Group Elements Whose Character Values are Roots of Unity

Abstract

We classify all finite groups G that possess an element x ∈ G such that every irreducible character of G takes a root of unity value at x.

1 Introduction

Let G be a finite group. Following [5], an element x ∈ G is called a nonvanishing element if χ(x)≠ 0 for all irreducible complex characters χ of G. This concept has been widely studied in recent years. In this paper, we consider nonvanishing elements of finite groups which satisfy certain minimal condition as follows. Given a nonvanishing element x of a finite group G, it is not hard to show that |C_G(x)|≥ k(G) and that the equality holds if and only if |χ(x)| = 1 for all irreducible characters χ of G (see Lemma 2.3), where C_G(x) is the centralizer of x in G and k(G) is the number of conjugacy classes of G. Note that if |χ(x)| = 1 for some character χ of G, then χ(x) is a root of unity (see, for example, Problem 3.2 in [4]). We will call an element x ∈ G a root of unity element if |χ(x)| = 1 for all irreducible characters of G. The condition |C_G(x)| = k(G) alone does not characterize root of unity elements. For example, if G = A₅, the alternating group of degree 5, and x ∈ G is an element of order 5, then |C_G(x)| = 5 = k(G) but x is not a root of unity element. We note that root of unity elements are called totally unitary or TU-elements by S. Ostrovskaya and E. M. Zhmud’ and they classify all finite metabelian groups with trivial center that contain a root of unity element in [1, Chapter XXII].

Write Irr(G) for the set of irreducible complex characters of G and F(G) for the Fitting subgroup of G, that is, the largest normal nilpotent subgroup of G. In our first result, we prove the following.

Theorem A

Let G be a finite group and let x ∈ G. If |χ(x)| = 1 for all irreducible characters χ of G, then x ∈F(G) and both F(G) and G/F(G) are abelian. In particular, G is abelian or metabelian.

Thus if a finite group G has a root of unity element, then it is abelian or metabelian. In particular, such a group is solvable. Theorem A confirms a conjecture proposed in [5] for root of unity elements. This conjecture states that every nonvanishing element of a finite solvable groups G must lie in F(G).

Our interest in root of unity elements stems from an observation that if χ ∈Irr(G) and g ∈ G such that |χ(g)| = 1, then the size of the conjugacy class g^G containing g is always divisible by χ(1) (see Lemma 2.1). Consequently, if x is a root of unity element of G, then |x^G| is divisible by χ(1) for all χ ∈Irr(G). This is related to Conjecture C in [6] asserting that if χ ∈Irr(G) is a primitive character of a finite group G, then χ(1) divides |g^G| for some g ∈ G. Thus the observation above gives us a way to locate the required element g ∈ G. However, not every primitive irreducible character admits a root of unity value. For example, if G is the sporadic simple group O’N, then G has a primitive irreducible character of degree 64790 which does not admit any root of unity value.

In the next result, we classify all finite groups with a root of unity element. Clearly, if G is abelian, then every element of G is a root of unity element. Let q > 2 be a prime power. We denote by Γ_q the unique doubly transitive Frobenius group with a cyclic complement of order q − 1 and degree q. So ${\varGamma }_{q}\cong \textrm {AGL}_{1}(\mathbb {F}_{q})=\mathbb {F}_{q}\rtimes \mathbb {F}_{q}^{\ast }$, where $\mathbb {F}_{q}$ is a finite field with q elements.

Theorem B

Let G be a finite group. Then G has a root of unity element x ∈ G if and only if one of the following holds:

• G is abelian;

• $\mathbf {F}(G)=G^{\prime }\mathbf {Z}(G)$ is abelian, $G^{\prime }\cap \mathbf {Z}(G)=1$; and $G/\mathbf {Z}(G)\cong {\varGamma }_{q_{1}}\times {\varGamma }_{q_{2}}\times \cdots \times {\varGamma }_{q_{m}}$, where each q_i > 2 is a prime power and m ≥ 1 is an integer.

