A New Representation of the Dihedral Groups - MAA Sections

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