On the torsion units in integral group rings of dihedral 2-groups

Alexander Zimmermann

Universität Stuttgart, Math. Inst. B, Pfaffenwaldring 57, D-70550 Stuttgart, Germany

A. Weiss proved in [16] that for a p-group G and a finite subgroup U of the units of augmentation 1 of the group ring ẐZ_{p}G over the p-adic integers there exists a group basis containing U. In [12] K. W. Roggenkamp and L. L. Scott proved that all group bases in ẐZ_{p}G are conjugate. We will be concerned with the analogous question in ZZG, the integral group ring of a p-group G: Are finite subgroups of the units of augmentation 1 of ZZG subgroups of group bases?

I. Hughes and K. E. Pearson showed in [8] that in the group of units of augmentation 1 of the integral group ring of the dihedral group of order 6 there is an involution not being part of any group basis. This involution however is not even part of a group basis in the group ring over the 2-adic integers. Therefore, they measure a local phenomenon rather than a global one. Furthermore, the dihedral group of order 6 is not a p-group and A. Weiss’ theorem is not available. We investigate this problem for dihedral groups of order a power of 2 such that A. Weiss’ theorem can be applied.

To go into details: Let D_{2^{n}} be the dihedral group of order 2^{n+1}. We denote by V (RG) the group of units of augmentation 1 in the group ring of the finite group G over the integral domain R. The augmentation is the map induced by the trivial representation. As announced in [18] we will show that in ZZD_{2^{n}} the number of conjugacy classes of involutions in V (ZZD_{2^{n}}) is equal to

1 + 2 · 2^{n−1} · ∏_{k=1}^{n} h_{2^{k}}^{+}

where h_{2^{k}}^{+} is the class number of the maximal real subfield of Q(ζ_{2^{k}}), ζ_{2^{k}} being a primitive 2^{k}-th root of unity over Q. We will show that the number of conjugacy classes of involutions in V (ZZD_{2^{n}}) consisting of elements that are part of a group basis is

1 + 2 · 2^{n−1}.

Therefore, for each involution x in V (ZZD_{2^{n}}) there is a group basis H such that x ∈ H if and only if the class number of ZZ[ζ_{2^{k}} +ζ_{2^{k}}^{−1}] is equal to 1 for all k ≤ n. H. Cohn conjectured that this always happens (cf. [1]). The problem is therefore reduced to a conjecture in number theory.

As proved by S. Endo, T. Miyata and K. Sekiguchi in [3] the number of conjugacy classes of group bases is equal to 2^{n−1} · 2^{n−2}. If we use this result — it will follow independently and shorter from our arguments — we are able to give generators for the group of outer central automorphisms of ZZD_{2^{n}} in terms of conjugation with group ring elements. Here we call an automorphism of a ring central if its restriction to the center is the identity.

In the integral group ring of the semidihedral group of order 16 there exists an involution that is not part of any group basis as is shown by the author in [19]. At the end of this paper we will give a sketch of the proof of the analogous statement for the semidihedral groups of order 2^{n+1}, n being an integral number greater than 3.

Our method is an application of the theory of K. W. Roggenkamp and L. L. Scott as reported in [11]. This theory is explained in detail in [13], so we restrict ourselves to give only some of the ideas of the theory.

Let G be a finite p-group and let U be a subgroup of G. We assume that ZZG satisfies the Eichler condition (cf. [2]) and that the centralizer of U in ZZG is commutative. We have to examine monomorphisms of U in V (ZZG) that are induced by conjugation with a unit x of QG. These give rise to ZZ(U × G) bimodules where U operates as multiplication by xUx^{−1} on the left and G acts as multiplication on the right, the so called twisted bimodule by conjugation x (cf. [5]). The isomorphism classes of those modules which are free ZZG modules of rank one at the right and that become trivial if they are tensored up to Q parametrize the embeddings modulo conjugation with a unit in ZZG. This is easily seen since the bimodule xZZG is isomorphic to ZZG if and only if modulo elements of the centralizer of U in QG the element x is a unit of ZZG. Each of those embeddings can be realized by conjugation in ẐZ_{q}G for all primes q by the result of A. Weiss [16]. Similar calculations as in the proof of A. Fröhlich’s localization sequence (cf. [5]) give us an element of the class group of C_{ZZG}(U) for each embedding modulo inner automorphisms of ZZG. This element of the class group lies in the kernel Cl_{ZZG}(C_{ZZG}(U)) of the induction map to the class group of ZZG by the above. Conversely, if an element of the class group maps to the neutral element of the class group of ZZG then it gives us a bimodule inducing an embedding. An element of Cl_{ZZG}(C_{ZZG}(U)) is in the image of the induction map from the class group of the center of ZZG to the class group of the centralizer of U in ZZG if and only if the corresponding conjugacy class of subgroups rationally conjugate to U consists of elements that are part of group bases. This is most easily seen by the fact that one obtains the bimodule for ZZ(U × G) from the bimodule for ZZ(G × G) by restricting the operation of G at the left to the subgroup U. This, however, corresponds to forming the induced module since also for Fröhlich’s localization sequence one gets the ZZ(G×G) bimodule from the ideal of the class group by inducing it to ZZG. Then the tensor product is associative and everything is done.

