Let G be a group, a ∈ G be any element. then the subset
N(a) = {x ∈ G : xa = ax} is called the Normalizer or Centralizer of a in G.
Theorem: (i) The Normalizer of an element in a group G is a subgroup of G.
(ii) Prove that Z(G) ⊆ N(a).
Proof: (i) Let a ∈ G be any element of a group G. Then the set
N(a) = {x ∈ G : xa = ax} is the normalizer of a in G.
To show that N(a) is a subgroup of G.
Clearly, N(a) ⊆ G.
Since ea = ae
⇒ e ∈ N(a).
Therefore N(a) is a non-empty subset of G.
Let x, y ∈ N(a) be any two elements
∴ xa = ax and ya = ay
⇒ ay-1 = y-1a
Now a(x y-1) = (ax)y-1 = x (ay-1)
= x(y-1a) = (xy-1) a
i.e., a(x y-1) = (xy-1) a
So, x y-1 ∈ N(a).
Hence N(a) is a subgroup of G.
(ii) ∀ x ∈ Z(G)
⇒ xg = gx ∀ g ∈ G
Also a ∈ G
⇒ xa = ax
⇒ x ∈ N(a).
∴ ∀ x ∈ Z(G)
⇒ x ∈ N(a).
⇒Z(G) ⊆ N(a).
Example: Let G = S3, the symmetric group on three numbers 1, 2 and 3.
Find (i) N(a), where a = (12).
(ii) Z(G), the centre of G.
Sol. Here G = S3 = {i, (12), (13), (23), (123), (132)}.
(i) N(a) = N(12) = {x ∈ S3: x(12) = (12)x}.
Since i(12) = (12)i
∴ i ∈ N(12).
Also (12)(12) = i = (12)(12)
∴ (12) ∈ N(12).
But
(13)(12) = (132) ≠ (123) = (12)(13)
∴ (13) ∉ N(12).
(23)(12) = (123) ≠ (132) = (12)(23)
∴ (23) ∉ N(12).
(123)(12) = (23) ≠ (13) = (12)(123)
∴ (123) ∉ N(12).
(132)(12) = (13) ≠ (23) = (12)(132)
∴ (132) ∉ N(12).
Thus N(12) = {i, (12)}.
(ii) Since Z(G) = Z(S3) = {x ∈ S3: xz = zx for all z ∈ S3}
We know
Z(G) ⊆ N(a), for each a ∈ G.
∴ Z(G) ⊆ {i, (12)}.
But (12)(13) (123) ≠ (132) = (13)(12)
∴ (12) ∉ Z(G)
Hence Z(G) = (i). [ix = x = xi, ∀ x S3].
Example: Cent(G) = 4 if, and only if, G/Z ≅ Z2 ⊕ Z2;
i.e., G modulo its center id isomorphic to the Klein four group.
Proof: If G/Z ≅ Z2 ⊕ Z2, then there are non central elements p, q and s of G such that
G = Z ∪ Zp ∪ Zr ∪ Zs. It follows that the three proper subgroups of G containing Z are
P = Z ∪ Zp, R = Z ∪ Zr and S = Z ∪ Zs.
Let x be one of p, r and s and
Let X be the corresponding subgroup.
Notice that for zx ∈ ZX, G ⊃ C(zx) ⊇ X.
So, [G : X] = [G : C(zx)][C(zx) : X] = 2 and [G : C(zx)] ≠ 1,
thus C(zx) = X.
Therefore the proper centralizers of G are precisely P, R and S; i.e., Cent(G) = 4.
For the converse,
it is sufficient to show that [G : Z] = 4 because then
either G/Z ≅ Z2 ⊕ Z2 or G/Z ≅ Z4.
As G is non-abelian, G/Z can not be cyclic which means the latter case is impossible.
So, suppose Cent(G) = 4. and
Let P = C(p), R = C(r) and S = C(s) be the three proper centralizers of G.
Since G cannot be written as the union of two proper subgroups and since an element must belong to its centralizer,
we may choose ip, r and s in G - (R ∪ S), G - (P ∪ S) and
G - (P ∪ R) respectively.
