ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC SPACES
OF DIHEDRAL GROUPS
arXiv:1205.3207v1 [math.GR] 14 May 2012
K. K. A. CUNNINGHAM, T. J. EDGAR, A. G. HELMINCK, B. F. JONES, H. OH, R. SCHWELL,
J. F. VASQUEZ
Abstract. We initiate the study of analogues of symmetric spaces for the family of finite
dihedral groups. In particular, we investigate the structure of the automorphism group,
characterize the involutions of the automorphism group, and determine the fixed-group and
symmetric space of each automorphism.
Introduction
Let G be a group and θ ∈ Aut(G) such that θm = id. We then define the following two
sets:
H = Gθ = {g ∈ G | θ(g) = g}
Q = {g ∈ G | g = xθ(x)−1 for some x ∈ G}.
The set H is the fixed-point subgroup of θ and Q is known as a Generalized Symmetric
Space. If θ is an involution and G is a real reductive Lie group, then the set Q is a reductive
symmetric space. If G is a reductive algebraic group defined over an algebraically closed field
k, then Q is also known as a symmetric variety and if G is defined over a non-algebraically
closed field k, then the set Qk := {xθ(x)−1 | x ∈ Gk } is called a symmetric k-variety. Here
Gk denotes the set of k-rational points of G. Reductive symmetric spaces and symmetric kvarieties are well known for their role in many areas of mathematics. They are probably best
known for their fundamental role in representation theory [5]. The generalized symmetric
spaces as defined above are of importance in a number of areas as well, including group
theory, number theory, and representation theory.
One of the first questions that arises in the study of these generalized symmetric spaces is
the classification of the automorphisms up to isomorphy, where isomorphy is given by either
conjugation by inner automorphisms, outer automorphisms, or both depending on which
makes the most sense.
To analyze the structure of these generalized symmetric spaces one looks at orbits of both
the group itself and the fixed point group. Both G and H act on Q by θ-twisted conjugation
which we denote by ∗. For g ∈ G and q ∈ Q we have
g ∗ q = gqθ(g)−1 .
Since H is the set of fixed points, this action is simply conjugation when restricted to H.
Also, if θ is an involution, then Q ∼
= G/H under the map τ : G → Q given by τ (g) = gθ(g)−1 .
We are interested in classifying G and H orbits in Q and describing how the G-orbits
decompose into H-orbits. In particular, if θ is an involution it is known that H\Q ∼
= H\G/H.
These orbits and double cosets play an important role in representation theory.
The authors thank the American Institute of Mathematics in Palo Alto, CA for their generous support.
1
This paper focuses on understanding the structures described above when G is the dihedral
group of order 2n, that is G = Dn . In particular, we determine the sets H and Q explicitly,
provide a procedure for enumerating the involutions, and give a closed formula for counting
the equivalence classes of involutions of Dn .
The paper is organized as follows. In Section 1, we introduce the necessary preliminaries
including our choices for notation and the well known description of the automorphism group
of Dn as well as some relevant examples. In Section 2, we investigate and describe all the
automorphisms of Dn of a fixed order. We recall the notion of equivalence of automorphisms
and give simple conditions for two automorphisms of Dn to be equivalent. Furthermore, we
provide a formula for computing the total number of equivalence classes of automorphisms
of a fixed order. In Section 3, we give full descriptions of the sets H and Q as well as the
orbits of Q under the action by H and by G. In Section 4, we find stronger results for the
involutions in the automorphism group. Our main technical result is to partition the set of
involutions in such a way that there are at most two distinct equivalence classes of involutions
in each piece of the partition (cf. Theorem 4.5). We also show that in the involution case,
Q is always a subgroup of G, and we describe the subgroup structure of Q. Furthermore,
we introduce the set of twisted involutions, R, and give a characterization of this set and
its relation to the generalized symmetric space. Using our results on involutions of Dn ,
we provide a counterexample to a standard theorem on equivalence of involutions which
holds for algebraic groups. Finally, in Section 5, we complete the discussion by applying our
methods to the infinite dihedral group in order to understand the automorphisms of finite
order, as well as H and Q in that context.
1. Preliminaries
1.1. The Dihedral Group and its Automorphism Group. Throughout the paper we
denote by Dn the group of symmetries of the regular n-gon. More specifically, Dn is a finitely
generated group given by the presentation
Dn = hr, s | rn = s2 = 1, sr = r−1 si.
We use this presentation, instead of the presentation as a Coxeter group (see [2, Example
1.2.7], [6, §1.1, §4.2], [3, §1.2]), because it is convenient for describing the automorphism
group of Dn . From the presentation it is clear that
Dn = {1, r, r2 , . . . , rn−1 , s, rs, r2 s, . . . , rn−1 s}
and we say that an element of Dn is presented in normal form if it is written as rk sm for
some integers 0 ≤ k < n and s ∈ {0, 1}.