The major tool in calculating the class groups is Reiner-Ullom’s version of Milnor’s Mayer-Vietoris-sequence. For this purpose we write the various orders as pullbacks corresponding to the Wedderburn decomposition of ZZD_{2^{n}}. Afterwards we calculate the images of the global unit groups in several quotients to obtain the order of the class groups inductively. The group structure and the homomorphisms among the class groups are obtained by idèle theoretic arguments. The automorphism group of ZZD_{2^{n}} is then easily obtained in a very explicit way by the previous results.

Acknowledgment: I want to express my thanks towards my academical teacher Professor Dr. K. W. Roggenkamp not only for his patient help but also for his encouragement and steady interest towards the progress of my thesis.

1. Preparations

We have to calculate some class groups and use several pullback diagrams for this topic. In general, a central idempotent e in the Q-algebra A induces a pullback diagram for the ZZ-order Λ as follows:

Λ  ⟶  Λ · e
↓  ↓
Λ · (1 − e) ⟶ (Λ · e) /(Λ ∩ (Λ · e))

Since e ∈ QΛ, we can form Λ · e ⊂ QΛ and intersect it with Λ ⊂ QΛ. The intersection is a twosided ideal of Λ · e. The quotient is denoted by (Λ · e) /(Λ ∩ (Λ · e)). For D_{2^{n}} :=< a, b | a^{2^{n}}, b^{2}, baba > we let

e_{n−1}^{n} := \frac{1}{2}(1 + a^{2^{n−1}})

and if we define

C(n) := C_{ZZD_{2^{n}}}(b) := {x ∈ ZZD_{2^{n}}|xb = bx}
= < 1, b, a^{i} + a^{−i}, b(a^{i} + a^{−i}), a^{2^{n−1}}, a^{2^{n−1}}b|i = 1, ..., 2^{n−1} − 1 >_{ZZ}

we get pullback diagrams for C(n), ZZD_{2^{n}} and B(n−1) := C(n)·e_{n−1}^{n}. Since

ZZD_{2^{n}} · e ≃ ZZD_{2^{n−1}}

we may identify B(n − 1) with a subring of ZZD_{2^{n−1}}:

B(n − 1) =< 1, b, a^{i} + a^{−i}, b(a^{i} + a^{−i}), 2a^{2^{n−1}}, 2a^{2^{n−1}}b|i = 1, ..., 2^{n−2} − 1 >_{ZZ} .

Throughout the whole paper we abbreviate ω_{n}^{+} := ζ_{2^{n}} + ζ_{2^{n}}^{−1} and ω_{n}^{+}(i) := ζ_{2^{n}}^{i} + ζ_{2^{n}}^{−i}.

Since ZZD_{2^{n}} · (1 − e_{n−1}^{n}) is a suborder of the 2 × 2 matix ring over ZZ[ω_{n}^{+}] (cf. [10]), we may identify C(n) · (1 − e_{n−1}^{n}) with ZZ[ω_{n}^{+}] < b >≃ ZZ[ω_{n}^{+}]C_{2} while denoting the cyclic group of order m by C_{m}. We observe that

B(n) · (1 − e_{n−1}^{n}) = C(n) · (1 − e_{n−1}^{n})

with the above identification. We see that

C(n) · (1 − e_{n−1}^{n}) ∩ C(n) = 2ZZ[ω_{n}^{+}] < b >

and the quotient ZZ[ω_{n}^{+}] < b > /(2ZZ[ω_{n}^{+}] < b >) is denoted by \overline{B}(n − 1). Analogously,

B(n) · (1 − e_{n−1}^{n}) ∩ B(n) = 2ω_{n}^{+}ZZ[ω_{n}^{+}] < b >

and the quotient ZZ[ω_{n}^{+}] < b > /(2ω_{n}^{+}ZZ[ω_{n}^{+}] < b >) is denoted by B̃(n − 1). Furthermore, we observe that B(n) · e_{n−1}^{n} ≃ B(n − 1).

We have now written the above defined orders B(n) and C(n) as pullbacks over smaller orders. It is now possible to apply Mayer-Vietoris sequences for the determination of the class groups of those orders.

2. Calculating the class number

We now give an upper bound for the class number of C(n). This will be done by discussing the index of the image of the unit group of ZZ[ω_{n}^{+}] < b > in \overline{B}(n − 1) and B̃(n − 1).

In [7] Gustafson and Roggenkamp gave generators for the image of the unit group of Z[ω_{n}^{+}] modulo 2.

