AATA The Class Equation

Section 14.2 The Class Equation

Let

X

be a finite

G

-set and

X_{G}

be the set of fixed points in

X;

that is,

X_{G} = {x \in X : g x = x for all g \in G} .

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Since the orbits of the action partition

X,

| X | = | X_{G} | + \sum_{i = k}^{n} | O_{x_{i}} |,

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where

x_{k}, \dots, x_{n}

are representatives from the distinct nontrivial orbits of

X .

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Now consider the special case in which

G

acts on itself by conjugation,

(g, x) \mapsto g x g^{- 1} .

The center of

G,

Z (G) = {x : x g = g x for all g \in G},

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is the set of points that are fixed by conjugation. The nontrivial orbits of the action are called the conjugacy classes of

G .

x_{1}, \dots, x_{k}

are representatives from each of the nontrivial conjugacy classes of

G

and

| O_{x_{1}} | = n_{1}, \dots, | O_{x_{k}} | = n_{k},

then

| G | = | Z (G) | + n_{1} + \dots + n_{k} .

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The stabilizer subgroups of each of the

x_{i}

’s,

C (x_{i}) = {g \in G : g x_{i} = x_{i} g},

are called the centralizer subgroups of the

x_{i}

’s. From Theorem 14.11, we obtain the class equation:

| G | = | Z (G) | + [G : C (x_{1})] + \dots + [G : C (x_{k})] .

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One of the consequences of the class equation is that the order of each conjugacy class must divide the order of

G .

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Example 14.12.

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It is easy to check that the conjugacy classes in

S_{3}

are the following:

{(1)}, {(1 2 3), (1 3 2)}, {(1 2), (1 3), (2 3)} .

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The class equation is

6 = 1 + 2 + 3 .

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Example 14.13.

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The center of

D_{4}

{(1), (1 3) (2 4)},

and the conjugacy classes are

{(1 3), (2 4)}, {(1 4 3 2), (1 2 3 4)}, {(1 2) (3 4), (1 4) (2 3)} .

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Thus, the class equation for

D_{4}

8 = 2 + 2 + 2 + 2 .

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Example 14.14.

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For

S_{n}

it takes a bit of work to find the conjugacy classes. We begin with cycles. Suppose that

σ = (a_{1}, \dots, a_{k})

is a cycle and let

τ \in S_{n} .

By Theorem 6.16,

τ σ τ^{- 1} = (τ (a_{1}), \dots, τ (a_{k})) .

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Consequently, any two cycles of the same length are conjugate. Now let

σ = σ_{1} σ_{2} \dots σ_{r}

be a cycle decomposition, where the length of each cycle

σ_{i}

r_{i} .

Then

σ

is conjugate to every other

τ \in S_{n}

whose cycle decomposition has the same lengths.

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The number of conjugate classes in

S_{n}

is the number of ways in which

n

can be partitioned into sums of positive integers. In the case of

S_{3}

for example, we can partition the integer

3

into the following three sums:

\begin{aligned} 3 & = 1 + 1 + 1 \\ 3 & = 1 + 2 \\ 3 & = 3; \end{aligned}

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therefore, there are three conjugacy classes. There are variations to problem of finding the number of such partitions for any positive integer

n

that are what computer scientists call NP-complete. This effectively means that the problem cannot be solved for a large

n

because the computations would be too time-consuming for even the largest computer.

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Theorem 14.15.

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Let

G

be a group of order

p^{n}

where

p

is prime. Then

G

has a nontrivial center.

Proof.

We apply the class equation

| G | = | Z (G) | + n_{1} + \dots + n_{k} .

Since each

n_{i} > 1

and

n_{i} ∣ | G |,

it follows that

p

must divide each

n_{i} .

Also,

p ∣ | G |;

hence,

p

must divide

| Z (G) | .

Since the identity is always in the center of

G,

| Z (G) | \geq 1 .

Therefore,

| Z (G) | \geq p,

and there exists some

g \in Z (G)

such that

g \neq 1 .

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Corollary 14.16.

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Let

G

be a group of order

p^{2}

where

p

is prime. Then

G

is abelian.

Proof.

By Theorem 14.15,

| Z (G) | = p

p^{2} .

Suppose that

| Z (G) | = p .

Then

Z (G)

and

G / Z (G)

both have order

p

and must both be cyclic groups. Choosing a generator

a Z (G)

for

G / Z (G),

we can write any element

g Z (G)

in the quotient group as

a^{m} Z (G)

for some integer

m;

hence,

g = a^{m} x

for some

x

in the center of

G .

Similarly, if

h Z (G) \in G / Z (G),

there exists a

y

Z (G)

such that

h = a^{n} y

for some integer

n .

Since

x

and

y

are in the center of

G,

they commute with all other elements of

G;

therefore,

g h = a^{m} x a^{n} y = a^{m + n} x y = a^{n} y a^{m} x = h g,

and

G

must be abelian. Hence,

| Z (G) | = p^{2} .

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