For each i with 1 ≤ i ≤ m, write ${\varGamma }_{q_{i}}=V_{i}\ltimes A_{i}$, where V_i is the Frobenius kernel and A_i is the Frobenius complement. Let $U:={\prod }_{i=1}^{m} (V_{i}-\{1\})$ and let $\mathcal {U}=\pi ^{-1}(U)$ where $\pi :G\rightarrow G/\mathbf {Z}(G)$ is the natural homomorphism. Then every element of U is a root of unity element of G/Z(G) by Lemma 3.17, Chapter XXII of [1] and from the proof of Theorem B, every element of $\mathcal {U}$ is a root of unity element of G.

Our notation is standard and we follow [4] for the character theory of finite groups.

2 Preliminaries

We collect some properties of root of unity elements in the next lemmas.

Lemma 2.1

Let G be a finite group and let g ∈ G. If χ ∈Irr(G) and |χ(g)| = 1, then χ(1) divides |g^G|. In particular, if x ∈ G is a root of unity element, then χ(1) divides |x^G| for all χ ∈Irr(G).

Proof

Assume that χ ∈Irr(G) and g ∈ G such that |χ(g)| = 1. Let K be the class sum of the conjugacy class g^G, that is, $K={\sum }_{y\in g^{G}}y$. Then

$$\omega_{\chi}(K)=\frac{|g^{G}|\chi(g)}{\chi(1)}$$

is an algebraic integer by [4, Theorem 3.7]. Since χ(g) is an algebraic integer,

$$ \frac{|g^{G}|}{\chi(1)}=\omega_{\chi}(K)\overline{\chi(g)} $$

is a rational algebraic integer, so it is an integer and hence χ(1) divides |g^G|.

If x ∈ G is a root of unity element, then for any χ ∈Irr(G), we have |χ(x)| = 1 and hence χ(1) divides |x^G| as wanted. □

Lemma 2.2

Let G be a finite group and let x ∈ G be a root of unity element. Then

• (a) k(G) = |C_G(x)|.

• (b) $G^{\prime }\leq \langle x^{G}\rangle$.

• (c) If $N\unlhd G$, then xN is a root of unity in G/N.

• (d) If z ∈Z(G), then xz is also a root of unity element.

Proof

From the Second Orthogonality relation, we have

$$ |\mathbf{C}_{G}(x)| = \sum\limits_{\chi\in\text{Irr}(G)}|\chi(x)|^{2} = \sum\limits_{\chi\in\text{Irr}(G)}1=k(G). $$

Let L = 〈x^G〉. For any χ ∈Irr(G/L), we see that $x\in L\subseteq \text {Ker}(\chi )$. Hence 1 = |χ(x)| = χ(1) and thus all characters χ ∈Irr(G/L) are linear which implies that G/L is abelian and so $G^{\prime }\leq L$. Since $\text {Irr}(G/N)\subseteq \text {Irr}(G)$ whenever $N\unlhd G$, if x is a root of unity of G then xN is a root of unity of G/N.

Finally, let z ∈Z(G) and χ ∈Irr(G). Then χ_Z(G) = χ(1)λ for some λ ∈Irr(Z(G)). We have χ(xz) = λ(z)χ(x) and thus if x is a root of unity element, then so is xz as |λ(z)| = 1. □

The next lemma follows from the proof of Lemma 3.17 in [1, Chapter XXII] and the previous lemma. For completeness, we include the proof here.

Lemma 2.3

Let G be a finite group and let x ∈ G be a nonvanishing element of G. Then |C_G(x)|≥ k(G); and the equality holds if and only if x is a root of unity element.

Proof

We first claim that if x ∈ G is a nonvanishing element, then |C_G(x)|≥ k(G). Let $\alpha ={\prod }_{\chi \in \text {Irr}(G)}\chi (x)$. Let n be the exponent of G and let $\mathbb {Q}_{n}=\mathbb {Q}(\xi )$, where ξ is a primitive n th-root of unity. Let $\mathcal {G}$ be the Galois group of $\mathbb {Q}_{n}$ over $\mathbb {Q}$. Then $\mathcal {G}$ acts on Irr(G) and we see that χ^σ ∈Irr(G) if and only if χ ∈Irr(G) for all $\sigma \in \mathcal {G}$. Hence α^σ = α for all $\sigma \in \mathcal {G}$. It follows that $\alpha \in \mathbb {Q}$. Since α is an algebraic integer, we must have that $\alpha \in \mathbb {Z}$. As x is nonvanishing, α≠ 0 and so |α|≥ 1.