Proposition 1 Let π_{2} be the natural epimorphism obtained by factoring the ideal of ZZ[ω_{n}^{+}] < b > generated by 2 and let π_{2ω_{n}^{+}} be the natural epimorphism obtained by factoring the ideal generated by 2ω_{n}^{+}. Then

1. π_{2}(U(ZZ[ω_{n}^{+}] < b >)) = U(\overline{B}(n − 1))

2. |U(B̃(n − 1)) : π_{2ω_{n}^{+}}(U(ZZ[ω_{n}^{+}] < b >))| = 2. 2 − b is not an element of π_{2ω_{n}^{+}}(U(ZZ[ω_{n}^{+}] < b >)).

Before proving Proposition 1 we establish two little lemmas:

Lemma 1 1 + ω_{n}^{+}(i) and b + ω_{n}^{+}(i) are units in ZZ[ω_{n}^{+}] < b > for all i = 1, ..., 2^{n−1} − 1.

Proof: The elements in the first series are Galois conjugates of cyclotomic units of ZZ[ω_{n}^{+}] and the second series can be mapped to elements of the first by the fact that

(b + ω_{n}^{+}(i)) · (b − ω_{n}^{+}(i)) = −(1 + ω_{n}^{+}(2i)).

Lemma 2 Let n ≥ 2 and k ≥ 2 be natural numbers. If (1+2^{k}x) ∈ U(ZZ[ω_{n}^{+}]) for an x ∈ ZZ[ω_{n}^{+}] then x ∈ ω_{n}^{+}ZZ[ω_{n}^{+}].

Proof: We use induction over n. For n = 2 the statement is clear. The field extension Q[ω_{n}^{+}] : Q[ω_{n−1}^{+}] is quadratic. Let

x = α + ∑_{i} α_{i} · ω_{n}^{+}(2i) + ∑_{j} β_{j} · ω_{n}^{+}(2j + 1) ∈ ZZ[ω_{n}^{+}]

be a generic element with integral coefficients. Then the nontrivial element σ of the Galois group of Q[ω_{n}^{+}] : Q[ω_{n−1}^{+}] acts on x as

x^{σ} = α + ∑_{i} α_{i} · ω_{n}^{+}(2i) − ∑_{j} β_{j} · ω_{n}^{+}(2j + 1) ∈ ZZ[ω_{n}^{+}]

and the norm nr down to Q[ω_{n−1}^{+}] of 1 + 2^{k}x calculates as

nr(1 + 2^{k}x) = 1 + 2^{k}tr(x) + 2^{2k}nr(x)
= 1 + 2^{k+1}(α + ∑_{i} α_{i} · ω_{n}^{+}(2i) + 2^{k−1}nr(x))
∈ U(ZZ[ω_{n−1}^{+}]) = U(ZZ[ω_{n}^{+}(2)])

tr being the trace. By induction we have

α + ∑_{i} α_{i} · ω_{n}^{+}(2i) + 2^{k−1}nr(x) ∈ ω_{n}^{+}(2)ZZ[ω_{n}^{+}(2)],

If n = 3 all summands following α are even integers. The sum has to be even, and therefore, α is even, proving the assertion. If n ≥ 4 then, since 2 and ω_{n}^{+}(2i) are multiples of ω_{n}^{+}(2), we conclude that α ∈ ω_{n}^{+}ZZ[ω_{n}^{+}]. This proves the lemma.

Now we are able to prove that 2 − b ∈ U(B̃(n − 1)) is not an image of a unit of ZZ[ω_{n}^{+}] < b > modulo 2ω_{n}^{+}.

The preimage of 2 − b is

2 − b + 2ω_{n}^{+}ZZ[ω_{n}^{+}] < b > .

This is in the Wedderburn components of the rational group algebra equal to

P_{2−b} := {(1 + 2ω_{n}^{+}(x + y), 3 + 2ω_{n}^{+}(x − y))|x, y ∈ ZZ[ω_{n}^{+}]}.

Let

ε : ZZ[ω_{n}^{+}] < b >⟶ ZZ[ω_{n}^{+}]

and

\overline{ε} : B̃(n − 1) ⟶ B̃(n − 1)/(b − 1)B̃(n − 1)

be the usual augmentation maps. Now \overline{ε}(2 − b) = 1 and ε together with \overline{ε} fit in a commutative square with respect to the projections modulo 2ω_{n}^{+}. Of course ε as well as \overline{ε} are split by x ⟶ x · 1. Therefore, if u ∈ P_{2−b} also u · ε(u)^{−1} ∈ P_{2−b} and u · ε(u)^{−1} has augmentation 1. Assuming this to be done then in the set P_{2−b} the element x + y = 0 and the second component turns out to be 3 + 4ω_{n}^{+}y with y ∈ ZZ[ω_{n}^{+}]. Since

3 + 4ω_{n}^{+}y = −1 + 4(1 + ω_{n}^{+}y)

we are in the situation of the lemma. In fact 1 + ω_{n}^{+}y cannot belong to the prime ideal above 2 as requested and u cannot be a unit.