Throughout the paper, Zn denotes the additive group of integers modulo n and Un denotes
the multiplicative group of units of Zn . Recall that a ∈ Un if and only if (ȧ, n) = 1 for any
representative integer ȧ of a ∈ Zn .
We also find it useful to have terminology for the k-th roots of unity in Zn (equivalently
in Un ), which can be described by
Rkn := {a ∈ Un | ak = 1}.
The automorphism group of Dn is well known (see [9, Theorem A]) and is described in
the following lemma.
2
Lemma 1.1. The automorphism group of Dn is isomorphic to the group of affine linear
transformations of Zn :
Aut(Dn ) ∼
= Aff(Zn ) = {ax + b : Zn → Zn | a ∈ Un , b ∈ Zn }
and the action of ax + b on elements of Dn in normal form is given by:
(ax + b).(rk sm ) = rak+bm sm .
Proof. Observe that θ(r) must have order n since r does. The elements of Dn of order n are
precisely the {ra | a ∈ Un }. So let θ(r) = ra for some a relatively prime to n. Similarly,
θ(s) = rb s for some 0 ≤ b < n since s has order 2 and the only elements of Dn of order 2 are
of the form rb s. Putting these together, we see that if θ ∈ Aut(Dn ), then for any rk sm ∈ Dn
we must have θ(rk sm ) = rak+bm sm . There is now a clear map from Aut(Dn ) → Aff(Zn ) and
it is easily checked to be an isomorphism.
Throughout the paper we abuse notation and write θ = ax + b for elements θ ∈ Aut(Dn )
according to Lemma 1.1.
Example 1.2. Consider θ = conj(g), conjugation by an element g ∈ Dn . If θ = ax + b
from Lemma 1.1, and if g = rk , then it is easy to see that a = 1 and b = 2k (in Un and
Zn respectively). Also, if g = rk s, then we have a = n − 1 and b = 2k. In particular, if
θ = ax + b is an inner automorphism, then (abusing notation) we have a ≡ ±1 (mod n).
Also, we see that when n is even, there are n distinct inner automorphisms (corresponding
to a ≡ ±1 (mod n) and b ∈ h2i ≤ Zn , and when n is odd, there are 2n distinct inner
automorphisms corresponding to a ≡ ±1 (mod n) and b ∈ h2i = Zn . Finally, note that the
identity automorphism of Dn corresponds to x ∈ Aff(Zn ).
Example 1.3. It is well known from the theory of Coxeter groups that when n is even, there
is an outer automorphism of Dn corresponding to the interchanging of the two conjugacy
classes of reflections, also known as the diagram automorphism (cf. [3, §4.2]). This automorphism is described by θ = ax + b where a = n − 1 and b = n − 1. The previous example
shows that this automorphism is inner when n is odd, and outer when n is even.
2. Automorphisms of Dn
In this section we describe the action of the automorphism group on Dn , we characterize
the automorphisms of fixed order, and we discuss the notion of equivalent automorphisms.
For any c ∈ Zn , we define ZDiv(c) = {y ∈ Zn | cy ≡ 0 mod n}. It is trivial to check
that ZDiv(c) is a subgroup of Zn . In particular, ZDiv(c) = ker πc where πc : Zn → Zn
n
given by πc (x) = cx. Note that im(πc ) = hci and so | im(πc )| = gcd(c,n)
. Since im(πc ) ∼
=
∼
Zn / ker(πc ) = Zn /ZDiv(c), it follows that |ZDiv(c)| = gcd(c, n). We use this notation to
describe the automorphisms of finite order dividing k.
Proposition 2.1. Let k ≥ 1 be an integer and θ ∈ Aut(Dn ) with θ = ax + b. Then θk = id
if and only if a ∈ Rkn and b ∈ ZDiv(ak−1 + ak−2 + · · · + a + 1).
Proof. Straightforward computation in Aut(Dn ) gives us that
θk = ak x + (ak−1 + ak−2 + · · · + a + 1)b
3
Since we have identified the identity automorphism with x ∈ Aff(Zn ) the following two
equations hold:
(2.1)
ak ≡ 1
(mod n)
(ak−1 + ak−2 + · · · + 1)b ≡ 0
(mod n).
In the notation described above, these are evidently equivalent to a ∈ Rkn and
b ∈ ZDiv(ak−1 + ak−2 + · · · + a + 1).
In what follows, we let Autk (Dn ) = {θ ∈ Aut(Dn ) | θk = id} ⊆ Aut(Dn ).
Proposition 2.2. For any n and k ≥ 1, we have
X
| Autk (Dn )| =
gcd(ak−1 + ak−2 + · · · + a + 1, n).
a∈Rkn
Proof. This follows from Proposition 2.1: for any a ∈ Rkn there are
|ZDiv(ak−1 + ak−2 + · · · + 1)| = gcd(ak−1 + ak−2 + · · · + 1, n)
elements b such that (ak−1 + ak−2 + · · · + 1)b ≡ 0 (mod n), and every automorphism θ ∈
Autk (Dn ) must be of this form.
