Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 34 (1993), No. 2, 157-161.

Is Whaples’ Theorem a Group Theoretical Result?

Arne Ledet

Mathematical Institute, University of Copenhagen Universitetsparken 5, DK - 2100 Copenhagen

A theorem by Whaples (see [2] or [3]) states, that if a field admits a cyclic extension of degree p, where p is an odd prime, it admits a pro-cyclic extension of degree p^{∞}. Similarly, the existence of a cyclic extension of degree 4 implies the existence of a procyclic extension of degree 2^{∞}.

This can be formulated as a statement about the absolute Galois group G of the field: If G has the cyclic group Z/pZ as a factor, it has the pro-cyclic group Ẑ_{p} of p-adic integers as a factor, and similarly for Z/4Z and the 2-adic integers Ẑ_{2}.

This property does not hold for pro-finite groups in general. (With Z/pZ and Z/4Z as obvious counter-examples.) But unless the field is formally real the absolute Galois group is torsion free, and it is therefore natural to ask whether Whaples’ result generalizes to torsion free pro-finite groups.

It does not.

Let p be a prime. For h ≥ 0, k, m > 0, h + k ≥ m, we define

𝔐_{h,k,m} = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ M_{2}(Ẑ_{p}) | \begin{matrix}ad − bc = 1, d ≡ 1 (mod p^{m}), \\ b ≡ 0 (mod p^{h}), c ≡ 0 (mod p^{k}) \end{matrix} \}.

It is easily seen that 𝔐_{h,k,m} is a closed p-subgroup of

SL_{2}(Ẑ_{p}) = lim_{⟵} SL_{2}(Z/p^{n}Z).

For odd primes p these groups are investigated in [1, III.§17], where the following theorems are proved:

Theorem 1. Let

B(x) = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) , C(x) = ( \begin{matrix}1 & 0 \\ x & 1 \end{matrix} ) , D(x) = ( \begin{matrix}x^{−1} & 0 \\ 0 & x \end{matrix} ) ,

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1993 Heldermann Verlag, Berlin

158 A. Ledet: Is Whaples’ Theorem a Group Theoretical Result? and consider the following subgroups of 𝔐_{h,k,m}:

𝔅_{𝔥} = {B(x) | x ≡ 0 (mod p^{h})},
ℭ_{𝔨} = {C(x) | x ≡ 0 (mod p^{k})},
𝔇_{𝔪} = {D(x) | x ≡ 1 (mod p^{m})}.

Every element x ∈ 𝔐_{h,k,m} can be written in one and only one way as x = bcd, b ∈ 𝔅_{𝔥}, c ∈ ℭ_{𝔨}, d ∈ 𝔇_{𝔪}.

[𝔐_{h,k,m} : 𝔐_{h+h′,k+k′,m+m′}] = p^{h′+k′+m′}.

An easy consequence of the decomposition 𝔐_{h,k,m} = 𝔅_{𝔥}ℭ_{𝔨}𝔇_{𝔪} is the following: 𝔐_{h+h′,k+k′,m+m′} ◁ 𝔐_{h,k,m}, if h′ ≤ m + m′ ≤ h + k + k′ and k′ ≤ m + m′ ≤ k + h + h′. Theorem 2. 𝔐′_{h,k,m} = 𝔐_{h+m,k+m,h+k}. Theorem 1 is also valid for p = 2, whereas Theorem 2 holds for odd primes only. The analogue of Theorem 2 for p = 2 (and m > 1) is Theorem 4 below. Lemma 3. Let p = 2 and m > 1. Then

\begin{aligned}[𝔅_{𝔥}, 𝔇_{𝔪}] & = 𝔅_{𝔥+𝔪+1} and \\ [ℭ_{𝔨}, 𝔇_{𝔪}] & = ℭ_{𝔨+𝔪+1}. \end{aligned}