By the inequality between arithmetic and geometric means, we have that

$$ 1\leq |\alpha|^{2}=\prod\limits_{\chi\in\text{Irr}(G)}|\chi(x)|^{2} \leq \left( \frac{{\sum}_{\chi\in\text{Irr}(G)}|\chi(x)|^{2}}{k(G)}\right)^{k(G)}= \left( \frac{|\mathbf{C}_{G}(x)|}{k(G)}\right)^{k(G)}. $$

It follows that |C_G(x)|≥ k(G) as wanted.

Next, assume that x is a nonvanishing element of G and |C_G(x)| = k(G). Then |α| = 1 from the inequality above. Hence

$$ \prod\limits_{\chi\in\text{Irr}(G)}|\chi(x)|^{2}= \left( \frac{{\sum}_{\chi\in\text{Irr}(G)}|\chi(x)|^{2}}{k(G)}\right)^{k(G)}=1. $$

Therefore, |χ(x)| = 1 for all χ ∈Irr(G). So x ∈ G is a root of unity element.

Conversely, if x is a root of unity, then clearly x is nonvanishing and |C_G(x)| = k(G) by Lemma 2.2(a). □

A consequence of the previous lemma is that if x ∈ G and k(G) > |C_G(x)| or equivalently |x^G| > |G|/k(G), the average of the conjugacy class size of G, then x is a vanishing element of G, that is, χ(x) = 0 for some χ ∈Irr(G). Also, if G has a root of unity element x, then the commuting probability

$$ \text{cp}(G)=\frac{|\{(a,b)\in G\times G: ab=ba\}|}{|G|^{2}}=\frac{k(G)}{|G|} $$

is equal to 1/|x^G|.

In the next two lemmas, we quote some results in Chapter XXII of [1].

Lemma 2.4

Let G be a finite group and suppose that x ∈ G is a root of unity element.

• (a) If x ∈F(G), then F(G) is abelian and $G^{\prime }\leq \mathbf {F}(G)$. In particular, G is abelian or metabelian.

• (b) Conversely, if G is metabelian, then x ∈F(G).

Proof

This is Lemma 1.5 in [1, Chapter XXII]. □

The following is the main result of Chapter XXII in [1]. Recall that the socle of a finite group G, denoted by Soc(G), is a product of all minimal normal subgroups of G.

Lemma 2.5

Let G be a finite metabelian group with trivial center. Then G has a root of unity element if and only if $G\cong {\varGamma }_{q_{1}}\times {\varGamma }_{q_{2}}\times \cdots \times {\varGamma }_{q_{m}}$, where $q_{1},q_{2},\dots , q_{m}$ are prime power > 2. Moreover, if x ∈ G is a root of unity, then $\mathbf {F}(G)=\text {Soc}(G)=G^{\prime }=\langle x^{G}\rangle$ and C_G(x) = F(G).

Proof

The equivalent statements follow from Theorem 1.12 and Corollary 1.11 and the last claim follows from Lemmas 3.8 and 3.14 in [1, Chapter XXII]. □

For a finite group G, recall that F(G), the Fitting subgroup of G, is the largest nilpotent normal subgroup of G. The Fitting series of a finite group G is defined by F₁(G) := F(G) and for any integer i ≥ 1, F_i+ 1(G)/F_i(G) = F(G/F_i(G)). Similarly, the upper central series of G is defined by Z₁(G) := Z(G) and for i ≥ 1, we have Z_i+ 1(G)/Z_i(G) = Z(G/Z_i(G)). The last term of the upper central series of G is called the hypercenter (or hypercentral) of G and is denoted by $\mathbf {Z}_{\infty }(G)$.

The following results are well-known.

Lemma 2.6

Let G be a finite group and let N be a normal subgroup of G such that N ≤Z(G).

• (1) If G/N is nilpotent, then G is nilpotent.

• (2) F(G/N) = F(G)/N.

• (3) F(G/Z_i(G)) = F(G)/Z_i(G) for all i ≥ 1.

Proof

The first two claims are well-known. The last claim follows from the second claim and induction. □

The next result is Corollary 2.3 in [8].