Now we come to the proof of Proposition 1: If part 1 is proved part 2 also follows: In fact, π_{2ω_{n}^{+}} factors through π_{2} by the natural projection

π_{2,2ω_{n}^{+}} : B̃(n − 1) ⟶ \overline{B}(n − 1)

so that π_{2,2ω_{n}^{+}} ∘π_{2ω_{n}^{+}} = π_{2}. The kernel of π_{2,2ω_{n}^{+}} is equal to {1, 1+2b, −1, −1+ 2b} and has order 4 and therefore, the group index above is either 1, 2 or 4. But −1 is in the kernel of π_{2} and not in the kernel of π_{2ω_{n}^{+}} and hence the index is at most 2. This index is not equal to 1 because 2 − b is not an image of a unit.

Therefore, we turn to the proof of part 1: We define the ring X_{n} to be equal to \overline{B}(n)/((b + 1)\overline{B}(n)) and then we have, fixing the natural splitting X_{n} ⟶ \overline{B}(n) obtained by x ⟶ x · 1 and identifying the image with X_{n},

U(\overline{B}(n)) = U(X_{n}) + (1 + b) · X_{n}

by the nilpotence of (1 + b)X_{n}. For each x, y ∈ X_{n} we multiply:

(1 + (1 + b)x) · (1 + (1 + b)y) = 1 + (1 + b)(x + y).

We will prove that

Ỹ_{n} := \{1, ω_{n}^{+}(i) · (1 + ω_{n}^{+}(i))^{−1}|i = 1, ..., 2^{n−1} − 1\} ⊂ \overline{B}(n)

gives a ZZ/2ZZ basis for X_{n}. This fact will then prove Proposition 1 since {1+(b+1)x|x ∈ Ỹ_{n}} consists of images of global units and since by Gustafson and Roggenkamp (cf. [7]) U(X_{n}) is generated by global units.

Set X := X_{n}/X_{n−1} and let Y_{n} be the subvector space generated by Ỹ_{n}. Then let Y be the image of Y_{n} in X. X_{n} is a subring of the group ring of the cyclic group of order 2^{n} by the identification ζ_{2^{n}} ⟶ a with a being a fixed generating element of C_{2^{n}}. Therefore, X_{n} is a module over the automorphism group of C_{2^{n}}. This automorphism group is isomorphic to C_{2} × C_{2^{n−2}}. However, the first factor, it sends a to a^{−1}, acts trivially on X_{n} and therefore, X_{n} is an IF_{2}C_{2^{n−2}}-module. The projection X_{n} ⟶ X is split as module homomorphism and we want to show that Y_{n} = X_{n}. For this purpose we show that Y = X. Since only the elements in

{ω_{n}^{+}(i) · (1 + ω_{n}^{+}(i))^{−1}|i = 1, ..., 2^{n−1} − 1, i odd }

are nonzero in Y, we have by induction that the set Ỹ_{n} is linearly independent and therefore generates X_{n} as a vector space.

If Y < X we have that Y ≤ rad_{IF_{2}C_{2^{n−2}}} X is annihilated by ∑_{σ∈C_{2^{n−2}}} σ = Δ. But

ω_{n}^{+} · (1 + ω_{n}^{+})^{−1} = (ω_{n}^{+}) + (ω_{n}^{+})^{3} + (ω_{n}^{+})^{5} + ...

in X. The powers greater than 1 come in even number of summands in the natural basis of X and vanish if one multiplies by Δ. Therefore, multiplication by Δ yields Δ · ω_{n}^{+} in the result, which is obviously not equal to zero. Hence, Y = X and by induction Y_{n} = X_{n}. This proves Proposition 1.

If we now use the Mayer-Vietoris-sequences for the above pullback diagrams the following estimates can be given for the class groups of B(n) and C(n).

Cl(C(n)) = Cl(B(n − 1)) ⊕ Cl(ZZ[ω_{n}^{+}]C_{2})

and

|Cl(B(n))| ∈ {λ · |Cl(B(n − 1)) ⊕ Cl(ZZ[ω_{n}^{+}]C_{2})| : λ ∈ {1, 2}}.

We observe that C(n) and also B(n) are subrings of ZZ(C_{2^{n}} × C_{2}) since both rings are commutative. Therefore, f := \frac{1}{2}(1 + b) induces pullback diagrams as introduced above. For abbreviation we write:

Ĉ(n) := C(n) · f ≃ C(n) · (1 − f), B̂(n) := B(n) · f ≃ B(n) · (1 − f),

ˆ\overline{C}(n) := C(n) · f/((C(n) · f) ∩ C(n)) = Ĉ(n)/2Ĉ(n),
ˆ\overline{B}(n) := B(n) · f/((B(n) · f) ∩ B(n)) = B̂(n)/2B̂(n).

