A New Representation of the Dihedral Groups - MAA Sections

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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
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A New Representation of the Dihedral Groups - MAA Sections

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