TOTAL CHARACTERS OF DIHEDRAL GROUPS AND SHARPNESS

Eirini Poimenidou and Amy Cottrell

Abstract. We define the total character τ of a finite group G as the sum of all its irreducible characters. A question of K. W. Johnson asks whether the total character of a finite group can be expressed as a polynomial with integer coefficients in some irreducible character χ of G. We show that in the case of dihedral groups of twice odd order the question has an affirmative answer and we give the explicit polynomial.

1. Sharp Characters. It is a well known result of Burnside (Theorem 4.3 in [3]) that if χ is a faithful character of a finite group G that takes on k distinct values, then every irreducible character of G appears as a constituent of at least one of χ, χ^{2}, …, χ^{k−1}. A strengthening of this result is obtained in a modern context by Cameron and Kiyota as follows. Let G be a finite group, χ be a generalized character of G of degree n, and L = {χ(g) | g ≠ 1}. With this notation, Cameron and Kiyota proved in [1] in Theorem 1.1 that if f_{L}(x) = ∏_{l∈L}(x − l), then f_{L}(x) ∈ ℤ[x] and |G| divides f_{L}(n). In the special case that f_{L}(n) = |G|, the character χ is said to be sharp. Another way to characterize a sharp character is to notice that f_{L}(χ) = ∏_{l∈L}(χ − l1_{G}) = ρ, where ρ is the character afforded by the right regular representation of G and 1_{G} is the trivial character of G. In other words, every irreducible character of G appears as a constituent of f_{L}(χ). In the same spirit, we define the total character τ of G to be the sum of all irreducible characters of G:

τ = ∑_{χ∈Irr(G)} χ.

A natural question that was raised first by Kenneth W. Johnson in private correspondence [5] is the following:

\underline{Question 1.1}. For a finite group G does there necessarily exist an irreducible character χ and a monic polynomial f(x) ∈ ℤ[x] such that f(χ) = τ, where τ is the total character of G?

In this paper we show that if χ is any faithful irreducible character of a dihedral group of twice odd order, then Question 1.1 has an affirmative answer. In particular we prove the following theorem.

\underline{Theorem A}. Let G be a dihedral group of order 2n with n odd. If χ is any faithful irreducible character of G, then a monic polynomial g(x) ∈ ℤ[x] exists such that g(χ) = τ, where τ is the total character of G. Furthermore g(x) is minimal with this property.

In section 2 we compute the total characters of all dihedral groups. In section 3, we prove first in Theorem 3.2 that all faithful irreducible characters of D_{2n} with n odd are sharp and then we proceed to prove our main Theorem in 3.7. It turns out that if n ≡ 1 or 3 (mod 8) then g(x) has degree (n + 1)/2 but if n ≡ 5 or 7 (mod 8) then we can get a polynomial of degree (n − 1)/2.

\underline{Example 1.2}. Recall the character table for S_{3}:

\begin{matrix}g_{i} & (1) & (12) & (123) \\ h_{i} & 1 & 3 & 2 \\ χ_{1} & 1 & 1 & 1 \\ χ_{2} & 1 & −1 & 1 \\ χ_{3} & 2 & 0 & −1 \\ τ & 4 & 0 & 1 \end{matrix}

where g_{i} denotes a representative of the i–th conjugacy class and h_{i} is the size of the i–th conjugacy class.

Notice that χ_{3} is sharp since L = {0, −1}, f_{L}(x) = x(x + 1), and f_{L}(2) = 2(2 + 1) = 6 = |S_{3}|. Clearly, χ_{3}^{2} = τ. Hence, g(x) = x^{2}.

