Characters of Finite Abelian Groups

A First Course in Number Theory

4.2 Characters of Finite Abelian Groups

(e) Prove that if the finite abelianp-groupGis of type (p^r1, . . . , p^rk) and of type (p^s1, . . . , p^sk), thenri=si fori= 1, . . . , k.

Hint: Use induction on the cardinality of G. Let j and be the greatest integers such that r_j ≥2 and s≥2, respectively.

Apply the induction hypothesis to pG to show that j = and ri=si fori= 1, . . . , j.

4.2 Characters of Finite Abelian Groups 127

Since|χ(g)|= 1 for allg∈G, we have

(χχ)(g) =χ(g)χ)(g) =|χ(g)|²= 1 =χ0(g), and so

χ⁻¹(g) =χ(g) for every characterχand allg∈G.

It follows that the set of all characters of a finite abelian groupGis an abelian group, called thedual group orcharacter groupofG, and denoted by G. We shall prove that G ∼=G for every finite abelian group G. We begin with finite cyclic groups.

Lemma 4.2 The dual of a cyclic group of order n is also a cyclic group of ordern.

Proof. We introduce the exponential functions e(x) =e^2πix and

en(x) =e(x/n) =e^2πix/n.

Thenth roots of unity are the complex numbersen(a) fora= 0,1, . . . , n−1.

Let Gbe a finite cyclic group of order nwith generator g0. Then G= {jg0:j= 0,1, . . . , n−1}. For every integera, we defineψa∈Gby

ψa(jg0) =en(aj). (4.1) By Exercise 3, we haveψaψb =ψa+b, ψ⁻_a¹=ψ_−a,ψa =ψb if and only if a≡b (mod n). It follows that

ψ_a=ψ^a₁

for every integera. Ifχis a character inG, thenχis completely determined by its value ong₀. Sinceχ(g₀) is annth root of unity, we haveχ(g₀) =e_n(a) for some integera= 0,1, . . . , n−1, and soχ(jg₀) =e_n(aj) for every integer j. Therefore,χ=ψa and

G={ψa :a= 0,1, . . . , n−1}={ψ₁^a:a= 0,1, . . . , n−1}

is also a cyclic group of ordern, that is,G∼=G.2

It is a simple but critical observation that if gis a nonzero element of a cyclic groupG, then ψ1(g)= 1 (Exercise 4).

Lemma 4.3 Let G be a finite abelian group and let G1, . . . , Gk be sub- groups ofGsuch thatG=G1⊕ · · · ⊕Gk. For every characterχ∈Gthere exist unique characters χi ∈ !Gi such that if g ∈G and g=g1+· · ·+gk

with gi∈Gi fori= 1, . . . , k, then

χ(g) =χ1(g1)· · ·χk(gk). (4.2) Moreover,

G∼=G!1× · · · ×G!k.

Proof. Ifχi ∈G!i fori= 1, . . . , k, then we can construct a mapχ:G→ C^× as follows. Letg ∈G. There exist unique elementsgi ∈Gi such that g=g1+· · ·+gk.Define

χ(g) =χ(g1+· · ·+gk) =χ1(g1)· · ·χk(gk).

Thenχ is a character inG, and this construction induces a map

Ψ :G!₁× · · · ×G!_k →G. (4.3) By Exercise 5, the map Ψ is a one-to-one homomorphism. We shall show that the map Ψ is onto. Letχ∈G. We define the functionχi onGi by

χi(gi) =χ(gi) for allgi∈Gi.

Thenχ_i is a character inG!_i. Ifg∈Gandg=g₁+· · ·+g_k withg_i∈G_i, then

χ(g) =χ(g1+· · ·+gk) =χ(g1)· · ·χ(gk) =χ1(g1)· · ·χk(gk).

It follows that

Ψ(χ1, . . . , χk) =χ, and so Ψ is onto.2

Theorem 4.4 Let G be a finite abelian group. If g is a nonzero element of G, then there is a character χ∈Gsuch that χ(g)= 1.

Proof. We writeG=G1⊕ · · · ⊕Gkas a direct product of cyclic groups.

Ifg= 0,then there existg1∈G1, . . . , gk∈Gk such thatg=g1+· · ·+gk, and gj = 0 for some j. Since the group Gj is cyclic, there is a character χ_j ∈G!_j such that χ_j(g_j)= 1. For i= 1, . . . , k, i=j, let χ_i ∈G!_i be the character defined by χi(g1) = 1 for allgi∈Gi. If χ= Ψ(χ1, . . . , χk)∈G, thenχ(g) =χj(gj)= 1.2

4.2 Characters of Finite Abelian Groups 129

Theorem 4.5 A finite abelian groupGis isomorphic to its dual, that is, G∼=G.

Proof. By Lemma 4.2, the dual of a finite cyclic group of ordernis also a finite cyclic group of ordern. By Theorem 4.3, a finite abelian group G has cyclic subgroupsG₁, . . . , G_k such that

G=G1⊕ · · · ⊕Gk. By Lemma 4.3 and Exercise 5 in Section 4.1,

G∼=G!₁× · · · ×G!_k ∼=G₁× · · · ×G_k∼=G₁⊕ · · · ⊕G_k =G.

This completes the proof.2

LetGbe a finite abelian group of ordern. There is apairing , from G×Ginto the group ofnth roots of unity defined by

a, χ=χ(a).