Lemma 2.7

Let G be a finite solvable group and assume that the Sylow 2-subgroups of F_i+ 1(G)/F_i(G) are abelian for 1 ≤ i ≤ 9. Then every nonvanishing element of G lies in F(G).

3 Solvability of Finite Groups with a Root of Unity Element

We first prove Theorem A for finite solvable groups.

Proposition 3.1

Let G be a finite solvable group and suppose that x ∈ G is a root of unity element. Then G is abelian or G is metabelian, x ∈F(G) and both F(G) and G/F(G) are abelian.

Proof

If G is abelian, then we are done. So assume that G is nonabelian. If G is metabelian, then the conclusion follows from Lemma 2.4. Thus we only need to show that G is metabelian. We will prove this by induction on |G|.

Let N be a minimal normal subgroup of G. Let $\overline {G}=G/N$ and use the ‘bar’ notation. By Lemma 2.2 (c), $\bar {x}$ is a root of unity element in $\overline {G}$ and thus by induction, $\overline {G}$ is abelian or metabelian. If $\overline {G}$ is abelian, then G is metabelian. So assume that $\overline {G}$ is metabelian (but G is neither metabelian nor abelian). Then $\bar {x}\in \mathbf {F}(\overline {G})$ and both $\mathbf {F}(\overline {G})$ and $\overline {G}/\mathbf {F}(\overline {G})$ are abelian by Lemma 2.4. Furthermore, N is abelian since it is a minimal normal subgroup and G (and so also N) is solvable. Thus N ≤F(G).

Therefore, G/F(G) is also metabelian. Again by Lemma 2.4 we have that F(G/F(G)) = F₂(G)/F₁(G) and G/F₂(G) are abelian. It follows that F₃(G) = G and F_i+ 1(G)/F_i(G) is abelian for all i ≥ 1. Now by Lemma 2.7, we have x ∈F(G) as x is nonvanishing and so G is metabelian. This contradiction completes the proof. □

The following result follows from the proof of Theorem A in [7].

Lemma 3.2

Let G be a finite group. Assume that G has a unique minimal normal subgroup N. If N is non-abelian and G/N is solvable, then every element in G − N is a vanishing element.

Let n ≥ 2 be an integer and let $\lambda =(\lambda _{1},\lambda _{2},\dots ,\lambda _{r})$ be a partition of n. For 1 ≤ i ≤ k and 1 ≤ j ≤ λ_i, we denote by $h_{i,j}^{\lambda }$ the hook length of the node (i,j) of the Young diagram of λ. Let λ and μ be partitions of n. We use the notation $\chi _{\mu }^{\lambda }$ to denote the value of the irreducible character of S_n labeled by λ evaluated at the conjugacy class with cycle type μ.

Lemma 3.3

Let n ≥ 6 be an integer and let x ∈A_n. Then there exists a partition λ of n which is not self-conjugate such that χ^λ(x)≠ ± 1.

Proof

We will use the following fact which can follow easily from Murnaghan-Nakayama formula. If m ≥ 1 is an integer and $\gamma =(\gamma _{1},\gamma _{2},\dots ,\gamma _{r})$, $\beta =(\beta _{1},\beta _{2},\dots ,\beta _{s})$ are partition of m with $h_{2,1}^{\gamma }<\beta _{1}$ and γ₁ − γ₂ ≥ β₁, then

$$ \chi^{\gamma}_{\beta}=\chi^{(\gamma_{1}-\beta_{1},\gamma_{2},\dots,\gamma_{r})}_{(\beta_{2},\beta_{3},\dots,\beta_{2})}. $$

Let α ⊩ n be the cycle partition of x ∈A_n. Since n ≥ 6, from the proof of Lemma 1.6 in [9] we may assume that all parts of α are distinct, except possibly for the part 1, which may have multiplicity 2.

Write $\alpha =(\alpha _{1},\alpha _{2},\dots ,\alpha _{l})\vdash n$. We consider the following cases.

Case 1: l ≥ 2 and α_l− 1 = α_l. In this case, we have α_l− 1 = α_l = 1, l ≥ 3 (as n ≥ 6) and α_l− 2 > 1.