\underline{Example 1.3}. The complete character table of S_{4} is:

\begin{matrix}g_{i} & (1) & (12) & (123) & (12)(34) & (1234) \\ h_{i} & 1 & 6 & 8 & 3 & 6 \\ χ_{1} & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 1 & −1 & 1 & 1 & −1 \\ χ_{3} & 2 & 0 & −1 & 2 & 0 \\ χ_{4} & 3 & 1 & 0 & −1 & −1 \\ χ_{5} & 3 & −1 & 0 & −1 & 1 \\ τ & 10 & 0 & 1 & 2 & 0 \end{matrix}

Observe that both χ_{4} and χ_{5} are sharp with L = {−1, 0, 1}. Assume that g(x) exists such that g(χ_{4}) = τ. But that would mean that g(χ_{4})((12)(34)) = g(−1) = 2 and g(χ_{4})(1234) = g(−1) = 0. No polynomial will satisfy that condition. Similarly for χ_{5}. In fact, Johnson’s question has a negative answer for any irreducible character of S_{4} as one easily checks.

2. Total Characters of Dihedral Groups. Let G be the dihedral group D_{2n} of order 2n with n ≥ 3, so that

G = ⟨a, b : a^{n} = b^{2} = 1, b^{−1}ab = a^{−1}⟩.

Using the notation in [4] we consider the character table of G for n odd and n even. We use g_{i} and h_{i} to denote a representative and the size of the i–th conjugacy class respectively.

\underline{Case 1}. n odd.

The conjugacy classes of D_{2n} (n odd) are:

{1}, {a^{r}, a^{−r}} (1 ≤ r ≤ (n − 1)/2), {a^{s}b | 0 ≤ s ≤ n − 1}.

The character table of D_{2n} (n odd) where ϵ = e^{2πi/n} is as follows:

g_{i}  1 a^{r} (1 ≤ r ≤ (n − 1)/2) b
h_{i}  1  2  n
χ_{1}  1  1  1
χ_{2}  1  1  −1
ψ_{j}  2  ϵ^{jr} + ϵ^{−jr}  0
(1 ≤ j ≤ (n − 1)/2)

\underline{Case 2}. n even.

If n is even, say n = 2m, then the conjugacy classes of D_{2n} are:

{1}, {a^{m}}, {a^{r}, a^{−r}} (1 ≤ r ≤ m − 1), {a^{s}b | s even}, {a^{s}b | s odd}.

The character table of D_{2n} (n even, n = 2m, ϵ = e^{2πi/n}) is given below:

g_{i}  1  a^{m} a^{r} (1 ≤ r ≤ m − 1) b  ab
h_{i}  1  1  2  (n/2) (n/2)
χ_{1}  1  1  1  1  1
χ_{2}  1  1  1  −1  −1
χ_{3}  1 (−1)^{m}  (−1)^{r}  1  −1
χ_{3}  1 (−1)^{m}  (−1)^{r}  −1  1
ψ_{j}  2 2(−1)^{j} ϵ^{jr} + ϵ^{−jr}  0  0

(1 ≤ j ≤ m − 1)

We are now ready to compute the total character for dihedral groups.

\underline{Proposition 2.1}.

1. The total character τ of D_{2n}, n odd, is:

\begin{matrix}g_{i} & 1 & a^{r} (1 ≤ r ≤ (n − 1)/2) & b \\ h_{i} & 1 & 2 & n \\ τ & n + 1 & 1 & 0 \end{matrix}

2. The total character τ of G = D_{2n}, n = 2m, is:

g_{i}  1  a^{m}  a^{r}  b ab
τ  2(m + 1) \{ \begin{matrix}0 & m odd \\ 2 & m even \end{matrix} \{ \begin{matrix}0, & r odd \\ 2, & r even \end{matrix} 0 0

\underline{Proof}. When n is odd we clearly have that τ(1) = 2 + 2(n − 1)/2 = n + 1 and τ(b) = 0. An easy computation yields

τ(a^{r}) = 2 + ∑_{j=1}^{(n−1)/2}(ϵ^{jr} + ϵ^{−jr}) = \frac{1 − ϵ^{r(n+1)/2}}{1 − ϵ^{r}} + \frac{1 − ϵ^{−r(n+1)/2}}{1 − ϵ^{−r}} .

Recall that ϵ = e^{(2πi)/n} and note that e^{((n+1)irπ)/n} = (−1)^{r}e^{(irπ)/n} = (−1)^{r}e^{r/2} and similarly e^{−((n+1)irπ)/n} = (−1)^{r}e^{−(irπ)/n} = (−1)^{r}ϵ^{−r/2}.

