Rep#2a: Finite subgroups of multiplicative groups of fields

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Rep#2a: Finite subgroups of multiplicative groups of fields Darij Grinberg

[not completed, not proofread]

This note is mostly an auxiliary note for Rep#2. We are going to prove a fact which is used rather often in algebra:

Theorem 1. Let A be a field, and let G be a finite subgroup of the multiplicative groupA^×. Then, Gis a cyclic group.

This theorem generalizes the (well-known) fact that the multiplicative group of a finite field is cyclic. Most proofs of this fact can actually be used to prove Theorem 1 in all its generality, so there is not much need to provide another proof here. But yet, let us sketch a proof of Theorem 1 that requires only basic number theory. The downside is that it is very ugly. First, an easy number-theoretical lemma:

Lemma 2. Let i, g and a be three integers such that a is positive, such that g | a, and such that i is coprime to g. Then, there exists an integer I such that I ≡imodg and such that I is coprime toa.

Proof of Lemma 2. For every integer n, let us denote by PFn the set of all prime divisors of n. By the unique factorization theorem, for any positive integer n, the set PFn is finite and satisfies n = ^Q

p∈PFn

p^v^p⁽ⁿ⁾.

Clearly, a 6= 0 (since a is positive) and g 6= 0 (since a 6= 0 and g | a). Now, g | a yields PFg ⊆PFa. We have

a= ^Y

p∈PFa

p^v^p^(a) = ^Y

p∈PFg

p^v^p^(a)· ^Y

p∈PFa\PFg

p^v^p^(a) (since PFg ⊆PFa).

In other words, a=a₁a₂, where a₁ = ^Q

p∈PFg

p^v^p^(a) and a₂ = ^Q

p∈PFa\PFg

p^v^p^(a).

The number g is not divisible by any prime p ∈PFa\PFg (because if g is divisible by a prime p, then p ∈ PFg, so that p cannot lie in PFa\PFg). Hence, g is coprime to p^v^p^(a) for every p∈PFa\PFg. Consequently, g is coprime to the product

Q

p∈PFa\PFg

p^v^p^(a). In other words, g is coprime to a₂ (since ^Q

p∈PFa\PFg

p^v^p^(a) =a₂). Thus, by Bezout’s Theorem¹, there exist integers ρ₁ and ρ₂ such that ρ₁g+ρ₂a₂ = 1. Thus, 1−ρ1g = ρ2a2 ≡ 0 moda2. Now, let I = i−(i−1)ρ1g. Then, I =i−(i−1)ρ1g ≡ imodg. Hence, I is coprime to g (since i is coprime to g). Hence, I is not divisible by any prime p∈PFg. Thus, I is coprime to p^v^p^(a) for every p∈PFg. Consequently, I is coprime to the product ^Q

p∈PFg

p^v^p^(a). In other words, I is coprime to a₁ (since

Q

p∈PFg

p^v^p^(a)=a₁). On the other hand, I is coprime toa₂ (since I =i−(i−1)ρ₁g =i(1−ρ₁g)

| {z }

≡0 moda2

+ρ₁g ≡ρ₁g ≡ρ₁g+ρ₂a₂ = 1 moda₂

1Bezout’s theoremstates that if λ1 andλ2 are two coprime integers, then there exist integers ρ₁andρ₂ such thatρ₁λ₁+ρ₂λ₂= 1.

(2)

). Hence, I is coprime to a₁a₂ (since I is coprime to a₁ and to a₂). In other words, I is coprime toa (sincea1a2 =a). This proves Lemma 2.

Proof of Theorem 1. We first notice that

if α and β are two elements of G, then there exists γ ∈G such that

α∈ hγi and β ∈ hγi. (1)

Proof of (1). Let a be the order of α in G, and letb be the order of β in G. Let g be gcd (a, b). Then, g |a and g |b. Thus, (ag)|a and (bg)|b.

The order of α in G is a. Hence, the order of α^a^g in G is a

ag = g (since (ag) | a). Consequently, the elements αâ^g⁰, αâ^g¹, ..., αâ^g^g−1 are pairwise distinct, and we have αâ^g^g = 1. Now, for every i ∈ {0,1, ..., g−1}, we have

α^a^gⁱ

g

=







α^a^g^g

| {z }

=1







i

= 1, and thus the element αâ^gⁱ is a root of the polynomial X^g −1 ∈ A[X]. In other words, the elements αâ^g⁰, αâ^g¹, ...,

αâ^g^g−1 are roots of the polynomial X^g −1 ∈ A[X]. Since we know that these elements αâ^g⁰, αâ^g¹, ..., αâ^g^g−1 are pairwise distinct, we thus see that the elementsαâ^g⁰,αâ^g¹, ...,αâ^g^g−1 are pairwise distinct roots of the polynomial X^g −1 ∈ A[X]. But the polynomial X^g −1 ∈ A[X] can only have at most g roots (since any nonzero polynomial of degreeg over a field can only have at most g roots), so these roots αâg⁰, αâg¹, ..., αâg^g−1 must be all the roots of the polynomial X^g −1 ∈ A[X]. Consequently, the polynomial X^g −1 equals a constant times

X−α^a^g⁰ X−α^a^g¹

...