Assume first that α_l− 2 > 2. Since n ≥ 6, the partition (n − 2,1,1) of n is not self-conjugate and we have that

$$ \chi_{\alpha}^{(n-2,1,1)}=\chi^{(\alpha_{l-2},1,1)}_{(\alpha_{l-2},1,1)}=0. $$

The first equality holds by the observation above and the latter equality holds since α_l− 2 > 2 so

$$ \begin{array}{@{}rcl@{}} h_{1,1}^{(\alpha_{l-2},1,1)}&=&\alpha_{l-2}+2>\alpha_{l-2},\\ h_{1,2}^{(\alpha_{l-2},1,1)}&=&\alpha_{l-2}-1<\alpha_{l-2},\\ h_{2,1}^{(\alpha_{l-2},1,1)}&=&2<\alpha_{l-2}. \end{array} $$

Assume next that α_l− 2 = 2. Then l ≥ 4 and α_l− 3 ≥ 3. Then the partition (n − 2,2) of n is not self-conjugate and by the observation above, we have

$$ \chi_{\alpha}^{(n-2,2)}=\chi^{(2,2)}_{(2,1,1)}=0. $$

Case 2: l ≥ 2 and α_l− 1≠α_l.

Assume first that l ≥ 3 or α_l− 1≠α_l + 1. Then

$$ \chi^{(n-\alpha_{l},1^{\alpha_{l}})}_{\alpha}=\chi^{(\alpha_{l-1},1^{\alpha_{l}})}_{(\alpha_{l-1},\alpha_{l})}=0 $$

as

$$ \begin{array}{@{}rcl@{}} h_{1,1}^{(\alpha_{l-1},1^{\alpha_{l}})}&=&\alpha_{l-1}+\alpha_{l}>\alpha_{l-1},\\ h_{1,2}^{(\alpha_{l-1},1^{\alpha_{l}})}&=&\alpha_{l-1}-1<\alpha_{l-1},\\ h_{2,1}^{(\alpha_{l-1},1^{\alpha_{l}})}&=&\alpha_{l}<\alpha_{l-1}, \end{array} $$

and $(n-\alpha _{l},1^{\alpha _{l}})'=(\alpha _{l}+1,1^{n-\alpha _{l}-1})\neq (n-\alpha _{l},1^{\alpha _{l}})$ as

$$ n-\alpha_{l}\ge \alpha_{l-2}+\alpha_{l-1}\ge 2\alpha_{l}+3>\alpha_{l}+1 $$

if l ≥ 3; and n − α_l ≥ α_l− 1 > α_l + 1 if α_l− 1≠α_l.

Assume next that ℓ = 2 and α₁ = α₂ + 1. Then n = 2α₂ + 1. As n ≥ 6, we have α₂ ≥ 3. Again (n − 2,2)^′≠(n − 2,2).

If α₂ = 3, then n = 7 and $\chi ^{(5,2)}_{(4,3)}=0$. Assume that α₂ ≥ 4. Then

$$ \chi_{\alpha}^{(n-2,2)}=\chi^{(n-2,2)}_{(\alpha_{2}+1,\alpha_{2})}=\chi^{(\alpha_{2}-2,2)}_{(\alpha_{2})}=0 $$

as $h_{1,1}^{(\alpha _{2}-2,2)}=\alpha _{2}-1<\alpha _{2}$.

Case 3: l = 1. Then α = (n). Since n ≥ 6, (n − 2,2) is not self-conjugate and $\chi ^{(n-2,2)}_{(n)}=0$ as $h_{1,1}^{(n-2,2)}=n-1<n$. □

We are now ready to prove Theorem A.

Proof Proof of Theorem A

Let x ∈ G be a root of unity element. If G is solvable, then the theorem follows from Proposition 3.1. Thus it suffices to show that G is solvable. Suppose not and let G be a counterexample to the theorem with |G| minimal. Then x ∈ G is a root of unity but G is non-solvable. Let L = 〈x^G〉. Then $G^{\prime }\leq L\unlhd G$ by Lemma 2.2 (b).