For r odd we get

τ(a^{r}) = \frac{1 + ϵ^{r/2}}{1 − ϵ^{r}} + \frac{1 + ϵ^{−r/2}}{1 − ϵ^{−r}} = \frac{1 + ϵ^{r/2}}{1 − ϵ^{r}} − ϵ^{r/2} \frac{1 + ϵ^{r/2}}{1 − ϵ^{r}} = 1.

Similar calculation for r even yield τ(a^{r}) = 1 and we have completed the proof of part 1 of Proposition 2.1.

Part 2 of Proposition 2.1 follows easily as above by examining separately the cases for r odd and even.

3. Polynomials of Sharp Characters in Dihedral Groups. Our strategy in proving our main theorem will be to show first that all faithful irreducible characters of dihedral groups of twice odd order are sharp. Then we proceed to construct a polynomial p(x) such that p(χ) = τ on all non–identity elements where χ = ψ_{1} (using the notation of section 2 for D_{2n} with n odd). We notice first that if such a polynomial exists, then τ(b) = 0 = p(ψ(b)) = p(0) = a_{0} and hence, our polynomial will have zero constant term. In addition since τ(a^{r}) = 1 for all 1 ≤ r ≤ (n − 1)/2 we must also have that p(ψ(a^{r})) = p(2 cos((2πr)/n)) = 1. The arguments will easily extend to any faithful irreducible character ψ_{j} of D_{2n} with n odd. The following lemma is mentioned in [1] without proof. We include its proof for completeness.

\underline{Lemma 3.1}.

∏_{k=1}^{m−1} 2 sin(\frac{kπ}{m} ) = m.

\underline{Proof}. Let L = \{e^{(2kiπ)/m} | 1 ≤ k ≤ m − 1\}. Then,

f_{L}(x) = ∏_{k=1}^{m−1}(x − e^{(2kiπ)/m}) = x^{m−1} + ⋯ + x + 1.

Note that f_{L}(1) = m, which implies that

∏_{k=1}^{m−1}(1 − e^{(2kiπ)/m}) = m.

Recall that

2 sin(\frac{kπ}{m} ) = \frac{1}{i}(e^{(kiπ)/m} − e^{−(kiπ)/m}) = \frac{i(1 − e^{(2kiπ)/m})}{e^{(kiπ)/m}}.

This implies that

∏_{k=1}^{m−1} 2 sin(\frac{kπ}{m} ) = ∏_{k=1}^{m−1} \frac{i(1 − e^{(2kiπ)/m})}{e^{(kiπ)/m}} = \frac{mi^{m−1}}{∏_{k=1}^{m−1} e^{(kiπ)/m}} .

However,

∏_{k=1}^{m−1} e^{(kiπ)/m} = e^{(i(m−1)π)/2}

= (cos(\frac{mπ}{2} ) + i sin(\frac{mπ}{2} ))(cos(\frac{−π}{2} ) + i sin(\frac{−π}{2} )).

If m = 4b, then i^{m−1} = i^{4b−1} = i^{4b}i^{−1} = −i, and

(cos(\frac{mπ}{2} ) + i sin(\frac{mπ}{2} ))(cos(\frac{−π}{2} ) + i sin(\frac{−π}{2} )) = (1 + 0i)(0 − i) = −i.

Hence,

∏_{k=1}^{m−1} 2 sin(\frac{kπ}{m} ) = m.

If m = 4b + 1, then i^{m−1} = i^{4b} = 1, and

(cos(\frac{mπ}{2} ) + i sin(\frac{mπ}{2} ))(cos(\frac{−π}{2} ) + i sin(\frac{−π}{2} )) = (0 + i)(0 − i) = 1.

Hence,

∏_{k=1}^{m−1} 2 sin(\frac{kπ}{m} ) = m.

Proceeding similarly for the remaining cases yields the result.