X−α^a^g^g−1

. But the constant just mentioned must be 1 (since the polynomials X^g−1 and

X−α^ag⁰ X−α^ag¹

...

X−α^ag^g−1

have the same leading term); hence, this becomes

X^g−1 =

X−α^a^g⁰ X−α^a^g¹

...

X−α^a^g^g−1

.

In other words, X^g−1 =

g−1

Q

i=0

X−α^agⁱ

. Applying this identity toX =β^bg, we obtain β^b^g^g −1 =

g−1

Q

i=0

β^b^g −α^a^gⁱ

. Since β^b^g^g −1 = β^b−1 = 0 (since b is the order of β, and thus β^b = 1), this becomes 0 =

g−1

Q

i=0

β^bg−α^agⁱ

. Hence, there must exist some i ∈ {0,1, ..., g−1} such that β^b^g − α^a^gⁱ = 0 (because if a product of elements of a field is zero, then one of the factors must be zero).

Consequently, this i ∈ {0,1, ..., g−1} satisfies β^b^g =αâ^gⁱ. Similarly, there exists some j ∈ {0,1, ..., g−1} satisfying αâ^g = β^b^g^j. Thus, αâ^g =







β^b^g

| {z }

=(α^a^g)ⁱ







j

=

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α^a^gⁱ

j

= α^a^g^ij, so that 1 =

α^ag^ij

αâ^g = αâ^gîj−1. Since the order of the elementαâ^g isg, this yieldsg |ij−1, so thatij ≡1 modg. Hence,ij is coprime tog, so that i must also be coprime tog. Thus, by Lemma 2, there exists an integer I such that I ≡imodg and such thatI is coprime to a. Since I ≡imodg, we have g |I−i, and thus αâgÎ−i = 1 (since g is the order of αâg), so that

αâ^gÎ =αâ^g^(I−i)+i =αâ^gÎ−i

| {z }

=1

α^a^gⁱ =α^a^gⁱ =β^b^g. (2)

Now, the integersagandbgare coprime (since gcd (ag, bg) = gcd (a, b)

| {z }

=g

g = gg = 1); hence, by Bezout’s Theorem, there exist integers u and v such that u·ag+v·bg = 1. Now, let γ =α^Ivβ^u. Then, γ ∈G and

γ^b^g =αÎvβû^b^g =αÎv^b^g

| {z }

=α^Iv·b^g

(β^u)^bg

| {z }

=(^β^b^g)^u

=α^Iv·b^g







β^b^g

| {z }

=(^α^a^g)^I

(by (2))







u

=α^Iv·b^g

α^a^g^I

u

| {z }

=(αâ^g)Îu=αÎu·a^g

=αÎv·b^gαÎu·a^g =αÎv·b^g+Iu·a^g =αÎ

(since Iv·bg+Iu·ag =I(u·ag+v·bg)

| {z }

=1

=I). Since I is coprime to a, there exist integers xand y such thatxI+ya= 1 (according to Bezout’s theorem). Thus,

α=α¹ =α^Ix+ay (since 1 =xI+ya=Ix+ay)

= α^Ix

|{z}

=(α^I)^x

α^ay

|{z}

=(α^a)^y

=





 α^I

|{z}

=γ^b^g







x







α^a

|{z}

=1 (sinceais the order ofα)







y

=γ^b^g^x1^y =γ^b^g^x ∈ hγi.

On the other hand, since γ =α^Ivβ^u, we have

γâ^g =αÎvβûâg = αÎvâg

| {z }

=α^Iv·α^g=α^(a^g)·Iv

=(^αâ^g)Îv⁼(^αâ^g)Î^v

·(β^u)^a^g

| {z }

=β^u·(a^g)

=







α^a^g^I

| {z }

=β^b^g (by (2))







v

·β^u·(a^g)

= β^b^g^v

| {z }

=β^(bg)·v=β^v·(bg)

·β^u·(a^g) =β^v·(b^g)·β^u·(a^g) =β^v·(b^g)+u·(a^g)

=β¹ (since v·(bg) +u·(ag) =u·ag+v·bg = 1)

=β,

and therefore β =γ^a^g ∈ hγi.

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Altogether, we have proven thatγ ∈G, thatα ∈ hγiand thatβ ∈ hγi. This proves (1).

Now, let us finally prove Theorem 1: Clearly, there exists a subset P of the group G such that G= hPi (in fact, the whole group G is an example of such a subset P).

LetU be such a subset with the smallest number of elements.² Then, U is a subset of the groupGsuch thatG=hUi, but there is no subsetU⁰ ofG with less elements than U that satisfiesG=hU⁰i.