Definition 2.3. Let θ1 , θ2 ∈ Aut(Dn ). We say that θ1 is equivalent to θ2 and write θ1 ∼ θ2
if and only if they are conjugate to each other, i.e. if there is σ ∈ Aut(Dn ) with σθ1 σ −1 = θ2 .
For any θ ∈ Aut(Dn ) we let θ = {σ ∈ Aut(Dn ) | θ ∼ σ} be the equivalence class of θ.
Remark 2.4. This definition of equivalence is broad (since we allow conjugation by any
automorphism, not just the inner ones), but still useful. In particular, it simplifies the
statement of the following proposition and it allows us not to worry about the parity of n in
several places.
Proposition 2.5. Let θ1 = ax + b ∈ Aut(Dn ) and θ2 = cx + d ∈ Aut(Dn ). Then θ1 ∼ θ2 if
and only if a = c and f b − d ∈ ha − 1i ≤ Zn for some f ∈ Un .
Proof. Suppose that σ = f x + g ∈ Aut(Dn ) (so that f ∈ Un and g ∈ Zn ). It is easily
checked that σ −1 = f −1 x − f −1 g where f −1 ∈ Zn since f ∈ Un . Then σθ1 σ −1 = τ , where
τ = f (a(f −1 x − f −1 g) + b) + g = ax + f b − g(a − 1). Thus, we can conclude that σθ1 σ −1 = θ2
if and only if a = c and f b − g(a − 1) = d. The second equation can be written as
f b − d = g(a − 1), and thus f b − d ∈ ha − 1i.
The previous proposition allows us to describe the conjugacy classes of automorphisms
with order dividing k.
Proposition 2.6. Suppose n is fixed and let k ≥ 1.
(1) For any a ∈ Rkn , ha − 1i ≤ ZDiv(ak−1 + ak−2 + · · · + a + 1).
(2) For any a ∈ Rkn , Un acts on the cosets ZDiv(ak−1 + ak−2 + · · · + a + 1)/ha − 1i.
(3) The set Autk (Dn ) is partitioned into equivalence classes indexed by pairs (a, B) where
a ∈ Rkn and B is an orbit of Un on ZDiv(ak−1 + ak−2 + · · · + a + 1)/ha − 1i.
Proof. Part (1) is trivial since (a − 1)(ak−1 + ak−2 + · · · + a + 1) = ak − 1 ≡ 0. For part (2), we
simply recognize that Un = Aut(Zn ) acts on Zn by multiplication and since every subgroup
is cyclic, Un must stabilize the subgroups of Zn . Finally, part (3) is simply Proposition 2.5
in terms of the Un -action on ZDiv(ak−1 + ak−2 + · · · + a + 1)/ha − 1i.
4
Proposition 2.5 and Proposition 2.6 help us determine the total number of equivalence
classes of finite order automorphisms. For fixed a ∈ Rkn , we want to compute
Na := |{θ | θ = ax + b ∈ Autk (Dn )}|.
(2.2)
According to Proposition 2.6, given a ∈ Rkn , computing Na amounts to counting the
number of orbits of the Un -action on ZDiv(ak−1 + ak−2 + · · · + a + 1)/ha − 1i.
Theorem 2.7. Let n be fixed and let a ∈ Rkn . Then, the number of orbits of Un on
ZDiv(ak−1 + ak−2 + · · · + a + 1)/ha − 1i is equal to the number of divisors of
gcd(a − 1, n) gcd(ak−1 + ak−2 + · · · + a + 1, n)
.
n
Proof. The Un -orbits on Zn are indexed by the subgroups of Zn . Thus, the Un -orbits
on ZDiv(ak−1 + ak−2 + · · · + a + 1)/ha − 1i are indexed by subgroups L ≤ Zn such that
ha − 1i ≤ L ≤ ZDiv(ak−1 + ak−2 + · · · + a + 1). It is well known that the subgroup lattice of Zn is isomorphic to the divisor lattice of n. Under the previous lattice isomorphism, the subgroup ha − 1i corresponds to the divisor gcd(a − 1, n) and the subgroup
ZDiv(ak−1 + ak−2 + · · · + a + 1) corresponds to the divisor
n
,
k−1
gcd(a
+ · · · + a + 1, n)
and the subgroups between these groups correspond to the divisors of n between gcd(a−1, n)
n
and gcd(ak−1 +···+a+a,n)
in the divisor lattice. Finally, it is known that this sub-lattice of the
divisors of n is isomorphic to the divisor lattice of
gcd(a − 1, n)
n
gcd(ak−1 +···+a+1,n)
=
gcd(a − 1, n) gcd(ak−1 + ak−2 + · · · + a + 1, n)
.
n
Given a specific n, Theorem 2.7 and Proposition 2.6 allow us to compute all the equivalence
classes of automorphisms of order k using any standard computer algebra package. In section
4, we investigate Na when a ∈ R2n (i.e. θ is an involution).
3. Fixed Groups and Symmetric Spaces of Automorphisms
Recall from the introduction that we are interested in two different subsets of a group G.