We also, as usual, identify B̂(n) and Ĉ(n) with suitably chosen subrings of ZZC_{2^{n}}.

We apply a result of M. J. Taylor [14, 7.(2.10)] and Mayer-Vietoris-sequences of the pullback diagrams for B̂(n) and Ĉ(n) induced by f to obtain Cl(B̂(n)) = Cl(Ĉ(n)). Since C(n) resp. Ĉ(n) are overorders of B(n) rsp. B̂(n) in QC(n) rsp. QĈ(n), we have induced epimorphisms of class groups of the smaller order to the larger one. Therefore, we have the following commutative (since Mayer Vietoris sequences are functorial) diagram with exact rows:

U(Ĉ(n)) ⟶^{π_{C}} U( ˆ\overline{C}(n)) ⟶ Cl(C(n)) ⟶ (Cl(Ĉ(n))^{2} ⟶ 0
↑ ↑ α ∥
U(B̂(n)) ⟶^{π_{B}} U( ˆ\overline{B}(n)) ⟶ Cl(B(n)) ⟶ (Cl(B̂(n))^{2} ⟶ 0.

By the above, α is surjective. We are interested in the kernel. This is isomorphic to the kernel of

U( ˆ\overline{B}(n))/π_{B}(U(B̂(n)) ⟶ U( ˆ\overline{C}(n))/π_{C}(U(Ĉ(n)).

We observe that

U( ˆ\overline{B}(n)) = 1 + 2ZZ/4ZZa^{2^{n−1}} + ∑_{i} ZZ/2ZZ(a^{i} + a^{−i})

and that

U( ˆ\overline{C}(n)) = 1 + ∑_{i} ZZ/2ZZ(a^{i} + a^{−i}) ∪ a^{2^{n−1}} + ∑_{i} ZZ/2ZZ(a^{i} + a^{−i}).

Since a^{2^{n−1}} is a unit in Ĉ(n) the kernel of the epimorphism

β : U( ˆ\overline{B}(n)) ⟶ ⟶ U( ˆ\overline{C}(n))/ < a^{2^{n−1}} >

is cyclic of order 2. The sum of the coefficient of 1 and that of a^{2^{n−1}} is odd since a unit has to have odd augmentation. Therefore, we immediately obtain that U(Ĉ(n))/ < a^{2^{n−1}} >= U(B̂(n)). We now show Lemma 3 No unit of B̂(n) maps onto the generating element of ker(β).

Proof: We even show that there is no unit of B̂(n) that maps modulo 2^{l}B̂(n) onto 1+2^{l}a^{2^{n−1}} for l ≥ 1. The crucial point is that the ring automorphism of ZZC_{2^{n}} induced by σ : a ⟶ −a induces a ring automorphism of B̂(n). Note that σ is the Galois automorphism from Lemma 2.

We prove the above by induction. For n = 2 the statement is trivial. Let x = 1 + 2^{l}a^{2^{n−1}} + 2^{l}y ∈ U(B̂(n)) with y ∈ B̂(n). Since y + σ(y) ∈ 2B̂(n), x · σ(x) = 1 + 2^{l+1}a^{2^{n−1}} + 2^{l+1}z for a z ∈ B̂(n). But

H^{0}(< σ >, B̂(n)) = B̂(n − 1) ⊂ B̂(n)

and therefore, since σ(x · σ(x)) = x · σ(x) the element x · σ(x) is a preimage of 1 + a^{2^{n−1}} modulo 2^{l+1}. By induction x · σ(x) cannot be a unit, however, it is a unit by construction. We reached a contradiction and the lemma is proven.

Now we see that

|Cl(B(n))| = 2|Cl(C(n))|

and therefore, |Cl(C(n))| = 2^{n−1} ∏_{k≤n} |Cl(ZZ[ω_{k}^{+}] < b >)|. Since

ZZ[ω_{k}^{+}] < b > ·\frac{(1 + b)}{2} ≃ ZZ[ω_{k}^{+}] ≃ ZZ[ω_{k}^{+}] < b > ·\frac{(1 − b)}{2}

a result of Taylor ([14, 7.(2.10)]) implies that

|Cl(ZZ[ω_{k}^{+}] < b >)| = |(Cl(ZZ[ω_{k}^{+}]))|^{2} =: (h_{2^{k}}^{+})^{2}.

Therefore,

|Cl(C(n))| = 2^{n−1} · ∏_{k≤n}(h_{2^{k}}^{+})^{2}.

The main achievement of the above section is the determination of the size of the class group of C(n) as well as the knowledge of 2−b as generator of the non trivial coset in the Mayer-Vietoris sequence corresponding to B(n).