This map isnondegeneratein the sense thata, χ= 1 for all group elements a∈Gif and only ifχ=χ₀, and a, χ= 1 for all charactersχ∈Gif and only ifa= 0 (by Theorem 4.4).

For eacha∈G, the functiona, is a character of the dual groupG, that is,a, ∈G. The map ∆ : G→G defined bya−→ a, or, equivalently,

∆(a)(χ) =a, χ=χ(a), (4.4)

is a homomorphism of the groupGinto itsdouble dualG. Since the pairing is nondegenerate, this homomorphism is one-to-one. Since|G|=|G|=|G|, it follows that ∆ is a natural isomorphism ofGontoG.

Theorem 4.6 (Orthogonality relations) LetGbe a finite abelian group of ordern, and letGbe its dual group. Ifχ∈G, then

a∈G

χ(a) =

n if χ=χ₀, 0 if χ=χ₀. If a∈G, then

χ∈G

χ(a) =

n ifa= 0, 0 ifa= 0.

Proof. Forχ∈G, let

S(χ) =

a∈G

χ(a).

Ifχ=χ0, thenS(χ0) =|G|=n. Ifχ=χ0, thenχ(b)= 1 for someb∈G, and

χ(b)S(χ) = χ(b)

a∈G

χ(a)

=

a∈G

χ(ba)

=

a∈G

χ(a)

= S(χ), and so S(χ) = 0.

Fora∈G, let

T(a) =

χ∈G

χ(a).

If a= 0, then T(a) =|G| =n. If a= 0, then χ(a)= 1 for some χ ∈ G (by Theorem 4.4), and

χ(a)T(a) = χ(a)

χ∈G

χ(a)

=

χ∈G

χχ(a)

=

χ∈G

χ(a)

= T(a), and so T(a) = 0. This completes the proof. 2

Theorem 4.7 (Orthogonality relations) LetGbe a finite abelian group of ordern, and letGbe its dual group. Ifχ1, χ2∈G, then

a∈G

χ1(a)χ2(a) =

n if χ1=χ2, 0 if χ1=χ2. If a, b∈G, then

χ∈G

χ(a)χ(b) =

n if a=b, 0 if a=b.

4.2 Characters of Finite Abelian Groups 131

Proof. These identities follow immediately from Theorem 4.6, since χ1(a)χ2(a) =χ1χ⁻¹₂ (a)

and

χ(a)χ(b) =χ(a−b).

This completes the proof.2

Thecharacter table for a group has one column for each element of the group and one row for each character of the group. For example, if C₄ is the cyclic group of order 4 with generatorg0, then the characters ofC4are the functions

ψa(jg0) =e4(aj) =i^aj

fora= 0,1,2,3, and the character table is the following.

0 g0 2g0 3g0

ψ₀ 1 1 1 1

ψ1 1 i −1 −i ψ2 1 −1 1 −1 ψ₃ 1 −i −1 i

Note the that sum of the numbers in the first row is equal to the order of the group, and the sum of the numbers in each of the other rows is 0.

Similarly, the sum of the numbers in the first column is the order of the group, and the sum of the numbers in each of the other columns is 0. This is a special case of the orthogonality relations.

Exercises

1. LetC2 be the cyclic group of order 2.

(a) Compute the character table forC2.

(b) Compute the character table for the groupC₂×C₂. 2. Compute the character table for the cyclic group of order 6.

3. LetGbe a finite cyclic group of ordern. Define the charactersψa on Gby (4.1). Prove that

(a) ψaψb=ψa+b,

(b) ψ_a⁻¹=ψ₋a,

(c) ψ_a =ψ_b if and only ifa≡b (mod n).

4. Prove that if Gis cyclic andg∈G, g= 0, thenψ1(g)= 1.

5. Prove that the map Ψ defined by 4.3 is a one-to-one homomorphism.

6. Consider the map , :G×G→C^× defined by g, χ=χ(g).

Prove that

g+g, χ=g, χg, χ and g, χχ=g, χg, χ for allgg∈Gandχ, χ ∈G.

7. LetG=Z/mZ×Z/mZ. For integersaandb, we define the function ψa,b onGby

ψa,b(x+mZ, y+mZ) =e2πi(ax+by)/m=em(ax+by).

(a) Prove thatψa,b is well-defined.

(b) Prove thatψ_a,b=ψ_c,d if and only ifa≡c (mod m) andb≡d (mod m).

(c) Prove thatψa,b is a character of the groupG.

(d) Prove thatG={ψa,b:a, b= 0,1, . . . , m−1}.

8. Letpbe a prime number, and letG= (Z/pZ)^×be the multiplicative group of units in the fieldZ/pZ. Letgbe a primitive root modulop.

For every integer a, define the functionχa :G→ C^× as follows: If (x, p) = 1 andx≡g^y (modp), then

χa(x+pZ) =e^{2πay/(p−1)}=ep−1(ay).

(a) Prove thatχ_a is a character, that is,χ_a ∈G.

(b) Prove thatχa =χb if and only ifa≡b (mod p−1).

(c) Prove thatG={χa :a= 0,1, . . . , p−2}.

9. LetGbe a finite abelian group. For every integerr, let G^r={rg:g∈G}

and

G_r={χ∈G:χ^r=χ₀}.

(a) Prove thatG^ris a subgroup of Gand Gr is a subgroup ofG.