Namely, given an automorphism, θ, we want to compute
Hθ = Gθ = {x ∈ G | θ(x) = x} and
Qθ = {g ∈ G | g = xθ(x)−1 for some x ∈ G}.
When θ is understood to be fixed, we drop the subscript from our notation. The following
theorem characterizes these spaces in the case of dihedral groups.
Theorem 3.1. Let G = Dn and θ = ax + b ∈ Aut(Dn ) be of finite order, and let H and Q
be as defined above. Then
H = {rk | k(a − 1) ≡ 0 (mod n)} ∪ {rk s | k(a − 1) ≡ −b (mod n)} and
Q = {rk | k ∈ ha − 1i ∪ (−b + ha − 1i)}.
5
Proof. Recall that if θ = ax+b ∈ Aut(Dn ), then the formula θ(rk sm ) = rak+bm sm provides the
full description of the action of θ. All of the results of the theorem arise from the definitions
of H and Q using the action described above by θ. We demonstrate the computation for H
in what follows, and Q is similar. Suppose θ(rk ) = rk , then rak = rk and so ak − k ≡ 0 (mod
n), i.e. k(a − 1) ≡ 0 (mod n). On the other hand, if θ(rk s) = rk s, then rak+b s = rk s and so
ak − k + b ≡ 0 (mod n), i.e. k(a − 1) ≡ −b (mod n).
Using the descriptions of H and Q, we obtain further results as well.
Corollary 3.2. Let θ = ax + b ∈ Aut(Dn ) be of finite order. Then
(1) If b 6∈ ha − 1i, then H ∼
= ZDiv(a − 1) is cyclic.
(2) If b ∈ ha − 1i, then H ∼
= ZDiv(a − 1) o Z2 is dihedral.
Corollary 3.3. Let n, k, and θ = ax + b ∈ Autk (Dn ) be fixed. If b ∈ ha − 1i then Q is a
subgroup of Dn and Q ∼
= ha − 1i.
Proof. Since b ∈ ha − 1i, then it follows from Theorem 3.1 that Q = {rk | k ∈ ha − 1i}.
We will show in the next section that when θ is an involution, Q is always a subgroup of
Dn (see Corollary 4.8).
Corollary 3.4. Let θ = ax + b ∈ Aut(Dn ) be a fixed automorphism. If b 6∈ ha − 1i, then
HQ 6= Dn . If b ∈ ha − 1i, then the following are equivalent
(1) HQ = Dn
(2) H ∩ Q = {1} in Dn , and
n
(3) gcd(a − 1, n) is relatively prime to gcd(a−1,n)
.
Proof. From Corollary 3.2 and the fact that Q ⊆ hri, we see that if b 6∈ ha − 1i then
HQ ⊆ hri 6= Dn . If b ∈ ha − 1i then H contains a reflection and so HQ = Dn if and only
if r ∈ HQ. From Corollaries 3.2 and 3.3, we see that r ∈ HQ if and only if the subgroups
ZDiv(a − 1) and ha − 1i in Zn have relatively prime generators. This last condition is clearly
equivalent to both (2) and (3).
Example 3.5. Let G = D36 . We illustrate Theorem 3.1 and its corollaries for three different
automorphisms of D36 . First, consider the automorphism θ1 = 19x+18. Note that θ1 is an involution. Applying Theorem 3.1 yields that for θ1 , H = H1 = {1, r2 , r4 , . . . , r34 , r3 s, r5 s, . . . , r35 s}
and Q = Q1 = {1, r18 }, which gives that H1 ∩ Q1 = {1, r18 } = Q1 . Observe that the statement of Corollary 3.3 holds in this case: H1 Q1 6= D36 . This follows from the fact that
18 ∈ h18i, but gcd(18, 36) = 18 is not relatively prime to 36/ gcd(18, 36) = 2.
Now consider θ2 = 5x + 2 which is an automorphism of order 6. In this case, H2 =
{1, r9 , r18 , r27 } and Q2 = {1, r2 , r4 , . . . , r34 }. Thus, H2 ∩ Q2 = {1, r18 } and H2 Q2 = D36 .
However, in this case it is because 2 6∈ h4i.
Finally, consider the automorphism θ3 = 5x+4 of D36 . This is also an automorphism of order 6. In this case, H3 = {1, r9 , r18 , r27 , r8 s, r17 s, r26 s, r35 s} and Q3 = {1, r4 , r8 , r12 , . . . , r32 }.
Thus, H3 ∩ Q3 = {1} and this agrees with Corollary 3.3 since r9 r28 = r and so H3 Q3 = D36 .
From Theorem 3.1, we see that H is the disjoint union of {rk | k(a − 1) ≡ 0 (mod n)}
and {rk s | k(a − 1) ≡ −b (mod n)}. Notice that the second set may be empty if there is no
solution, i, to the equation i(a − 1) ≡ −b (mod n) for fixed a and b (i.e. if b 6∈ ha − 1i). In
fact, the two possibilities for the H-orbits on Q are determined by the existence of such a
solution.
