A New Representation of the Dihedral Groups - MAA Sections

A New Representation of the Dihedral Groups
Chris G. Devillier
Department of Mathematics
Southeastern Louisiana University
SLU 10687
Hammond, LA 70402
U.S.A.
email: cdevillier@selu.edu
December 18, 2002
Abstract: In [1] it was shown that solutions to the conditional Cauchy equation
of the cylindrical type can be used to form a group called the pointed automorphism
group. Several applications of this group were given; one of which was
that every group is a subgroup of some pointed automorphism group. In the
spirit of that result, this paper presents a representation of the Dihedral groups
as a set of functions, all of which are solutions to the conditional Cauchy equation
of the cylindrical type. This new representation has pedagogical appeal
in that every function in these permutation groups is a function over a cyclic
group. Moreover, every function is easily described, and every composition is
tantamount to addition in a cyclic group. Therefore this representation should
prove beneficial in an introductory undergraduate course in group theory.
From [1], we have the following three definitions and the following proposition.
Definition 1. Let G be a group. Let f ε M(G), which is the set of all functions
from G to G. If there exists s ε G such that f(x+s)= f(x) + f(s) for all x ε G,
f is said to be a solution to the conditional Cauchy equation of the cylindrical
type.
Definition 2. Let G be a group, and let f ε M(G). Then H r (f) = {s ε G |
f(x + s) = f(x) + f(s) for all x ε G}.
Definition 3. Let G be a group and S be a subgroup of G. Then PAut S G =
{fεM(G)|S ⊆ H r (f), f(S) = S, and f is a bijection }.
Proposition 4. Let G be a group and S be a subgroup of G. Then (PAut S G, ◦)
is a group.
We now follow with a representation of the finite Dihedral groups.
1

Proposition 5. Let n ≥ 2 be a natural number, let G = Z 2n , and let S be the
subgroup of Z 2n that is isomorphic to Z n . Define
{
x : x is even
f(x) =
x + 2 : x is odd
and
{
g(x) =
−x : x is even
−x + 2 : x is odd.
Then f, g ε PAut S G and for n > 2, 〈f, g〉 ∼ = D 2n .
Proof. In [2], we find that a non-Abelian group that is generated by two
elements σ and τ where τ 2 = e and τστ = σ −1 is isomorphic to a Dihedral
group. Moreover, the order of the non-Abelian group is twice the order of σ.
So we show that g 2 = e and that gfg = f −1 . First observe that
{
g 2 (x) =
Therefore, g 2 = e.
Next observe that
{
(gfgf)(x) =
x : x is even
−(−x + 2) + 2 : x is odd.
−(−x) : x is even
−[(−(x + 2) + 2) + 2] + 2 : x is odd.
Therefore gfgf = e and hence, gfg = f −1 .
To demonstrate that G is non-Abelian when n > 2, we have that
{
−x : x is even
(fg)(x) =
−x + 4 : x is odd
and
(gf)(x) =
{ −x : x is even
−x : x is odd.
Thus if G ≠ Z 4 , then 〈f, g〉 is a Dihedral group.
If x is even, then f m (x) = x for any positive integer m. If x is odd, then
f m (x) = x + (2m mod 2n) for any positive integer m. Therefore, |〈f〉| = n.
Hence 〈f, g〉 ∼ = D 2n .
Next we show that f, g ε PAut S G. First note that f and g are bijections.
Also, since f and g are the identity on S, f(S) = S and g(S) = S. To show that
S ⊆ H r (f), note that if x and s are both even, then x + s is even; and we have
that f(x) = x, f(s) = s, and f(x+s) = x+s. Therefore, f(x+s) = f(x)+f(s).
If x is odd and s is even, then x + s is odd; and therefore f(x + s) = x + s + 2 =
x + 2 + s = f(x) + f(s). Thus S ⊆ H r (f). A similar argument proves that
S ⊆ H r (g). Hence f,g ε PAut S G. So by Proposition 4, 〈f, g〉 is a group of
solutions to a conditional Cauchy equations of the cylindrical type.
Remark: We know that for all x that are even and for any positive integer m,
f m (x) = x and (f m g)(x) = −x. For x odd the elements of 〈f, g〉 are as follows:
2

f(x) = x + 2, f 2 (x) = x + 4, f 3 (x) = x + 6, · · ·, f n−1 (x) = x + 2n − 2, f n (x) = x
and
g(x) = −x + 2, fg(x) = −x + 4, f 2 g(x) = −x + 6, · · ·, f n−2 g(x) = −x + 2n − 2,
f n−1 g(x) = −x.
Proposition 6. Let G = Z and let S = 2Z. Define
{
x : x is even
f(x) =
x + 2 : x is odd
and
{
g(x) =
Then f, g ε PAut S G and 〈f, g〉 ∼ = D ∞ .
Proof. From [1], 〈f, g〉 = PAut S G ∼ = D ∞ .
−x : x is even
−x + 2 : x is odd.
Again observe that for x an even number, f(x) = x, gf m (x) = −x, and
f m g(x) = −x. For x an odd number,
f(x) = x + 2, f 2 (x) = x + 4, f 3 (x) = x + 6, · · · ,
g(x) = −x + 2, fg(x) = −x + 4, f 2 g(x) = −x + 6, · · · , and
gf(x) = −x, gf 2 (x) = −x − 2, gf 3 (x) = −x − 4, · · · .
REFERENCES
1. Devillier, C. Conditional Cauchy Equations and Applications, Aequationes Mathematicae
(to appear)
2. Rose, J. A Course on Group Theory, Cambridge University Press, 1978.
3