\underline{Theorem 3.2}. Every faithful irreducible character of D_{2n} (n odd) is sharp. \underline{Proof}. Consider ψ_{1} as defined previously in the character table of D_{2n}. We

show first that ψ_{1} is sharp. We must first discern if there is any r, (1 ≤ r ≤ (n−1)/2) such that 2 cos((2rπ)/n) = 2. Now, cos((2rπ)/n) = 1 whenever (2rπ)/n = 2kπ, k ∈ ℤ. Hence, we must have that r/n = k, k ∈ ℤ, but, 1 ≤ r ≤ (n − 1)/2. So, r < n implies r/n ∉ ℤ and we see that cos((2rπ)/n) ≠ 1 for any r such that 1 ≤ r ≤ (n − 1)/2. Hence, 2 cos((2rπ)/n) ≠ 2, and ψ_{1} is faithful. Now, we have

∏_{r=1}^{(n−1)/2}(2 − 2 cos(\frac{2rπ}{n} )) = ∏_{r=1}^{(n−1)/2} 4 sin^{2}(\frac{rπ}{n} )

= ∏_{r=1}^{(n−1)/2} 2 sin(\frac{rπ}{n} ) ∏_{r=1}^{(n−1)/2} 2 sin(\frac{rπ}{n} ).

Also recall that

sin(π − \frac{rπ}{n} ) = sin π cos(\frac{rπ}{n} )

− cos π sin(\frac{rπ}{n} ) = sin(\frac{rπ}{n} ).

Thus,

∏_{r=1}^{(n−1)/2} 2 sin(\frac{rπ}{n} ) ∏_{r=1}^{(n−1)/2} 2 sin(\frac{rπ}{n} ) = ∏_{r=1}^{(n−1)/2} 2 sin(\frac{rπ}{n} ) ∏_{r=1}^{(n−1)/2} 2 sin(\frac{(n − r)π}{n} ).

Reindex the latter product by letting j = n − r and by letting j run from (n + 1)/2 to n − 1. So, we now get:

∏_{r=1}^{(n−1)/2} 2 sin(\frac{rπ}{n} ) ∏_{j=(n+1)/2}^{n−1} 2 sin(\frac{jπ}{n} ) = ∏_{r=1}^{n−1} 2 sin(\frac{rπ}{n} ) = n.

This is the crucial step in proving that ψ_{1} is sharp since

f_{L}(x) = (x − 0) ∏_{r=1}^{(n−1)/2}(x − 2 cos(\frac{2rπ}{n} ))

and hence, we get f_{L}(2) = 2n = |G| and ψ_{1} must be sharp. Now consider ψ_{j} with 1 < j ≤ (n − 1)/2. It is not difficult to see that ψ_{j} is faithful if and only if j is relatively prime to n. In that case the set of character values of ψ_{j} is the same as that of ψ_{1} and the same arguments yield that all faithful irreducible characters of D_{2n} with n odd are sharp.

The following lemma is formula JO (570) in [2].

\underline{Lemma 3.3}. For n odd,

n sin(x) ∏_{k=1}^{(n−1)/2}(1 − \frac{sin^{2}(x)}{sin^{2}(\frac{kπ}{n} )}) = sin(nx).

\underline{Theorem 3.4}.

∏_{k=1}^{(n−1)/2} 2 cos(\frac{2kπ}{n} ) = \{ \begin{matrix}1, & n ≡ 1 (mod 8) \\ −1, & n ≡ 3 (mod 8) \\ −1, & n ≡ 5 (mod 8) \\ 1, & n ≡ 7 (mod 8). \end{matrix}

\underline{Proof}. Letting x = π/4 in the formula of Lemma 3.3 above, we get that

sin(\frac{nπ}{4} ) = n sin(\frac{π}{4} ) ∏_{k=1}^{(n−1)/2}(\frac{sin^{2}(\frac{kπ}{n} ) − \frac{1}{2}}{sin^{2}(\frac{kπ}{n} )} ),

from whence we get

\frac{sin(\frac{nπ}{4} )}{n sin(\frac{π}{4} )} = \frac{∏_{k=1}^{(n−1)/2}(\frac{1}{2}(2 sin^{2}(\frac{kπ}{n} ) − 1)}{∏_{k=1}^{(n−1)/2} sin^{2}(\frac{kπ}{n} )}