6
Proposition 3.6. Let G = Dn and θ = ax + b ∈ Autk (Dn ) for some k.
(1) If b 6∈ ha − 1i, then the H-orbits on Q are:
H\Q = {rj } | j ∈ ha − 1i ∪ (−b + ha − 1i) .
(2) If b ∈ ha − 1i, then the H-orbits on Q are:
H\Q = {rj , r−j } | j ∈ ha − 1i .
In either situation G\Q = {Q}, i.e. there is a single G-orbit on Q.
Proof. Since H is fixed by θ, the action of H on Q is simply by conjugation. We note that
Q ⊆ hri ≤ Dn and so we only need to describe the action of H on a general element, rj .
Then, let ri ∈ {rk | k(a−1) ≡ 0 (mod n)} ⊆ H. We see ri rj r−i = rj so hri fixes Q pointwise.
Now suppose ri s ∈ {rk s | k(a − 1) ≡ −b (mod n)} ⊆ H. We have ri srj (ri s)−1 = r−j and
so ri s takes rj to r−j . The result now follows since we will have {rj , r−j } as an orbit if and
only if there is ri s ∈ H, which is true if and only if b ∈ ha − 1i.
Finally, we demonstrate that every element of Q is in the G-orbit of 1 ∈ Q. Notice
that every element of Q is either of the form q1 = rp(a−1) ∈ Q or q2 = r−b+p(a−1) ∈ Q.
We have r−p ∈ G and r−p 1θ(r−p )−1 = rp(a−1) = q1 . Additionally, we have r−p s ∈ G and
r−p s1θ(r−p s)−1 = r−p rap−b = r−b+p(a−1) = q2 . Therefore, for any q ∈ Q, we can find g ∈ G
such that g1θ(g)−1 = q and so G ∗ 1 = Q. Hence G\Q = {Q}.
Example 3.7. Revisiting θ1 , θ2 , and θ3 of Example 3.5, we apply Proposition 3.6 to obtain
that for θ1 , since 18 ∈ h18i and r18 = r−18 ∈ D36 ,
H1 \Q1 = {{1}, {r18 }};
for θ2 , since 2 6∈ h4i,
H2 \Q2 = {{1}, {r2 }, {r4 }, . . . , {r34 }};
lastly, for θ3 , since 4 ∈ h4i,
H3 \Q3 = {{1}, {r4 , r−4 }, {r8 , r−8 }, . . . , {r32 , r−32 }}.
All of the results in this section hold for automorphism of arbitrary finite order k, but
in the next section we demonstrate that we can say more when we restrict our attention to
k = 2, the involutions. In Section 5, we describe which of these results hold in the infinite
dihedral group.
4. Involutions in Aut(Dn )
In this section, we utilize the results from sections 2 and 3 and expand on them in the
situation when θ is an involution in Aut(Dn ). For this entire section, we assume that
θ = ax + b ∈ Aut(Dn ) and θ2 = id. In particular, Proposition 2.1 forces a ∈ R2n and
b ∈ ZDiv(a + 1), i.e. (a+1)b ≡ 0 (mod n). The discussion prior to Proposition 2.1 describes
how to determine the total number of involutions in Aut(Dn ). Recall that Aut2 (Dn ) := {θ ∈
Aut(Dn ) | θ2 = id} is the collection of involutions along with the identity automorphism.
For our discussion of the involutions in Aut(Dn ), it is necessary to understand the structure
of R2n , the square roots of unity in Zn . The following results can be found in an elementary
number theory text (e.g. [7, Example 3.18]) but we include them here for completeness and
for the usefulness of the proof.
7
600
168 =23 ·3 ·7
500
120 =23 ·3 ·5
400
300
200
100
24 =23 ·3
50
100
150
200
Figure 1. The number of involutions in Aut(Dn ) for n ≤ 200. The dashed
lines represent the lines with slope 1, 2, and 3.
Theorem 4.1. Suppose that n ≥ 1 and n =
primes. Then,
k
2
|R2n | = 2k+1
2k+2
2m pr11 · · · prkk where the pi are distinct odd
if m = 0, 1
if m = 2
if m ≥ 3.
Remark 4.2. We include the proof of Theorem 4.1 in Appendix A because it demonstrates
exactly how to construct the square roots of unity in Zn provided we have the prime factorization of n. The construction only involves the Euclidean Algorithm and a map provided
by the Chinese Remainder Theorem. One can use the procedure described in the proof
to effectively compute the square roots of unity in Zn for any n. The construction of the
square roots of unity provided in the appendix also provides the understanding necessary for
Corollary 4.6 below.
The following corollary is the statement of Proposition 2.2 for involutions.
Corollary 4.3. For any n ≥ 1,
| Aut2 (Dn )| =
X
gcd(a + 1, n).
a∈R2n
Remark 4.4. Corollary 4.3 together with the proof of Theorem 4.1 (see Remark 4.2) gives
an easy way to compute the total number of involutions in Aut(Dn ). Figure 1 provides a
graph of the number of involutions in Dn for n ≤ 200. This plot was generated using Sage
[8].