Moreover, atleast one of the proper centralizers, say, P has index two in G. For otherwise
|G| ≤ |P| + |R| + |S| - 2|Z| ≤ |G|/3 + |G|/3 + |G|/3 - 2 < |G|.
Further,
P ∩ R = P ∩ R ∩ S = Z.
because if x ∈ (P ∩ R) - Z, then
(i) C(x) ≠ G because x ∉ Z,
(ii) C(x) ≠ P and C(x) ≠ R because p, r ∈ C(x),
(iii) C(x) ≠ S because x ∉ S,
which means that Cent(G) must be atleast 5.
Now we can compute |Z| using the fact that for subgroup x and Y of G,
|X ∩ Y| = (|X||Y|)/ |XY|
≥ (|X||Y|)/ |G|
Indeed,
|Z| = |P ∩ R|
≥ (|P||R|)/ |G|
= |R|/2
since |P| = |G|/2.
But Z ≠ R, So |Z| = |R|/2.
Similarly, |Z| = |S|/2.
Thus |G| = |P| + |R| + |S| -2|Z| = |G|/2 + 2|Z| + 2|Z| -2|Z|
= |G|/2 + 2|Z|
implies |G|/2 = 2|Z|; i.e, [G : Z] = 4, as desired.
Groups for which G/Z ≅ Z2 ⊕ Z2 are as abelian as an non-abelian group can be in the probabilistic sense also.
To see this, recall that the order of the conjugacy class of an element is the index of the centralizer of that element.
Thus, each conjugacy class of G is of order one or two.
Therefore the number of conjugacy classes in G is
k = |Z| + (|G| - |Z|)/2 = |G|/4 + 3|G|/8 = 5|G|/8
and prob(G) = 5/8.
This computation also suggests why prob(G) ≤ 5/8 for non-abelian groups: prob(G) is as large as possible when the centre is as large as possible and all of the non-central elements are in conjugacy classes of size two.
This occurs when [G : Z] = 4;
i.e., when G/Z ≅ Z2 ⊕ Z2.
The threshold of 'abelianness', as measured by Cent(G) or Prob(G), is occupied by the same class of groups.
N(a) = {x ∈ G : xa = ax} is called the Normalizer or Centralizer of a in G.
Theorem: (i) The Normalizer of an element in a group G is a subgroup of G.
(ii) Prove that Z(G) ⊆ N(a).
Proof: (i) Let a ∈ G be any element of a group G. Then the set
N(a) = {x ∈ G : xa = ax} is the normalizer of a in G.
To show that N(a) is a subgroup of G.
Clearly, N(a) ⊆ G.
Since ea = ae
⇒ e ∈ N(a).
Therefore N(a) is a non-empty subset of G.
Let x, y ∈ N(a) be any two elements
∴ xa = ax and ya = ay
⇒ ay-1 = y-1a
Now a(x y-1) = (ax)y-1 = x (ay-1)
= x(y-1a) = (xy-1) a
i.e., a(x y-1) = (xy-1) a
So, x y-1 ∈ N(a).
Hence N(a) is a subgroup of G.
(ii) ∀ x ∈ Z(G)
⇒ xg = gx ∀ g ∈ G
Also a ∈ G
⇒ xa = ax
⇒ x ∈ N(a).
∴ ∀ x ∈ Z(G)
⇒ x ∈ N(a).
⇒Z(G) ⊆ N(a).
Example: Let G = S3, the symmetric group on three numbers 1, 2 and 3.
Find (i) N(a), where a = (12).
(ii) Z(G), the centre of G.
Sol. Here G = S3 = {i, (12), (13), (23), (123), (132)}.
(i) N(a) = N(12) = {x ∈ S3: x(12) = (12)x}.
Since i(12) = (12)i
∴ i ∈ N(12).
Also (12)(12) = i = (12)(12)
∴ (12) ∈ N(12).
But
(13)(12) = (132) ≠ (123) = (12)(13)
∴ (13) ∉ N(12).
(23)(12) = (123) ≠ (132) = (12)(23)
∴ (23) ∉ N(12).
(123)(12) = (23) ≠ (13) = (12)(123)
∴ (123) ∉ N(12).