3. Embedding involutions into the unit group

We are interested in K_{n} := ker(Cl(C(n)) ⟶ Cl(ZZD_{2^{n}})), the kernel of the induction homomorphism.

By Fröhlich, Keating and Wilson (cf. [6]) we have the following isomorphism:

Cl(ZZD_{2^{n}}) = Cl(ZZD_{2^{n−1}}) ⊕ Cl(ZZ[ω_{n}^{+}])_{2×2}.

Fröhlich proved in [4] that Cl(ZZD_{2}) = 1. Induction behaves well with taking summands and so the kernel K_{n} is equal to a direct sum K′_{n} ⊕ K′′_{n} according to the summation above.

For the moment we define R_{n} := ZZ[ω_{n}^{+}]. The following diagram has exact rows and columns:

0  0  0
↓  ↓  ↓
0 → C_{2} →  K′_{n}  →  K′_{n−1} ⊕ K′′_{n−1}  → 0
↓  ↓  ↓
0 → C_{2} → Cl(B(n − 1)) → Cl(B(n − 2)) ⊕ Cl(R_{n−1} < b >) → 0
↓ ↓ ↓
0 → 0 → Cl(ZZD_{2^{n−1}}) → Cl(ZZD_{2^{n−2}}) ⊕ Cl((R_{n−1})_{2×2}) → 0

The surjectivity at the top row follows by the serpent lemma. Therefore, we essentially have to calculate K′′_{n} to get the order of K_{n}. For this we use the representation given by K. W. Roggenkamp in [10]:

a ⟶ ( \begin{matrix}−1 & −1 \\ 2 + ω_{n}^{+} & 1 + ω_{n}^{+} \end{matrix} ) , b ⟶ ( \begin{matrix}−1 & 0 \\ 2 + ω_{n}^{+} & 1 \end{matrix} ) .

By [2, Exercise 53.1] we see that an element of Cl(ZZ[ω_{n}^{+}] < b >) is mapped to an element of Cl((ZZ[ω_{n}^{+}])_{2×2}) according to the following rule: If we have a pair of ideals (A, B) in Cl((ZZ[ω_{n}^{+}])^{2}), it is represented by the idèle ∏_{℘∈Spec(ZZ[ω_{n}^{+}])}(2α_{℘}, 2β_{℘}), say, which is mapped to the idèle

(γ) := ∏_{℘} ( \begin{matrix}2α_{℘} & 0 \\ (2 + ω_{n}^{+})(β_{℘} − α_{℘}) & 2β_{℘} \end{matrix} )

of (ZZ[ω_{n}^{+}])_{2×2}. Since we may calculate the norm locally (cf. [9, (24.2)] in connection with [9, (8.5)]) and intersect the local results afterwards we see that the reduced norm of the induced ideal is isomorphic to AB. Therefore, we have that K′′_{n} ≃ Cl(ZZ[ω_{n}^{+}]) via the codiagonal in the direct sum of two copies of the class group of Cl(ZZ[ω_{n}^{+}]). Using now a theorem of H. Weber ([15, p. 244, Satz C]) the 2′-Hall subgroup of K_{n} is isomorphic to ∏_{k=1}^{n} Cl(ZZ[ω_{k}^{+}]).

On the whole we have

|K_{n}| = 2^{n−1} · ∏_{k=1}^{n} h_{2^{k}}^{+}.

This formula presents the main achievement of the above section.

4. Connections to group bases

As we already did in the introduction we use for a suborder Γ of ZZD_{2^{n}} the notation

Cl_{ZZD_{2^{n}}}(Γ) := ker(Cl(Γ) ⟶ Cl(ZZD_{2^{n}}))

Remind that in the previous sections we used the notation

Cl_{ZZD_{2^{n}}}(C_{ZZD_{2^{n}}}(b)) = Cl_{ZZD_{2^{n}}}(C(n)) = K_{n}.

We calculate the kernel and the image of the natural homomorphism

Cl_{ZZD_{2^{n}}}(Z(ZZD_{2^{n}})) ⟶ Cl_{ZZD_{2^{n}}}(C_{ZZD_{2^{n}}}(b)).

By S. Endo, T. Miyata and K. Sekiguchi ([3]) the first group is isomorphic to the outer central automorphism group and is furthermore the 2-Sylow subgroup of Cl(Z(ZZD_{2^{n}})) by Weber’s theorem. Hence, we have to deal with the 2-Sylow subgroup of Cl(Z(ZZD_{2^{n}})).

Let e_{k}^{n} be the central idempotent that maps ZZD_{2^{n}} onto ZZD_{2^{k}} induced by the epimorphism D_{2^{n}} ⟶ D_{2^{k}}. Then the idèle

(α_{k}(n)) := (((1 − e_{k}^{n}) + e_{k}^{n} · (1 + 2b)) × ∏_{p≠2} 1)

defines an element of Cl(C(n)). For k = 1 and k = 0 the idèle α_{k}(n) is central and hence defines an ideal of C(n) isomorphic to one induced by an ideal of Z(ZZD_{2^{n}}).