= \frac{∏_{k=1}^{(n−1)/2}(−1) cos(\frac{2kπ}{n} )}{∏_{k=1}^{(n−1)/2}(2 sin^{2}(\frac{kπ}{n} ))} = \frac{∏_{k=1}^{(n−1)/2}(−1)2 cos(\frac{2kπ}{n} )}{∏_{k=1}^{(n−1)/2} 4 sin^{2}(\frac{kπ}{n} )}

= \frac{(−1)^{(n−1)/2} ∏_{k=1}^{(n−1)/2}(2 cos(\frac{2kπ}{n} ))}{∏_{k=1}^{(n−1)/2} 2 sin(\frac{kπ}{n} )} = \frac{(−1)^{(n−1)/2} ∏_{k=1}^{(n−1)/2}(2 cos(\frac{2kπ}{n} ))}{n}.

This implies that

∏_{k=1}^{(n−1)/2} 2 cos(\frac{2kπ}{n} ) = (−1)^{(n−1)/2} \begin{matrix}sin & ( & \frac{nπ}{4} & ) \\ sin & ( & \frac{π}{4} & ) \end{matrix}.

If n = 8k + 1, then (n − 1)/2 = 4k, and (−1)^{(n−1)/2} = 1, and we have that

∏_{k=1}^{(n−1)/2} 2 cos(\frac{2kπ}{n} ) = \begin{matrix}sin & ( & \frac{(8k+1)π}{4} & ) \\ sin & ( & \frac{π}{4} & ) \end{matrix} = \frac{sin(2kπ + \frac{π}{4} )}{sin(\frac{π}{4} )} = \begin{matrix}sin & ( & \frac{π}{4} & ) \\ sin & ( & \frac{π}{4} & ) \end{matrix} = 1.

Similar arguments for n congruent to 3, 5, 7 (mod 8) complete the proof.

\underline{Proposition 3.5}. Let x_{r} = 2 cos \frac{2rπ}{n} for n odd and r = 1, …, (n − 1)/2. Let t = (n − 1)/2. The system

( \begin{matrix}x_{1}^{t} & x_{1}^{t−1} & ⋯ & x_{1} \\ x_{2}^{t} & x_{2}^{t−1} & ⋯ & x_{2} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ x_{t}^{t} & x_{t}^{t−1} & ⋯ & x_{t} \end{matrix} ) ( a_{t} ; a_{t−1} ; ⋮ ; a_{1} ) = ( 1 1 ⋮ 1 )

has a unique integer solution with

\begin{aligned}a_{t} & = \{ \begin{matrix}−1, & n = 1, 3 (mod 8) \\ 1, & n = 5, 7 (mod 8) \end{matrix} \\ a_{t−r} & = (−1)^{r}a_{t}S_{r}(x_{1}, …, x_{t}), \end{aligned}

where S_{r}(x_{1}, …, x_{t}) is the r–th elementary symmetric function in x_{1}, x_{2}, …, x_{t}.

\underline{Proof}. First observe that the uniqueness part follows easily since

\begin{aligned}det ( \begin{matrix}x_{1}^{t} & x_{1}^{t−1} & ⋯ & x_{1} \\ x_{2}^{t} & x_{2}^{t−1} & ⋯ & x_{2} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ x_{t}^{t} & x_{t}^{t−1} & ⋯ & x_{t} \end{matrix} ) & = x_{1}x_{2} ⋯ x_{t} det ( \begin{matrix}x_{1}^{t−1} & x_{1}^{t−2} & ⋯ & 1 \\ x_{2}^{t−1} & x_{2}^{t−2} & ⋯ & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ \\ x_{t}^{t−1} & x_{t}^{t−2} & ⋯ & 1 \end{matrix} ) \\ & = x_{1}x_{2} ⋯ x_{t} ∏_{1≤i<j≤t}(x_{i} − x_{j}) ≠ 0 \end{aligned}

using the well known formula for the Vandermonde determinant. Let p(x) = a_{t}x^{t} + ⋯ + a_{1}x such that p(x_{i}) = 1 for all 1 ≤ i ≤ t. Note that p(x_{i}) − 1 = 0 when 1 ≤ i ≤ t. We have p(x) − 1 = a_{t}(x − x_{1}) ⋯ (x − x_{t}). By equating the constant terms, we see that (−1)^{t}a_{t}x_{1} ⋯ x_{t} = −1. This in turn yields

a_{t} = \frac{−(−1)^{t}}{x_{1} ⋯ x_{t}} .

