The normalizer includes the elements of G that commute with a. The center includes only those elements of G that commute with all other elements of G.
Can someone help me with the definitions of centralizer and normalizer? My class doesn't use a book and I don't understand the definitions we were given.
Normalizer N(a)={x in G | xa=ax} if (G, *) is a group and a is in G.
Center of G=Z(G)={x in G | ax=xa , for all a in G}.
I need to prove that a is in Z(G) iff N(a)=G.
the normalizer of an element is also called the centralizer of an element (these concepts differ when you consider subsets of a group).
why the name normalizer?
because the defining relation, xa = ax, can also be written: xax^{-1} = a, which should remind you of the normality condition for a subgroup.
for a subset, S, of a group G, the normalizer of S is N(S) = {x in G: xSx^{-1} = S}, whereas:
the centralizer is C(S) = {x in G: xs = sx for all s in S}.
if S = G, then C(G) = Z(G), the center of G (the elements that commute with everything). this is always the entire group if G is abelian.
if S = H, a subgroup of G, then N(H) is the largest subgroup of G in which H is normal. if H already is a normal subgroup, N(H) = G.
i understand that this may be a bewildering array of subtly differing concepts, but each has its uses. sometimes we want to look at conjugation, sometimes we want to know "what commutes".
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anyway, with your problem:
suppose a is in Z(G). then ag = ga for EVERY g in G. this means that N(a) = {x in G: ax = xa} is all of G (everything commutes with a, because a commutes with everything).
on the other hand, if N(a) = G, then ax = xa for EVERY x in G, so ag = ga for every g in G (it doesn't matter whether we use "x" or "g" as a stand-in for a "typical" element of G), so a is in the center.
in general, Z(G) is always a subgroup of N(a) (those things that commute with everything certainly commutes with a. however, some elements might commute with a, but not with other elements).
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note that it might be "hard" for an element to be in the center of G (G might be a "mostly" non-abelian group. for example, the dihedral group of order 2n, where n is odd, has only the identity in the center. but if r is a rotation of order n, it should be clear that N(r) contains all of <r>, the cyclic subgroup generated by r (powers of r commutes with r). it should also be clear that the reflection s is NOT is N(r), since rs = sr^{-1} (s and r do NOT commute)).
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normalizers and conjugates are related: it turns out that the number of DISTINCT conjugates of a is the INDEX of the normalizer of a.
it is useful to distinguish between two cases:
1)a is in Z(G).
here, we see that xa = ax for all x in in G, so that xax^{-1} = a, that is: a has only ONE conjugate: itself.
as we saw above, if a is in Z(G), then N(a) = G, and [G:G] = 1 (G has index one in itself, the only coset of G is...G).
2)a is NOT in Z(G). let's show in this case that xN(a) = yN(a) if and only if xax^{-1} = yay^{-1}.
suppose xN(a) = yN(a). this means y^{-1}x is in N(a). thus:
(y^{-1}x)a = a(y^{-1}x)
y^{-1}(xax^{-1}) = ay^{-1} (multiplying on the right by x^{-1}).
xax^{-1} = yay^{-1} (multiplying on the left by y).
running the argument "in reverse" gives the other half of the iff.
this means if xN(a) is NOT equal to yN(a), then xax^{-1} and yay^{-1} are not the same, we have different conjugates of a.
this means that the number of different conjugates of a is the number of costs of N(a) in G, that is to say the index of N(a) in G, [G:N(a)].
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now this becomes really useful, because conjugacy is an equivalence relation on G, so it partitions G. so we can sometimes "count the equivalence classes" to get information about G. since the elements in the center of G all are only conjugate to themselves, we "lump them all together" and these account for |Z(G)| elements of G.
then we take an element a not in Z(G), and we get [G:N(a)] more elements from [a] = {x in G: x = gag^{-1} for some g in G}.
then we take another element b not in Z(G), or [a], and get [G:N(b)] more elements. if G is finite, we'll eventually account for all of them. sometimes this is written as:
this can be used to prove facts about G. one result you will almost always be exposed to is:
if |G| = p^{n} (for n > 0), then G has a non-trivial center.
if G is abelian, there is nothing to prove, since Z(G) = G is certainly non-trival.
so assume G is not abelian. this means for a not in Z(G), N(a) ≠ G, so [G:N(a)] > 1, so must be a positive power of p for each distinct conjugacy class [a].
this means:
p^{n} = |Z(G)| + kp (we don't even need to know what "k" is).
hence |Z(G)| = p(p^{n-1} - k), that is p divides |Z(G)|. since any subgroup is non-empty, |Z(G)| ≠ 0, so we have that |Z(G)| is at least p > 1.