Now we proceed to determine the equivalence classes of involutions. Let θ1 = ax + b
and θ2 = cx + d. We know from Proposition 2.5 that if θ1 ∼ θ2 , then a = c. Thus, to
8
determine the equivalence classes of involutions it suffices to fix an a ∈ R2n and determine
when ax+b ∼ ax+d. Furthermore, according to Proposition 2.6, with a fixed, we simply need
to describe the action of Un on ZDiv(a + 1)/ha − 1i in order to describe all the equivalence
classes.
Recall that we denote the equivalence class of an involution θ by θ and the number of
equivalence classes with fixed leading coefficient a by:
Na := θ | θ = ax + b ∈ Aut2 (Dn ) .
Theorem 4.5. Let a ∈ R2n . The following hold.
(1) ha − 1i ≤ ZDiv(a + 1) and |ZDiv(a + 1)/ha − 1i| ≤ 2.
(2) Na = |ZDiv(a + 1)/ha − 1i| ≤ 2.
(3) Na = 2 if and only if n = 2m · k where k ≥ 1 is odd, m > 0, and a ≡ ±1 (mod 2m ).
Proof. The first part of (1) is simply a restatement of Proposition 2.6 part (1). Now, suppose
n
and we have
that |ZDiv(a + 1)/ha − 1i| = j. It is well known that |ha − 1i| = gcd(a−1,n)
already observed that |ZDiv(a + 1)| = gcd(a + 1, n). Thus we have that
gcd(a+1,n)
n
gcd(a−1,n)
= j, or
gcd(a + 1, n) gcd(a − 1, n) = jn. Now, suppose that
n = 2m pr11 · · · prkk
with m ≥ 0, the pi distinct odd primes, and ri ≥ 1. Then, we must have that
0
gcd(a + 1, n) = 2m ps11 · · · pskk with m0 ≤ m and 0 ≤ si ≤ ri for all i, and
00
gcd(a − 1, n) = 2m pt11 · · · ptkk with m00 ≤ m and 0 ≤ ti ≤ ri for all i.
Now, we note that for all i ∈ {1, ..., k} either si = 0 or ti = 0. Indeed, if si > 0 and ti > 0
then pi divides both a − 1 and a + 1 which is impossible (since pi > 2). Similarly, either
0
00
m0 ≤ 1 or m00 ≤ 1; otherwise, 2min{m ,m } divides a − 1 and a + 1 which is impossible.
Since gcd(a + 1, n) gcd(a − 1, n) = jn, we have that
0
00
2m +m p1s1 +t1 · · · pskk +tk = j2m pr11 · · · prkk .
Now, since for all i either ti = 0 or si = 0, we necessarily have si + ti = ri . Dividing out,
0
00
we get 2m +m = j2m . We also know that m0 ≤ m and m00 ≤ m and since either m0 ≤ 1
0
00
or m00 ≤ 1, we have m ≤ m0 + m00 ≤ m + 1. Dividing out we have j = 2m +m −m and so
1 ≤ j ≤ 2 as required.
Now, according to Theorem 2.7, Na is the number of divisors of gcd(a−1,n)ngcd(a+1,n) , and
thus Na is the number of divisors of j computed above. Therefore, Na = 1 if j = 1 and
Na = 2 if j = 2. In either case, Na = |ZDiv(a + 1)/ha − 1i| ≤ 2 proving Claim (2).
Finally, according to the proof of (1), we have j = 2 if and only if m0 = 1 and m00 = m or
m0 = m and m00 = 1. In the first case, we have that 2m divides a − 1, and in the second case
we have that 2m divides a + 1. Therefore, a ≡ ±1 (mod 2m ).
Theorem 4.5, Theorem 4.1 and Remark 4.2 combined together allow us to compute the
number of distinct equivalence classes of involutions in Aut2 (Dn ) for fixed n if we have the
prime factorization of n. In fact, we get closed formula for the number of equivalence classes
elements in Aut2 (Dn ) as follows.
9
Cn
12
10
m =0
m =1
m =2
m =3
m =4
m =5
8
6
4
2
5
10
15
20
25
30
35
40
45
50
n
Figure 2. Number of equivalence class of involutions in Aut(Dn ) vs. n for
n ≤ 50.
Corollary 4.6. Suppose that n ≥ 1 and n = 2m pr11 · · · prkk where the pi are distinct odd
primes. Then, the number of equivalence classes, Cn , of Aut2 (Dn ) is given by
2k
if m = 0
2k+1
if m = 1
Cn =
k+2
2
if m = 2
k+3
k+1
2
−2
if m ≥ 3.