Recall that t = (n − 1)/2. If n ≡ 1 (mod 8), then (−1)^{t} = 1 and x_{1} ⋯ x_{t} = 1. Hence, a_{t} = −1. If n ≡ 3 (mod 8), then (−1)^{t} = −1 and x_{1} ⋯ x_{t} = −1. Hence, a_{t} = −1. If n ≡ 5 (mod 8), then (−1)^{t} = 1 and x_{1} ⋯ x_{t} = −1. Hence, a_{t} = 1. If n ≡ 7 (mod 8), then (−1)^{t} = −1 and x_{1} ⋯ x_{t} = 1. Hence, a_{t} = 1.

The remaining coefficients are clearly given by a_{t−r} = (−1)^{r}a_{t}S_{r}(x_{1}, …, x_{t}), where S_{r}(x_{1}, …, x_{t}) is the r–th elementary symmetric function in x_{1}, x_{2}, …, x_{t}. Recall that x_{r} = ϵ^{r} + ϵ^{−r} where ϵ is a primitive n–th root of unity. Therefore all of the x_{r} and hence, S_{r}(x_{1}, …, x_{t}) are algebraic integers. Note also that S_{r}(x_{1}, …, x_{t}) remains invariant under the Galois automorphisms of ℚ(ϵ) over ℚ and is therefore a rational integer. Hence, we have as desired that a_{t−r} ∈ ℤ.

As immediate consequences of the above, we get:

\underline{Corollary 3.6}. Let n be an odd integer, x_{r} = 2 cos((2rπ)/n) where 1 ≤ r ≤ (n − 1)/2. Let t = (n − 1)/2.

1. The polynomial

p(x) = −(x − x_{1}) ⋯ (x − x_{t}) + 1, n ≡ 1, 3 (mod 8)

has the property that p(0) = 0 and p(x_{r}) = 1 for 1 ≤ r ≤ t.

2. The polynomial

p(x) = (x − x_{1}) ⋯ (x − x_{t}) + 1, n ≡ 5, 7 (mod 8)

has the property that p(0) = 0 and p(x_{r}) = 1 for 1 ≤ r ≤ t.

We are now ready to prove our main Theorem.

\underline{Theorem 3.7}. Let G be a dihedral group of order 2n with n odd. If χ is any faithful irreducible character of G, then a monic polynomial g(x) ∈ ℤ[x] exists such that g(χ) = τ, where τ is the total character of G. Furthermore g(x) is minimal with this property.

\underline{Proof}. We need to examine separately the following two cases.

\underline{Case 1}. n ≡ 5, 7 (mod 8).

Let χ = ψ_{1} be the irreducible character of D_{2n} as labeled in its character table previously. Let x_{r} = ϵ^{r} + ϵ^{−r} = 2 cos((2rπ)/n), 1 ≤ r ≤ (n − 1)/2 and t = (n − 1)/2. We show that g(x) = (x − x_{1}) ⋯ (x − x_{t}) + 1 is the right polynomial. We have established in Corollary 3.6 part 2, that g(x) ∈ ℤ[x] and that g(χ(a^{r})) = g(x_{r}) = 1 = τ(a^{r}) and g(χ(b)) = g(0) = 0 = τ(b). We need only show that g(χ(1)) = g(2) = τ(1) = n + 1. Notice that since χ is sharp

(2 − 0) ∏_{j=1}^{(n−1)/2}(2 − 2 cos(\frac{2jπ}{n} )) = 2n

and hence, we have that

∏_{j=1}^{(n−1)/2}(2 − 2 cos(\frac{2jπ}{n} )) = n.

Now it follows easily that

g(2) = ∏_{j=1}^{(n−1)/2}(2 − 2 cos(\frac{2jπ}{n} )) + 1 = n + 1 = τ(1)

as desired.