Let N be a minimal normal subgroup of G. By Lemma 2.2 (c), G/N has a root of unity xN. Since |G/N| < |G|, G/N is solvable. As G is non-solvable, N is non-solvable. If G has two distinct minimal normal subgroups, say N₁≠N₂, then N₁ ∩ N₂ = 1 and thus G embeds into G/N₁ × G/N₂, where the latter group is solvable by the argument above. Therefore, G is solvable, which is a contradiction. It follows that G has a unique minimal normal subgroup N, which is non-solvable and G/N is solvable. Hence N = S₁ × S₂⋯ × S_k, where each S_i≅S for some non-abelian simple group S. By Lemma 3.2, x ∈ N as every element in G − N is a vanishing element. It follows from Lemma 2.2 (b) that $G^{\prime }=\langle x^{G}\rangle =N$, and so G/N is abelian.

Write $x=(x_{1},x_{2},\dots ,x_{k})\in N$, where x_i ∈ S_i for 1 ≤ i ≤ k. As G is nonabelian, x is nontrivial and so o(x), the order of x, is divisible by some prime p ≥ 2. Clearly, o(x_i) is divisible by p for some i ≥ 1. Assume that S has an irreducible character 𝜃 of p-defect zero. Then λ = 𝜃 × 𝜃 ×⋯ × 𝜃 ∈Irr(N) has p-defect zero. Clearly, every G-conjugate of λ also has p-defect zero and hence if χ ∈Irr(G) lying over λ, then χ_N is a sum of G-conjugates of λ so that χ(x) = 0 since every conjugate of λ vanishes at x as o(x) is divisible by p. Therefore, we can assume that S has no p-defect zero character. By [3, Corollary 2], one of the following cases holds.

• (i) p = 2 and S is isomorphic to M₁₂, M₂₂, M₂₄, J₂, HS, Suz, Ru, Co₁, Co₃ or A_n for some integer n ≥ 7; or

• (ii) p = 3 and S is isomorphic to Suz, Co₃ or A_n for some integer n ≥ 7.

We make the following observation. Assume that S has a rational-valued irreducible character 𝜃 ∈Irr(S) which is extendible to Aut(S). Then φ = 𝜃 × 𝜃 ×⋯ × 𝜃 ∈Irr(N) extends to χ ∈Irr(G) and

$$ 1=|\chi(x)|=|\varphi(x)|=\prod\limits_{i=1}^{k}|\theta(x_{i})|. $$

Since 𝜃 is rational, 𝜃(x_i) is a non-zero integer and thus |𝜃(x_i)|≥ 1 for all i. The previous equation now implies that |𝜃(x_i)| = 1 for all i.

(a) Assume first that S is one of the sporadic simple groups in (i) but not in (ii). Then x and hence $x_{i}^{\prime }s$ must be a 2-element. Using [2], we can find an irreducible rational-valued character 𝜃 which is extendible to Aut(S) and does not take root of unity values on any 2-elements. So this case cannot occur.

Similarly, if S is one of the sporadic simple groups in (ii), then x and hence $x_{i}^{\prime }s$ are {2,3}-elements. Again, by using [2], we can find an irreducible rational-valued character 𝜃 which is extendible to Aut(S) and does not take root of unity values on any {2,3}-elements.

(b) Assume that S≅A_n, where n ≥ 7 is an integer and that A_n has no block of p-defect zero for p = 2,3 or both.

By the observation above, if λ is a partition of n which is not self-conjugate, then χ^λ, the irreducible character of S_n labeled by λ, remains irreducible upon reduction to A_n and thus |χ^λ(x_i)| = 1. Note that χ^λ is rational-valued. Now Lemma 3.3 provides a contradiction.

Therefore, G must be solvable as wanted. The proof is now complete. □

4 Finite Metabelian Groups with a Root of Unity Element

In this section, we will characterize finite metabelian groups with a root of unity element. Such a group with trivial center was classified by S. Ostrovskaya and E. M. Zhmud’. Recall that if q > 2 is a prime power, then Γ_q is a doubly transitive Frobenius group with a cyclic complement of order q − 1 and degree q. Note that Γ_q has a root of unity element and every root of unity element of Γ_q lies in F(Γ_q), which is an elementary abelian p-group, where q is a power of a prime p. Moreover, if ${\varGamma }={\varGamma }_{q_{1}}\times {\varGamma }_{q_{2}}\times \cdots \times {\varGamma }_{q_{m}}$, where each q_i > 2 are prime powers, then Γ has a root of unity element and furthermore, all Sylow subgroups of Γ are abelian.