Lemma 4 Let J and J′ be ideals of B(n) with e_{k}^{n} · J ≃ e_{k}^{n} · J′ in B(k). Then there is an element u ∈< (α_{n−1}(n)), ..., (α_{k}(n)) > with corresponding ideal U such that J · U ≃ J′ as B(n)-modules.

Proof: This is trivial for k = n−1 and if it is true for all k ∈ {n−1, ..., k_{0}+1}, then by [2, Exercise 53.1] and Proposition 1.2. it is also true for k = k_{0} and the lemma is proven.

Lemma 5 The 2-Sylow subgroup of Cl(C_{ZZD_{2^{n}}}(b)) is cyclic, it is generated by (α_{0}(n)) and it lies in the image of Cl(Z(ZZD_{2^{n}})) under the induction map.

Proof: We show that (α_{k}(n)) has order 2^{n−k} in Cl(B(n)). One calculates easily, using [2, Exercise 53.1], that e_{k+2}^{n} · (α_{k}(n))^{2} is equivalent to e_{k+2}^{n}(α_{k+1}(n)) in Cl(B(k + 2)) and therefore has order 2^{n−k−1} by induction. The remark following the definition of (α_{k}(n)) completes the proof of the lemma.

We reached now the goal of the present section, namely to determine the structure of the class group of C_{ZZD_{2^{n}}}(b), the kernel of the homomorphism of that class group to the class group of ZZD_{2^{n}} as well as the image of the natural homomorphism of the class group of the centre of the integral group ring in the class group of the centralizer of b in the integral group ring.

If we use that by [16] every involution in V (ZZG) is 2-adically conjugate to either a^{2^{n−1}}, b or to ab and those can be distingished in the commutative quotient ZZD_{2} we may summarize the above results in

Theorem 1 Let D_{2^{n}} =< a, b|a^{2^{n}}, b^{2}, baba > be the dihedral group of order 2^{n+1}. Let h_{2^{k}}^{+} be the class number of the maximal real subfield of the field of 2^{k}-th roots of unity. Then

1. Cl(C_{ZZD_{2^{n}}}(b)) ≃ C_{2^{n−1}} × ∏_{k=1}^{n}(Cl(ZZ[ω_{k}^{+}]))^{2},

2. Cl_{ZZD_{2^{n}}}(C_{ZZD_{2^{n}}}(b)) ≃ C_{2^{n−1}} × ∏_{k=1}^{n} Cl(ZZ[ω_{k}^{+}]).

3. In V (ZZD_{2^{n}}) there are 2^{n−1} · ∏_{k=1}^{n} h_{2^{k}} conjugacy classes of involutions that are rationally conjugate to b.

4. In V (ZZD_{2^{n}}) there are 2^{n−1} conjugacy classes of involutions that are rationally conjugate to b and that are part of group bases.

5. Every involution in V (ZZD_{2^{n}}) is part of a group basis if and only if h_{2^{k}}^{+} = 1 for every k ≤ n.

Remark 1 One of the crucial points in [3] is to prove that there are elements of order 2^{n−1} in Cl(Z(ZZD_{2^{n}})), an immediate consequence of our theorem.

5. Application to the automorphism group

We may use our result to give an explicit description of the outer automorphism group as conjugations with given group ring elements.

We first note that if a unit u of QD_{2^{n}} normalizes a maximal order Λ of QD_{2^{n}} containing ZZD_{2^{n}} and if furthermore u normalizes ẐZ_{2}D_{2^{n}}, then u normalizes ZZD_{2^{n}}. This follows by the fact that we have to show it only locally and there it is obvious.

1 + b · (a + a^{−1}) is a unit in ẐZ_{2}D_{2^{n}}. It normalizes the maximal order

Γ_{n} := ZZ × ZZ × ZZ × ZZ × ∏_{k=2}^{n}(ZZ[ω_{k}^{+}])_{2×2}.

This is seen most easily if one looks at its representation in Γ_{n} as given above. A. Whitcomb showed in [17] that conjugation by the rational unit u_{2} = 1 + a + ba yields a non inner automorphism of ZZD_{4}.

This result combined with our main theorem implies

Theorem 2 The automorphism group of ZZD_{2^{n}} is generated by the inner automorphisms, conjugation by u_{n} = 1 + a + ba and conjugation by v_{n} = 1 + b · (a + a^{−1}).