Also [B(b), C(c)] = B(b)C(c)B(−b)C(−c) = B(−b^{2}ce^{−1})C(bc^{2}e)D(e), where e = 1 − bc. Proof. We have

[B(b), D(d)] = B(b)D(d)B(b)^{−1}D(d)^{−1}

= ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ) ( \begin{matrix}d^{−1} & 0 \\ 0 & d \end{matrix} ) ( \begin{matrix}1 & −b \\ 0 & 1 \end{matrix} ) ( \begin{matrix}d & 0 \\ 0 & d^{−1} \end{matrix} )

= ( \begin{matrix}1 & b(1 − d^{−2}) \\ 0 & 1 \end{matrix} ) = B(b(1 − d^{−2})).

If d ≡ 1 (mod 2^{m}), we have 1 − d^{−2} = (1 + d^{−1})(1 − d^{−1}) ≡ 0 (mod 2^{m+1}). Hence [𝔅_{𝔥}, 𝔇_{𝔪}] = 𝔅_{𝔥+𝔪+1}. [ℭ_{𝔨}, 𝔇_{𝔪}] = ℭ_{𝔨+𝔪+1} is proved similarly. Theorem 4. Let p = 2 and m > 1. If h + k > m,

𝔐′_{h,k,m} = 𝔐_{h+m+1,k+m+1,h+k}.

If h + k = m,

𝔐′_{h,k,m} = 𝔐_{h+m+1,k+m+1,m+1} ⟨B(2^{h+m})C(2^{k+m})D(1 + 2^{m})⟩ .

Proof. h + k > m: If b ≡ 0 (mod 2^{h}) and c ≡ 0 (mod 2^{k}), we have −b^{2}ce^{−1} ≡ 0 (mod 2^{2h+k}) and bc^{2}e ≡ 0 (mod 2^{h+2k}). Since 2h+k ≥ h+m+1 and h+2k ≥ k +m+1, we have D(e) ∈ 𝔐′_{h,k,m} by Lemma 3. 𝔇_{𝔥+𝔨} is generated by these elements D(e), and we conclude, that 𝔇_{𝔥+𝔨} ⊆ 𝔐′_{𝔥,𝔨,𝔪}. Hence

𝔐_{h+m+1,k+m+1,h+k} ⊆ 𝔐′_{h,k,m}.

Since 𝔐_{h+m+1,k+m+1,h+k} ◁ 𝔐_{h,k,m} and 𝔐_{h,k,m}/𝔐_{h+m+1,k+m+1,h+k} is abelian, we have

𝔐′_{h,k,m} = 𝔐_{h+m+1,k+m+1,h+k}.

h + k = m: We still have 𝔅_{𝔥+𝔪+1} ⊆ 𝔐′_{𝔥,𝔨,𝔪} and ℭ_{𝔨+𝔪+1} ⊆ 𝔐′_{𝔥,𝔨,𝔪}. If b ≡ 0 (mod 2^{h+1}) and c ≡ 0 (mod 2^{k}), we get D(e) ∈ 𝔐′_{h,k,m} and e ≡ 1 (mod 2^{m+1}). Hence

𝔐_{h+m+1,k+m+1,m+1} ⊆ 𝔐′_{h,k,m}

These subgroups are both normal.

We know that B(−2^{2h}2^{k}(1 − 2^{m})^{−1})C(2^{h}2^{2k}(1 − 2^{m}))D(1 − 2^{m}) ∈ 𝔐′_{h,k,m}, and B(2^{h+m})C(2^{k+m})D(1 + 2^{m}) ≡ B(−2^{2h}2^{k}(1 − 2^{m})^{−1})C(2^{h}2^{2k}(1 − 2^{m}))D(1 − 2^{m})

(mod 𝔐_{h+m+1,k+m+1,m+1}). If we let B = B(2^{h+m}), C = C(2^{k+m}) and D = D(1 + 2^{m}) this gives

𝔐_{h+m+1,k+m+1,m+1} ⟨BCD⟩ ⊆ 𝔐′_{h,k,m}.

