, we tend to focus on 2 things:

1) Conjugates ---leads to normalizers

2) Commutation ---leads to centralizers (and, at a more advanced level, commutators)

Groups are not always abelian, which is too bad, because for abelian groups, all of this stuff goes away. As a consequence, the algebra gets dicey, and being the mindless and lazy creatures we are, we'd like short-cuts to the stuff we want to know, without going through a lot of arduous computation. In fact, it turns out that with groups, we aren't really even interested in the groups per se, but in the mappings BETWEEN them that preserve multiplication (homomorphisms)...just as with sets, the really "cool stuff" all happens with functions. Unfortunately, it turns out that the set of automorphisms of even an abelian group, isn't necessarily itself abelian (this is because composition of functions is "one-way", the order matters). So, even if we were only going to study abelian groups to begin with, eventually we have to consider non-abelian groups when we start studying groups of automorphisms (and it turns out that this concept of group of automorphisms applies to other algebraic structures as well, such as vector spaces, or directed graphs, or just about anything that has isomorphisms of some sort).

So, we're sort of "stuck" with non-abelian groups, and we're going to need short-cuts, or else...I dunno, maybe go crazy.

********

So, the short answer to the question about "normalizers" is: it arises from the term "normal subgroup". This does, a bit, beg the question. Why do we call "normal subgroups", "normal"? The answer is, however, a bit complicated. Evariste Galois, when he was first studying permutations of roots of polynomials, realized certain field extensions had desirable qualities: certain polynomials in the original field COMPLETELY factored in the larger field. He was thus led to the idea of a "normal" field extension, which corresponded to normal subgroups of the Galois group of the polynomial under consideration. For example, it turns out that the Galois group of the polynomial (over the rational field):

$\displaystyle x^3 - 2$

is the permutation group S_{3}.

We can factor this as:

$\displaystyle x^3 - 2 = (x - \sqrt[3]{2})(x - \sqrt[3]{2}\omega)(x - \sqrt[3]{2}\omega^2) = (x - r_1)(x - r_2)(x - r_3)$

where $\displaystyle \omega$ is the complex number:

$\displaystyle \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$

If we were to extend the rationals to a field which included $\displaystyle \sqrt[3]{2}$, but was contained in the reals, the only root permutations that do this, are:

(2 3) (here we are permuting the indices on the roots, just to be clear), or the null (identity) permutation (it turns out that this is just complex conjugation for the first permutation). But this means that $\displaystyle \mathbb{Q}(\sqrt[3]{2})$ isn't "big enough" to contain all the roots of $\displaystyle x^3 - 2$, so we say the extension isn't "normal". This turns out to mean that {e, (2 3)} isn't a normal subgroup of S_{3} (it turns out that the sole normal extension of the rationals contained in the smallest subfield of the complex numbers containing all the roots (that is to say, a splitting field for $\displaystyle x^3 - 2$) is given by $\displaystyle \mathbb{Q}(\omega)$, which contains all the roots of $\displaystyle x^3 - 1$, and thus of $\displaystyle x^2 + x + 1$ the minimal polynomial of this primitive cube root of unity, which is why it is a normal extension. This extension corresponds to the subgroup of the identity permutation, and the 2 3-cycle permutations of the roots on our cubic $\displaystyle x^3 - 2$, which in fact IS a normal subgroup of S_{3}).

Much later, it was discovered that normal subgroups are precisely those that are the kernels of quotient mappings. Whether this was genius or accident on Galois' part is open to debate, although the historical literature supports the former.

So, that's the "abbreviated" actual reason....for an amusing "fictional" take on this state of affairs, see: Normal subgroups and quotient groups | Gowers's Weblog

Re: Meanings of centralizer, center, and normalizer

I glanced at a text on group theory and saw the following fact that can be relevant to the etymology of the term "center". Consider the conjugation action of a group G on itself, i.e., g(x) = gxg^{-1}. Then the orbit of x has size 1 iff x is in Z(G). Now, the name "orbit" probably originates from a group of rotations acting on points on the plane, in which case orbits are circles. Then which elements have degenerate orbit? The ones in the center!

Re: Meanings of centralizer, center, and normalizer

Quote:

Originally Posted by

**emakarov** I glanced at a text on group theory and saw the following fact that can be relevant to the etymology of the term "center". Consider the conjugation action of a group G on itself, i.e., g(x) = gxg^{-1}. Then the orbit of x has size 1 iff x is in Z(G). Now, the name "orbit" probably originates from a group of rotations acting on points on the plane, in which case orbits are circles. Then which elements have degenerate orbit? The ones in the center!

Another possibility: if one considers the "central line" of a regular polygon as extending from the point (1,0) (the first vertex) through the center, then the center of the full symmetry group of the polygon are those rotational symmetries that map the central line to itself..so we have:

$\displaystyle Z(D_n) = \{1\}$ if n is odd, and $\displaystyle Z(D_n) = \{1,r^{\frac{n}{2}}\}$ if n is even.