(132)(12) = (13) ≠ (23) = (12)(132)
∴ (132) ∉ N(12).
Thus N(12) = {i, (12)}.
(ii) Since Z(G) = Z(S3) = {x ∈ S3: xz = zx for all z ∈ S3}
We know
Z(G) ⊆ N(a), for each a ∈ G.
∴ Z(G) ⊆ {i, (12)}.
But (12)(13) (123) ≠ (132) = (13)(12)
∴ (12) ∉ Z(G)
Hence Z(G) = (i). [ix = x = xi, ∀ x S3].
Example: Cent(G) = 4 if, and only if, G/Z ≅ Z2 ⊕ Z2;
i.e., G modulo its center id isomorphic to the Klein four group.
Proof: If G/Z ≅ Z2 ⊕ Z2, then there are non central elements p, q and s of G such that
G = Z ∪ Zp ∪ Zr ∪ Zs. It follows that the three proper subgroups of G containing Z are
P = Z ∪ Zp, R = Z ∪ Zr and S = Z ∪ Zs.
Let x be one of p, r and s and
Let X be the corresponding subgroup.
Notice that for zx ∈ ZX, G ⊃ C(zx) ⊇ X.
So, [G : X] = [G : C(zx)][C(zx) : X] = 2 and [G : C(zx)] ≠ 1,
thus C(zx) = X.
Therefore the proper centralizers of G are precisely P, R and S; i.e., Cent(G) = 4.
For the converse,
it is sufficient to show that [G : Z] = 4 because then
either G/Z ≅ Z2 ⊕ Z2 or G/Z ≅ Z4.
As G is non-abelian, G/Z can not be cyclic which means the latter case is impossible.
So, suppose Cent(G) = 4. and
Let P = C(p), R = C(r) and S = C(s) be the three proper centralizers of G.
Since G cannot be written as the union of two proper subgroups and since an element must belong to its centralizer,
we may choose ip, r and s in G - (R ∪ S), G - (P ∪ S) and
G - (P ∪ R) respectively.
Moreover, atleast one of the proper centralizers, say, P has index two in G. For otherwise
|G| ≤ |P| + |R| + |S| - 2|Z| ≤ |G|/3 + |G|/3 + |G|/3 - 2 < |G|.
Further,
P ∩ R = P ∩ R ∩ S = Z.
because if x ∈ (P ∩ R) - Z, then
(i) C(x) ≠ G because x ∉ Z,
(ii) C(x) ≠ P and C(x) ≠ R because p, r ∈ C(x),
(iii) C(x) ≠ S because x ∉ S,
which means that Cent(G) must be atleast 5.
Now we can compute |Z| using the fact that for subgroup x and Y of G,
|X ∩ Y| = (|X||Y|)/ |XY|
≥ (|X||Y|)/ |G|
Indeed,
|Z| = |P ∩ R|
≥ (|P||R|)/ |G|
= |R|/2
since |P| = |G|/2.
But Z ≠ R, So |Z| = |R|/2.
Similarly, |Z| = |S|/2.
Thus |G| = |P| + |R| + |S| -2|Z| = |G|/2 + 2|Z| + 2|Z| -2|Z|
= |G|/2 + 2|Z|
implies |G|/2 = 2|Z|; i.e, [G : Z] = 4, as desired.
Groups for which G/Z ≅ Z2 ⊕ Z2 are as abelian as an non-abelian group can be in the probabilistic sense also.
To see this, recall that the order of the conjugacy class of an element is the index of the centralizer of that element.
Thus, each conjugacy class of G is of order one or two.
Therefore the number of conjugacy classes in G is
k = |Z| + (|G| - |Z|)/2 = |G|/4 + 3|G|/8 = 5|G|/8
and prob(G) = 5/8.
This computation also suggests why prob(G) ≤ 5/8 for non-abelian groups: prob(G) is as large as possible when the centre is as large as possible and all of the non-central elements are in conjugacy classes of size two.
This occurs when [G : Z] = 4;
i.e., when G/Z ≅ Z2 ⊕ Z2.
The threshold of 'abelianness', as measured by Cent(G) or Prob(G), is occupied by the same class of groups.
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