Proof. According to Theorem 4.5, we know Na for every a ∈ R2n . When m = 0, n is odd
and we know that Na = 1 for all a. Thus, there is one equivalence class for every element
of R2n proving the first formula. If m = 1, 2, then every a ∈ R2n is odd and so satisfies
a ≡ ±1 (mod m). Thus for every element a ∈ R2n , we get two equivalence classes for a total
of 2 · |R2n | equivalence classes. Finally, when m ≥ 3, using the construction of the elements
of R2n found in the proof in the appendix, we see that exactly half of these elements satisfy
the condition a ≡ ±1 (mod m). Therefore, for half the elements in R2n we have Na = 1 and
for the other half we have Na = 2 giving us a total of 2k+2 + (2k+2 − 21 · 2k+2 ) = 2k+3 − 2k+1
equivalence classes, proving the result.
Remark 4.7. According to the Online Encyclopedia of Integer Sequences, the sequence
described in Corollary 4.6 also counts the number (up to isomorphism) of groups of order 2n
10
that have a subgroup isomorphic to Zn [1]. To be more precise, we include a brief discussion
of the natural bijection between these two objects. Suppose θ = ax+b is an involution. Then
wa : Z2 → Aut(Zn ) = Un given by wa (0) = 1 and wa (1) = −a is an injective homomorphism
and induces a Z2 -action on Zn . Next, we define f : Z2 × Z2 → Zn by f (c, d) = 0 if c = 0 or
d = 0 and f (1, 1) = b; it is easily shown that f is a 2-cocycle if and only if b ∈ ZDiv(a + 1).
Finally, it is well known (see [10, Chapter 6.6])) that 2-cocyles give rise to group extensions
and two different 2-cocycles give isomorphic group extensions if they are cohomologous. A
simple calculation shows that if θ = ax + b and θ0 = ax + b0 , then the corresponding 2cocycles are cohomologous if and only if b − b0 ∈ ha − 1i, which matches the characterization
of automorphisms being equivalent above.
We now use our previous results from Section 3 to fully describe the sets Hθ and Qθ when
θ is an involution; moreover, in this situation, we are also interested in the set of twisted
involutions
R = Rθ = {x ∈ G | θ(x) = x−1 }.
If θ = ax + b is an involution, a quick calculation gives
(4.1)
R = {rk | k ∈ ZDiv(a + 1)} ∪ {rk s | k(a − 1) ≡ −b (mod n)}.
Following the results from Theorem 3.1 we describe Q and R \ Q.
Corollary 4.8. If θ = ax + b is an involution, then Q is a subgroup of hri and under the
∼
=
natural isomorphism ψ : hri −→ Zn we have
(
ha − 1i,
if b ∈ ha − 1i
ψ(Q) =
ZDiv(a + 1), otherwise.
Furthermore, R \ Q = {rk | k ∈ ZDiv(a + 1) \ ha − 1i} ∪ {rk s | k(a − 1) ≡ −b (mod n)}
Proof. We proceed using Theorem 3.1 and applying Theorem 4.5. Indeed, by Theorem 3.1
ψ(Q) consists of the union of the two cosets ha−1i and −b+ha−1i. These cosets are the same
if −b ∈ ha−1i (which means b ∈ ha−1i) and then ψ(Q) = ha−1i ≤ ZDiv(a + 1), or they are
distinct and by Theorem 4.5 part (1) there are only 2 cosets total so ha−1i∪(−b+ha−1i) =
ZDiv(a + 1). In either case, Q is a subgroup of hri and ψ(Q) is as described in the statement
of the corollary. Finally, the result about R \ Q is immediate from equation (4.1).
Corollary 4.9. If b 6∈ ha − 1i then R = Q.
Remark 4.10. We note that due to Theorem 4.5, ψ(Q) given in Corollary 4.8 is almost
always ZDiv(a + 1) for involutions. The only instance where ψ(Q) 6= ZDiv(a + 1) occurs
when |ZDiv(a + 1)/ha − 1i| = 2 and b ∈ ha − 1i. We revisit θ1 = 19x + 18 from Example 3.5,
which is an involution, noting that this is in fact a case where b ∈ ha−1i. Recalling that Q1 =
{1, r18 }, we see that ψ(Q1 ) is indeed isomorphic to ha − 1i = h18i < Z36 . We also have that
R = R1 = {1, r, r3 , r5 , . . . , r17 , r18 , r19 , r21 , . . . , r35 }, and thus R1 \ Q1 = {r, r3 , r5 , . . . , r35 }.
Corollary 4.8 is interesting because in the setting of algebraic groups, the symmetric space
Q is almost never a subgroup; whereas here Q is always a subgroup.
In the setting of algebraic groups, it is known that involutions θ1 ∼ θ2 if and only if
Hθ1 ∼
= Hθ2 (see [4]). Now we show that this result does not hold for finite groups.
Corollary 4.11. There exists n and involutions θ1 , θ2 ∈ Aut(Dn ) such that Hθ1 = Hθ2 but
θ1 6∼ θ2 .
11
Proof. Let n = 8 and θ1 = 7x and θ2 = 3x. According to Proposition 2.5, θ1 6∼ θ2 because 7 6=
3 (in U8 ). However, according to Theorem 3.1, Hθ1 = {1, r4 , s, r4 s} and Hθ2 = {1, r4 , s, r4 s}.