𝔐_{h,k,m} is generated by b = B(2^{h}), c = C(2^{k}) and d = D, so to prove normality of 𝔐_{h+m+1,k+m+1,m+1} ⟨BCD⟩ one directly verifies that [x, BCD] ∈ 𝔐_{h+m+1,k+m+1,m+1} for x = b, c, d. This is easily done, and we conclude

𝔐_{h+m+1,k+m+1,m+1} ⟨BCD⟩ ◁ 𝔐_{h,k,m}.

Since 𝔐_{h,k,m}/𝔐_{h+m+1,k+m+1,m+1} ⟨BCD⟩ is abelian, we have

𝔐_{h+m+1,k+m+1,m+1} ⟨BCD⟩ = 𝔐′_{h,k,m}.

Proposition 5. Let p be an odd prime. Then

𝔐_{h,k,m}/𝔐′_{h,k,m} ≃ Z/p^{m}Z × Z/p^{m}Z × Z/p^{h+k−m}Z.

Let p = 2 and m > 1. Then

𝔐_{h,k,m}/𝔐′_{h,k,m} ≃ Z/2^{m+1}Z × Z/2^{m+1}Z × Z/2^{h+k−m}Z.

Proof. p odd: 𝔐_{h,k,m} is generated by B(p^{h}), C(p^{k}) and D(1 + p^{m}). It follows that 𝔐_{h,k,m}/𝔐′_{h,k,m} is the direct product of the subgroups generated by B(p^{k})𝔐′_{h,k,m}, C(p^{h})𝔐′_{h,k,m} and D(1 + p^{m})𝔐′_{h,k,m}.

p = 2 and h + k > m: The structure of 𝔐_{h,k,m}/𝔐′_{h,k,m} is obvious, since 𝔅_{𝔥}, ℭ_{𝔨} and 𝔇_{𝔪} are pro-cyclic with generators B(2^{h}), C(2^{k}) and D(1 + 2^{m}).

p = 2 and h + k = m: Let the notation be as in the proof of Theorem 4. From BCd, d^{2} ∈ 𝔐′_{h,k,m} we get d ≡ BC (mod 𝔐′_{h,k,m}), so 𝔐_{h,k,m}/𝔐′_{h,k,m} is generated by b and c. This gives

𝔐_{h,k,m}/𝔐′_{h,k,m} ≃ Z/2^{m+1}Z × Z/2^{m+1}Z.

Proposition 6. i) 𝔐_{h,k,m} is torsion free for p > 3. ii) For p = 3, 𝔐_{h,k,m} is torsion free unless (h, k, m) = (0, 1, 1). If (h, k, m) = (0, 1, 1), every torsion element has order 3 and is of the form

( \begin{matrix}(x − 1)/2 & b \\ c & −(x + 1)/2 \end{matrix} ) , where x^{2} = 1 − 4(1 + bc).

iii) For p = 2, 𝔐_{h,k,m} is torsion free unless m = 1, in which case −E is a torsion element. If m = 1 and h + k > 1, −E is the only torsion element. If m = 1 and h + k = 1, there are in addition torsion elements of order 4, and any such torsion element is of the form

( \begin{matrix}x & b \\ c & −x \end{matrix} ) , where bc ≡ 6 (mod 8) and x^{2} = −(1 + bc).

Proof. Let A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(Ẑ_{p}), and assume that A^{p^{k}} = E for some k ∈ N. The

eigenvalues of A must be p^{k th} roots of unity, and also roots of the characteristic polynomial λ^{2} − (a + d)λ + 1.

If 1 is the only eigenvalue, we have λ^{2} − (a + d)λ + 1 = λ^{2} − 2λ + 1, i.e., a + d = 2 and ad = 1 + bc. We can assume that

a = 1 + \sqrt{−bc} and d = 1 − \sqrt{−bc}.