The proof of Theorem 2 will fill the rest of section 5. We assume that conjugation with v_{n} is inner. Then there is a unit v′ of ZZD_{2^{n}} such that v^{′−1} · v_{n} is contained in the center of ZZD_{2^{n}} and is also a unit in QD_{2^{n}}. First we examine the case n = 3. Then in Γ_{3} we have

v_{3} = 3 ⊕ −1 ⊕ −1 ⊕ 3 ⊕ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) ⊕ ( \begin{matrix}1 − \sqrt{2} & 0 \\ 2 + \sqrt{2} & 1 + \sqrt{2} \end{matrix} ) .

v_{2} is central and therefore since the center of ZZD_{4} contains only the units ±1 and ±a^{2} (it is a subring of ZZ(C_{4} × C_{2})) we immediately see that no central element of ZZD_{8} carries v_{3} to a unit v′.

Similar arguments can also be applied to conclude that conjugation by u_{2} is not inner.

Since Outcent(ZZD_{2^{n}}) ≃ Cl_{ZZD_{2^{n}}}(Z(ZZD_{2^{n}})) we see by an application of Mayer-Vietoris sequences and the fact that induction to overorders in the same algebra yields epimorphisms of the class groups that the natural map ZZD_{2^{n}} ⟶ ⟶ ZZD_{2^{n−1}} induces an epimorphism

Outcent(ZZD_{2^{n}}) ⟶ ⟶ Outcent(ZZD_{2^{n−1}}).

Now S. Endo, T. Miyata and K. Sekiguchi show ([3]) that

Outcent(ZZD_{2^{n}}) ≃ C_{2^{n−1}} × C_{2^{n−2}} =:< α_{n} > × < β_{n} > .

Theorem 1 tells us that we may assume that β_{n} centralizes b for all n. This is the point where we use the theorem! Under this assumption (α_{n})^{j} does not map b to a conjugate for all j =

1, ..., 2^{n−1} − 1. Let β′_{n} be a preimage of β_{n−1} such that β′_{n} centralizes b. Since β_{n−1} has no square root, β′_{n} has also none. Therefore, β′_{n} is an odd power of β_{n}.

Let α′_{n} be a preimage of α_{n−1}. Since β_{n} is in the kernel of the epimorphism Outcent(ZZD_{2^{n}}) ⟶ Outcent(ZZD_{4}), α_{n−1}, and hence also α′_{n} maps to α_{2}. Therefore, α′_{n} has order 2^{n−1}. This proves Theorem 2 since we may choose α_{n}(x) = u_{n}xu_{n}^{−1} and β_{n}(x) = v_{n}xv_{n}^{−1} for all x ∈ ZZD_{2^{n}}.

6. The semidihedral groups

Let S_{n} be the semidihedral group of order 2^{n+1} with the presentation

S_{n} =< a, b|a^{2^{n}} = b^{2} = 1, bab = a^{2^{n−1}−1} > .

Factoring its center S_{n} maps onto D_{2^{n−1}} and if one chooses e := (1+a^{2^{n−1}})/2 then ZZS_{n} · (1 − e) embeds into the 2 × 2 matrix ring over R_{n}, the ring of algebraic integers in the fixed field of Q[ζ_{n}] under the Galois automorphism sending ζ to −ζ^{−1}. We observe that C_{ZZS_{n}}(b)·e = B(n−1) and that C_{ZZS_{n}}(b)· (1 − e) = R_{n} < b >. The quotient (C_{ZZS_{n}}(b) · e)/(C_{ZZS_{n}}(b) · e ∩ C_{ZZS_{n}}(b)) is isomorphic to R_{n} < b > /2R_{n} < b >. Since 2ZZ is totally ramified in ZZ[ζ_{n}] and hence also in R_{n} and since every unit besides −1 in R_{n} is a real unit, we can prove that

1 ⟶ C_{2} ⟶ Cl(R_{n} < b >) ⟶ (Cl(R_{n}))^{2} ⟶ 1

is exact. Let \overline{A} be the nontrivial ideal in the kernel. Then we shall show that there is an ideal of C_{ZZS_{n}}(b) that, firstly, maps multiplied by e to an ideal isomorphic to B(n − 1), secondly, maps multiplied by 1 − e to an ideal isomorphic to \overline{A}. This yields an ideal that, thirdly, maps, induced to an element of Cl(ZZS_{n}), to the identity in Cl(ZZS_{n}) and, fourth, that is not induced by an ideal of Z(ZZS_{n}). Let A be an arbitrary ideal satisfying the first two conditions. The techniques for the dihedral groups may be applied to prove that \overline{A} maps to a principal ideal of the 2 × 2 matrix ring over R_{n}. Now \overline{A} is not induced by an ideal of Z(ZZS_{n}) · (1 − e) as is easily seen. Therefore, by a theorem of Endo as quoted in [14, 3. 2.5] A maps either to 1 or to a Swan module which itself is induced by an ideal of Z(ZZS_{n}). Elementary diagram chasing gives us the ideal satisfying the last two conditions and hence proving

Theorem 3 In the group of units of augmentation 1 of the integral group ring of the semidihedral group S_{n} of order 2^{n+1} greater than 16 there is an involution that is not contained in any group basis of ZZS_{n}.

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