Notice that the result is true for any faithful irreducible character of G since they all share the same set of character values and τ(a^{r}) = 1 for all r.

\underline{Case 2}. n ≡ 1, 3 (mod 8).

Let p(x) = −(x − x_{1}) ⋯ (x − x_{t}) + 1 and g(x) = (p(x) − 1)(−x + 1) + 1. Notice that by Corollary 3.6 part 1, we have that p(0) = 0 and p(x_{r}) = 1 and hence, g(0) = 0 and g(x_{r}) = 1. In addition

p(2) = − ∏_{j=1}^{(n−1)/2}(2 − 2 cos(\frac{2jπ}{n} )) = −n + 1.

Hence, g(2) = (p(2) − 1)(−2 + 1) + 1 = (−n + 1 − 1)(−1) + 1 = n + 1 = τ(1) as desired. As before the argument can be shown to be independent of the sharp character we choose.

Notice that by construction both polynomials g(x) are minimal with the required properties.

Using GAP and Mathematica we can easily compute the polynomials for the first few dihedral groups

D_{6} : x^{2}
D_{10} : x^{2} + x
D_{14} : x^{3} + x^{2} − 2x
D_{18} : x^{5} − 4x^{3} + x^{2} + 3x
D_{22} : x^{6} − 5x^{4} + x^{3} + 6x^{2} − 2x
D_{26} : x^{6} + x^{5} − 5x^{4} − 4x^{3} + 6x^{2} + 3x
D_{30} : x^{7} + x^{6} − 6x^{5} − 5x^{4} + 10x^{3} + 6x^{2} − 4x.

4. Final Remarks and Some Questions. It appears that sharpness is instrumental in answering Johnson’s question. In fact we are not aware of any instances of a non–sharp character giving an affirmative answer to his question.

In the case of dihedral groups of twice even order we are not always guaranteed an irreducible sharp character. For example, if we look at the character table of D_{12}

\begin{matrix}g_{i} & 1 & a^{3} & a & a^{2} & b & ab \\ h_{i} & 1 & 1 & 2 & 2 & 3 & 3 \\ χ_{1} & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 1 & 1 & 1 & 1 & −1 & −1 \\ χ_{3} & 1 & −1 & −1 & 1 & 1 & −1 \\ χ_{4} & 1 & −1 & −1 & 1 & −1 & 1 \\ χ_{5} & 2 & −2 & 1 & −1 & 0 & 0 \\ χ_{6} & 2 & 2 & −1 & 1 & 0 & 0 \\ τ & 8 & 0 & 0 & 4 & 0 & 0 \end{matrix}

The only possibility for a sharp character is χ_{5}. However, (2−(−2))(2−1)(2− (−1))(2 − 0) = 4(1)(3)(2) = 24 ≠ 12 and D_{12} has no irreducible sharp characters. It is easy to show that Johnson’s question has a negative answer on D_{12}.

It appears however, that if a sharp character χ exists in a dihedral group of twice even order then a polynomial g(x) ∈ ℤ[x] may exist such that g(χ) = τ as is the case in D_{8} where if χ is the faithful irreducible character of degree 2, then χ^{2} + χ = τ.

A natural question arising is therefore the following:

\underline{Question 4.1}. If χ is an irreducible character of a finite group G such that for some monic polynomial g(x) ∈ ℤ[x] we have g(χ) = τ, is χ necessarily sharp?

\underline{References}

1. P. J. Cameron and M. Kiyota, “Sharp Characters of Finite Groups,” Journal of Algebra, 115 (1988), 125–143.

2. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, 1980.

3. I. M. Isaacs, Character Theory of Finite Groups, Academic Press, 1976.

4. G. James and M. Liebeck, Representations and Characters of Groups, Cambridge Mathematical Textbooks, Cambridge, 1995.

5. K. W. Johnson, private correspondence, 1997.

Eirini Poimenidou Division of Natural Sciences New College of USF Sarasota, FL 34243 email: poimenid@virtu.sar.usf.edu

Amy Cottrell Division of Natural Sciences New College of USF Sarasota, FL 34243 email: cottrell@virtu.sar.usf.edu