An induction argument gives

A^{n} = ( \begin{matrix}1 + n\sqrt{−bc} & nb \\ nc & 1 − n\sqrt{−bc} \end{matrix} ) ,

so b = c = 0 and A = E.

p > 3: The p^{i th} cyclotomic polynomial is irreducible in Ẑ_{p}[X] of degree p^{i−1}(p−1) > 2 for i > 0, so the only possible eigenvalue is 1, and we get A = E by the above argument.

p = 3: If 1 is the only eigenvalue, we get A = E as above. If the eigenvalues are ζ and ζ^{−1}, where ζ is a primitive third root of unity, we get λ^{2} − (a + d)λ + 1 = λ^{2} + λ + 1, i.e., a + d = −1 and ad = 1 + bc. We can assume that

a = \frac{−1 + \sqrt{1 − 4(1 + bc)}}{2} and d = \frac{−1 − \sqrt{1 − 4(1 + bc)}}{2}.

Since a ≡ 1 (mod 3^{m}) we get bc ≡ −3 (mod 3^{m}). bc ≡ 0 (mod 3^{h+k}) and h + k ≥ m, so m = 1. If h + k > 1 we have 1 − 4(1 − bc) ≡ −3 (mod 9), and since −3 is not a square in Z/9Z, 1 − 4(1 − bc) is not a square in Ẑ_{3}. Therefore h + k = 1, i.e., h = 0 and k = 1.

Now let x = \sqrt{1 − 4(1 + bc)}. Then

A = ( \begin{matrix}(x − 1)/2 & b \\ c & −(x + 1)/2 \end{matrix} )

and direct calculation shows A^{3} = E.

Example: A = ( \begin{matrix}1 & 1 \\ −3 & −2 \end{matrix} ).

p = 2: The eigenvalues are either 1, −1 or primitive fourth roots of unity. If 1 is the only eigenvalue, we have A = E. If −1 is the only eigenvalue, −A ∈ SL_{2}(Ẑ_{2}) has 1 as its only eigenvalue, hence −A = E and A = −E. This is obviously only possible for m = 1.

Assume now, that the eigenvalues are primitive fourth roots of unity. Then λ^{2} − (a + d)λ + 1 = λ^{2} + 1, i.e., a + d = 0. Since a ≡ d ≡ 1 (mod 2^{m}), we have m = 1. We can assume

a = \sqrt{−(1 + bc)}, d = −\sqrt{−(1 + bc)},

hence A^{2} = −E and A has order 4.

Since −(1+bc) ≡ −1 (mod 2), −(1+bc) is a square in Ẑ_{2}, if and only if −(1+bc) ≡ 1 (mod 8), if and only if bc ≡ 6 (mod 8). Since bc ≡ 0 (mod 2^{h+k}), we get h + k = 1.

Example: A = ( \begin{matrix}1 & 1 \\ −2 & −1 \end{matrix} ).

It is now clear, that Whaples’ Theorem does not generalize to arbitrary torsion free pro-finite groups: For an odd prime p and a natural number n we get examples of torsion free pro-p-groups with Z/p^{n}Z, but not Z/p^{n+1}Z, as a factor. For p = 2 and n ≥ 3 we get examples of torsion free pro-2-groups with Z/2^{n}Z, but not Z/2^{n+1}Z, as a factor.

Unfortunately, these groups do not give us an example of a torsion free pro-2-group with Z/4Z, but not Z/8Z, as a factor.

References

[1] Huppert, B.: Endliche Gruppen I, Springer 1967.

[2] Kuyk, W.; Lenstra, H.W. Jr.: Abelian extensions of arbitrary fields. Math. Ann. 216 (1975), 99–104.

[3] Whaples, G.: Algebraic extensions of arbitrary fields, Duke Math. J. 24 (1957), 201–