5. The Infinite Dihedral Group D∞
So far we have discussed the finite dihedral groups Dn . However, it turns out that there
are similar results for the infinite dihedral group
D∞ = hr, s | s2 = 1, rs = sr−1 i.
In this case, the automorphisms are the affine linear transformations of Z, so are of the form
ax + b where b ∈ Z and a ∈ {±1}. Then, the congruences given in equation (2.1) become
equations over the integers. In particular, it easy to show that the only automorphisms of
finite order, besides the identity, are the involutions and they have the form −x + b where
b ∈ Z.
Proposition 5.1. The following hold:
(1) If θ ∈ Aut(D∞ ) has finite order, then θ = −x + b for some integer b.
(2) −x + b ∼ −x + d if and only if b ≡ d (mod 2).
Proof. Part (1) follows from the equations (2.1) given in the proof of Proposition 2.1. For
part (2), observe that Proposition 2.5 holds for D∞ and says that −x + b ∼ −x + d if and
only if either b − d ∈ 2Z or −b − d ∈ 2Z, equivalently: b ≡ d (mod 2).
We see that in D∞ the situation is simple: the only automorphisms of finite order are
involutions and there are only two distinct equivalence classes of involutions represented by
χ0 = −x = conj(s) (inner) and χ1 = −x + 1 (outer). Note that χ1 represents the class of
the diagram automorphism discussed in Example 1.3.
The fixed group and symmetric space of an involution θ = χi is similarly easy to compute.
(5.1)
Hχ0 = {1, s}, Qχ0 = hr2 i, Rχ0 = hr2 i ∪ {s}
Hχ1 = {1},
Qχ1 = hri, Rχ1 = hri.
We note that the descriptions in equation (5.1) show that Corollary 4.8 holds in the infinite
case as well. However, in this case, the description is simpler because the description of Q
depends only on whether b is even or odd. We also note that despite the fact that the two
cases are different when viewing Q as a subgroup of D∞ , we always have Q ∼
= Z.
Appendix A. Proof of Theorem 4.1
Proof. Our goal is to determine the size of |R2n | for all n. First, suppose that n = pr where p
is an odd prime and r ≥ 1. Then, suppose a ∈ Zpr and that a2 − 1 ≡ 0 (mod pr ). Then we
have that (a + 1)(a − 1) ≡ 0 (mod pr ) and so (a + 1)(a − 1) = kpr for some k ∈ Z. Since p is
an odd prime, it is clear that both a + 1 and a − 1 cannot divide pd for any d ≥ 1 and so we
have that a+1 = pr or a−1 = pr . This implies that a = 1 or a = n−1, thus |R2pr | = 2. Next,
it is clear that |R22 | = 1 and |R24 | = 2. Suppose that n = 2m with m ≥ 3. Next, suppose
that a ∈ Zn \ {1, n − 1} with a2 ≡ 1 (mod n). Since n is even, we must have that a is odd,
so assume that a = 2k + 1 for some 0 ≤ k ≤ 2m−1 . Then 1 ≡ a2 ≡ (2k + 1)2 ≡ 4k 2 + 4k + 1
(mod n) so that 4k(k + 1) = l · 2m for some l ≥ 1. In particular, since m ≥ 3, we have
k(k + 1) = l · 2m−2 for some m. Either k or k + 1 is odd and thus cannot divide 2m−2 .
12
Case 1: k is even. So k = h · 2m−2 for some 1 ≤ h ≤ 2 (since k ≤ 2m−1 ). Then, if
h = 1, a = 2m−1 + 1. If h = 2, then a ≡ 1 (mod n).
Case 2: k is odd. So k + 1 is even and k + 1 = h · 2m−2 with 1 ≤ h ≤ 2. Then, if
h = 1, a = 2m−1 − 1. If h = 2, then a = 2m − 1 ≡ n − 1 ≡ −1 (mod n).
So, there are four possibilities for a: 1, −1, 2m−1 + 1, 2m−1 − 1. It is easy to check that
these are all in fact square roots of 1, showing that |R22m | = 4 if m ≥ 3.
Finally, the result holds due to the fact that the Chinese Remainder Theorem guarantees
that if n = 2m pr11 · · · prkk then
Zn ∼
= Z2m × Z r1 × · · · × Z rk .
p1
pk
Then, it is clear that any square root of (1, ..., 1) (the identity from the right hand side)
must be of the form (a0 , a1 , ..., ak ) where a0 ∈ R22m and ai ∈ Rnpri for i ∈ {1, ..., k}. We
i
have counted the possibilities for ai above and so we can multiply these together to get all
the possible choices for a ∈ Zn with a2 ≡ 1 (mod n). In particular, we get |R2n | = 2k if
m ∈ {0, 1}, |R2n | = 2 · 2k if m = 2 and |R2n | = 22 · 2k if m ≥ 3 as required.
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