Lemma 4.1

Let G be a finite group and let x ∈ G be a root of unity element. Let $K=\mathbf {Z}_{\infty }(G)$ be the hypercenter of G. Assume that G is nonabelian. Then

• (1) $G/K\cong {\varGamma }_{q_{1}}\times {\varGamma }_{q_{2}}\times \cdots \times {\varGamma }_{q_{m}}$, where each q_i > 2 is a prime power and m ≥ 1 is an integer. Moreover, $\mathbf {F}(G)=G^{\prime }K$ is abelian, and C_G/K(xK) = F(G/K) = F(G)/K.

• (2) F(G) = C_G(x) and k(G) = |F(G)| = |C_G(x)|.

• (3) If N = Z_i(G) for some i ≥ 1 or N ≤Z(G), then C_G/N(xN) = F(G/N) = F(G)/N and k(G/N) = |F(G) : N|.

Proof

By Theorem A, x ∈ F := F(G), F is abelian and $G^{\prime }\leq F$. Since $K\unlhd G$ is nilpotent, we have Z(G) ≤ K ≤ F. Now the center of G/K is trivial by the definition of K. Moreover F(G/K) = F/K by Lemma 2.6 (3). Since G/K has a root of unity element xK, part (1) follows from Lemma 2.5.

Since x ∈ F and F is abelian, we have K ≤ F ≤C_G(x). Let $\overline {G}=G/K$. From part (1), we have $\mathbf {C}_{\overline {G}}(\overline {x})=\mathbf {F}(\overline {G})=\overline {F}$. Hence

$$ \overline{F} \leq \overline{\mathbf{C}_{G}(x)} \leq \mathbf{C}_{\overline{G}}(\overline{x})=\overline{F}. $$

Thus $\overline {F}=\overline {\mathbf {C}_{G}(x)}$ and hence F = C_G(x). As k(G) = |C_G(x)| by Lemma 2.2 (a), part (2) follows.

Finally, let N = Z_i(G) for some i ≥ 1 or N ≤Z(G). Then G/N has a root of unity and it is not nilpotent. By Lemma 2.6, F(G/N) = F/N. Now part (3) follows by applying part (2) to G/N. □

Following P. Hall, a finite solvable group G is called an A-group if every Sylow subgroup of G is abelian. The next lemma shows that any finite group with a root of unity element is an A-group.

Proposition 4.2

If a finite group G has a root of unity element, then G is an A-group.

Proof

Let G be a finite group with a root of unity element x ∈ G. Clearly, if G is abelian, then G is an A-group. So, we can assume that G is nonabelian and hence by Theorem A, x ∈ F := F(G) and both F and G/F are abelian so G is solvable. We prove by induction on |G| that all Sylow subgroups of G are abelian.

Notice first that if $1<N\unlhd G$, then G/N has a root of unity element xN and so by induction, every Sylow subgroup of G/N is abelian. Now let N be a minimal normal subgroup of G. Since G is solvable, N is an elementary abelian p-group for some prime p. If Q is a Sylow r-subgroup of G for some prime r≠p, then QN/N≅Q/Q ∩ N≅Q is abelian. Thus it remains to show that every Sylow p-subgroup of G is abelian. Let P ∈Syl_p(G). Then $N\unlhd P$ and P/N is abelian as P/N ∈Syl_p(G/N). Hence $P^{\prime }\leq N$. Now if G has another minimal normal subgroup, say M≠N, then $P^{\prime }\leq N\cap M=1$ and hence P is abelian as wanted if M is also a p-group. If instead M is not a p-group then we can conclude as in the r≠p case that P is abelian. Therefore, we may assume that N is the unique minimal normal subgroup of G.

Since F is abelian and N ≤ F is the unique minimal normal subgroup of G, F must be a p-group and so F = O_p(G). Since P/F is a Sylow p-subgroup of an abelian group G/F, we deduce that $P/F\unlhd G/F$ which implies that $P\unlhd G.$ Hence P ