Lecture 9. Permutation Representations

Permutation representations The regular representation The number of irreducibles

Lecture 9. Permutation Representations

Daniel Bump

May 5, 2020


Permutation representations

Let G be a finite group, and X a set with a group actionG× X −→ X. Let V be the free complex vector space on X.Then G acts on V by linear transformations, extending the givenaction on the basis X. Thus we have a representationπX : G −→ GL(V).

For example, let G = S3 with its usual action on X = {1, 2, 3} bypermutations. With respect to this basis, πX(g) becomes amatrix:

πX((12)) =

11

1

, πX((123)) =

11

1

.


The character of the permutation representations

Diagonal entries in the matrix of πX(g) with respect to the basisX of V are then fixed points: that is, there is 1 in the x-thdiagonal if gx = x, that is, if x is a fixed point of g. From this, wesee that the character χX(g) is the number of fixed points of g:

χX(g) = |{x ∈ X|gx = x}| .

For example, if G = S3 and X = {1, 2, 3}:

g 1 (123) (12)χX(g) 3 0 1

We will call πX and χX the permutation representation andpermutation character associated with the group action on X.


The reduced character

The permutation representation πX is reducible (unless |X| = 1).Indeed, it contains a copy of the trivial representation, namelythe span of the basis element ξ =

∑x∈G x. Another submodule

(which may or may not be irreducible) consist of the elements

V◦ =

{∑x∈X

axx∣∣∣ ∑

x∈X

ax = 0

}.

Let π◦X be the representation of G on V◦. The dimension of V◦ is|X| − 1 and

V = Cξ ⊕ V◦.

The character χ◦X of the G-module V◦ equals the number offixed points minus 1, and this character may or may not beirreducible.


Example

For S3, χ◦X(g) is the number of fixed points, minus one:

g 1 (123) (12)χ◦X(g) 2 −1 0

We will call π◦X and χ◦X the reduced permutation representationand character respectively.

In this example, the reduced character is irreducible.

The reduced character may or may not be irreducible.


Irreducibility of the reduced permutation character

Let us consider whether π◦X is an irreducible representation ornot. First, if the action on X is not transitive, it can never beirreducible. This is because if there are more than one orbit:

X = X1 ∪ X2 ∪ · · · ∪ Xh (disjoint)

then for each orbit Xi the vector ξi that is the sum of theelements in Xi is an invariant vector, so π◦X will contain h copiesof the trivial representation.

But even if the action on X is transitive, the reducedpermutation representation may or may not be irreducible. Wewill look at couple of examples.


A test for irreducibility

First let us describe an easy way of checking whether arepresentation is irreducible.

PropositionLet χ be the character of a representation π. Then π isirreducible if and only if 〈χ, χ〉 = 1.

By Schur orthogonality, if χ is irreducible, then 〈χ, χ〉 = 1. Soprove the converse, we may decompose π into distinctirreducible representations:

π ∼=⊕

i

niπi.

Here πi are the distinct irreducibles.


Proof, continued

The character is:χ =

∑i

niχi.

Now using orthogonality

〈χ, χ〉 =∑

i

n2i .

So if 〈χ, χ〉 = 1 then∑

n2i = 1, which implies that exactly one of

the ni is > 0, and that ni = 1. This means that π ∼= πi isirreducible.


Example: S4

With this in hand, we consider a couple of permutationrepresentations. First, there is the action of S4 onX = {1, 2, 3, 4}. (We considered the case of S3 last week.) Hereare the character values. The first row gives representatives ofthe conjugacy class. The second row gives the number ofelements of the conjugacy class.

1 (123) (12)(34) (12) (1234)1 8 3 6 6

χ◦X 3 0 −1 1 −1

〈χ, χ〉 = 124

(1 · 32 + 8 · 02 + 3 · (−1)2 + 6 · 12 + 6 · (−1)2) = 1

So this character is irreducible.


Example: D8

Now let us restrict this permutation representation to thedihedral group D8. It is still a transitive action.

1 (1234) (13)(24) (12)(34) (13)1 2 1 2 2

χ◦X 3 −1 −1 −1 1

〈χ, χ〉 = 18(1 · 32 + 2 · (−1)2 + 1 · (−1)2 + 2 · (−1)2 + 2 · 12) = 2.

So this representation is not irreducible.


Example: D8 (continued)

Still we can use this information to construct an irreduciblerepresentation of D8. The derived group is the centerZ(D8) = 〈(13)(24)〉 of order 2. The quotient D8/Z(D8) isisomorphic to Z2 × Z2 so we find four linear characters, and wehave part of the character table:

1 (1234) (13)(24) (12)(34) (13)1 2 1 2 2

χ1 1 1 1 1 1χ2 1 1 1 −1 −1χ3 1 −1 1 1 −1χ4 1 −1 1 −1 1χ◦X 3 −1 −1 −1 1

From these values, 〈χ◦X, χ4〉 = 1. This means we can subtractχ4 from χ◦X and obtain another character which we will call χ5.


D8 (concluded)

1 (1234) (13)(24) (12)(34) (13)1 2 1 2 2

χ1 1 1 1 1 1χ2 1 1 1 −1 −1χ3 1 −1 1 1 −1χ4 1 −1 1 −1 1χ5 2 0 −2 0 0χ◦X 3 −1 −1 −1 1

QuestionCan you think of another way to compute χ5 once χ1, χ2, χ3, χ4have been found?


The regular representation

These two examples show that the reduced character of atransitive permutation representation may or may not beirreducible.

A particular important permutation representation is the actionof G on itself by left translation. The correspondingrepresentation is called the regular representation πreg. Let χregbe its character. The free vector space on G is C[G] so we maythink of the regular representation as being the representationon the C[G]-module C[G].


The character of the regular representation

PropositionThe character of the regular representation is

χreg(g) ={|G| if g = 1,0 otherwise.

To prove this, note that x ∈ G is a fixed point of g if and only ifgx = x, which means g = 1. So g = 1 has |G| fixed points, andany other element has none.


Decomposing the regular reprsentation

Now let us decompose the regular representation intoirreducibles. We recall that di = dim(Vi), so di = χi(1).

TheoremThe multiplicity of πi in a decomposition of C[G] into irreduciblesis di. We have

χreg =∑

i

diχi

and ∑d2

i = |G|.


Proof

To prove this, we note that by Maschke’s theorem, there doesexist a decomposition of C[G] into irreducibles:

C[G] =⊕

j

Mj

where Mj are simple left ideals. If ki is the number of Mj

isomorphic to πi then we may also write

C[G] ∼=⊕

kiVi.

We are claiming that the multiplicity ki = di, the dimension of Vi.


Proof, continued

Consider the character

χreg =∑

i

kiχi.

By Schur orthogonality, the χi are orthonormal, so

ki = 〈χreg, χi〉.

This may be calculated since we know the character of χreg.Remember:

χreg(g) ={|G| if g = 1,0 otherwise.

So

〈χreg, χi〉 =1|G|∑g∈G

χreg(g)χi(g) =1|G||G|χi(1) = χi(1) = di.


Proof, concluded

We have proved that

C[G] =∑

i

diπi, χreg =∑

i

diχi.

Evaluating χreg at the identity gives

|G| = χreg(1) =∑

i

diχi(1) =∑

i

d2i .


Example

The identity |G| =∑

i d2i gives one way of knowing if we have

found all the irreducible characters of the group, and it is usefulinformation that can tell us in advance what the degrees of theirreducible characters will be. Let us check this for S3.

For S3, we quickly find three irreducible characters, namely twolinear characters (the trivial and sign character) and thereduced character of the permutation representation (numberof fixed points minus 1). So the di are 1, 1, 2 and∑

d2i = 12 + 12 + 22 = 6, confirming the numerology in the

theorem.

Similarly for D8 we found three linear characters and one ofdegree 2, and 12 + 12 + 12 + 12 + 22 = 8.


The number of irreducibles (I)

PropositionLet G be a finite group. Let h be the number of conjugacyclasses of G. Then G has at most h nonisomorphic irreduciblerepresentations.

Indeed, let πi : G −→ GL(Vi) be nonisomorphic irreduciblerepresentations (i = 1, 2, 3, · · · ) and let χi be their characters.We saw that χi are elements of L2

class(G), which is the vectorspace of class functions with the inner product

〈f1, f2〉 =1|G|∑g∈G

f1(g)f2(g),

making L2class(G) a finite-dimensional Hilbert space.


Proof, concluded

The dimension of L2class(G) is the number h of conjugacy

classes, since if C1, · · · , Ch are the conjugacy classes theircharacteristic functions are obviously a basis of L2

class(G). Onthe other hand the χi are orthonormal by Schur orthogonality:

〈χi, χj〉 = δij

so the number of πi is at most h.

There is a loose end.We have proved that there are at most h irreduciblerepresentations of G, where h is the number of conjugacyclasses. We will show later that actually there are exactly hirreducible representations.We need to know that the irreducible characters spanL2

class(G). We will prove that later.

Lecture 9. Permutation Representations

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