We let k =|U|, and we write the set U asU ={u1, u2, ..., uk}, whereu1, u2, ...,uk

are the k (pairwise distinct) elements of U. Assume now that k >1. Then, u₁ and u₂ are well-defined. Now, there exists an element γ ∈ G such that u₁ ∈ hγi and u₂ ∈ hγi (by (1), applied to α = u1 and β =u2), and therefore ui ∈ hγ, u3, u4, ..., uki for every i∈ {1,2, ..., k} ³. Hence,hu₁, u₂, ..., u_ki ⊆ hγ, u₃, u₄, ..., u_ki, so that

G=hUi=h{u₁, u₂, ..., u_k}i=hu₁, u₂, ..., u_ki ⊆ hγ, u₃, u₄, ..., u_ki=h{γ, u₃, u₄, ..., u_k}i=hU⁰i, where U⁰ denotes the subset {γ, u₃, u₄, ..., u_k} of G. But clearly, also G⊇ hU⁰i. Thus,

G = hU⁰i. Besides, the subset U⁰ of G has less elements than U (because U⁰ = {γ, u₃, u₄, ..., u_k} has at most k −1 elements, while U has |U| = k elements). This contradicts to the fact that there is no subset U⁰ of G with less elements than U that satisfies G = hU⁰i. This contradiction shows that our assumption k > 1 was wrong.

Hence, k ≤ 1, so that k = 1 or k = 0. If k = 0, then |U| = k = 0 and thus U = ∅, which leads toG=h∅i= 1, so thatG is a cyclic group. Ifk= 1, then |U|=k = 1, so that U ={u}for some u∈G, and therefore G=hUi=h{u}i=hui is a cyclic group.

Hence, in both cases, G is a cyclic group. This proves Theorem 1.

Here is an easy consequence of Theorem 1:

Lemma 3. Let A be a field. Let n be a positive integer, and for every i∈ {1,2, ..., n}, let ξ_i be a root of unity in A. Then, there exists some root of unity ζ of A and a sequence (k₁, k₂, ..., k_n) of nonnegative integers such that ξ_i =ζ^kⁱ for every i∈ {1,2, ..., n} and gcd (k₁, k₂, ..., k_n) = 1.

Proof of Lemma 3. LetGbe the subgrouphξ₁, ξ₂, ..., ξ_niof the multiplicative group A^×. Then, the map

Φ :hξ₁i × hξ₂i ×...× hξ_ni → hξ₁, ξ₂, ..., ξ_ni defined by (x1, x2, ..., xn)7→x1x2...xn

is surjective (because every element ofhξ₁, ξ₂, ..., ξ_nihas the form ^Qⁿ

i=1

ξ_i^fⁱfor somen-tuple (f₁, f₂, ..., f_n) of integer, and thus is Φξ₁^f¹, ξ₂^f², ..., ξ_n^fⁿ), and the sethξ₁i×hξ₂i×...×hξ_ni is finite (since the sethξ_iiis finite for everyi∈ {1,2, ..., n},becauseξ_i is a root of unity).

Hence, the set hξ₁, ξ₂, ..., ξ_ni is finite. Thus, G = hξ₁, ξ₂, ..., ξ_ni is a finite subgroup of

2Indeed, such aU exists, because the set of all subsets of the group Gis finite (since G itself is finite).

3In fact, three cases are possible: eitheri= 1, ori= 2, ori≥3. Ifi= 1, thenui∈ hγ, u3, u4, ..., uki follows fromu1∈ hγi ⊆ hγ, u3, u4, ..., uki. Ifi= 2, thenui∈ hγ, u3, u4, ..., ukifollows fromu2∈ hγi ⊆ hγ, u3, u4, ..., uki. Finally, if i≥3, then ui ∈ hγ, u3, u4, ..., ukiis trivial. Thus, ui ∈ hγ, u3, u4, ..., uki holds in all cases.

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A^×. Hence, by Theorem 1, this groupGis cyclic, so that there exists some τ ∈Gsuch that G=hτi. Now, if u is the order of τ in the group G, thenhτi={τ⁰, τ¹, ..., τ^u−1}.

Hence, for every i ∈ {1,2, ..., n}, there exists some nonnegative integer `_i such that ξ_i = τ^`ⁱ (since ξ_i ∈ G = hτi = {τ⁰, τ¹, ..., τ^u−1}). Now, let ` = gcd (`₁, `₂, ..., `_n).

Let ζ = τ^`, and let ki = `i` for every i ∈ {1,2, ..., n}. Then, `i = `ki for every i∈ {1,2, ..., n}.

Now we know that ζ is a root of unity (since ζ ∈G, and thus Lagrange’s theorem yieldsζ^|G| = 1), and for everyi∈ {1,2, ..., n}we haveξ_i =τ^`ⁱ =τ^`kⁱ =



 τ^`

|{z}

=ζ





ki

=ζ^kⁱ. Finally, recall that k_i = `_i` for every i ∈ {1,2, ..., n}. Thus, gcd (k₁, k₂, ..., k_n) = gcd (`₁`, `₂`, ..., `_n`) = gcd (`₁, `₂, ..., `_n)

| {z }

=`

` = 1. Thus, Lemma 3 is proven.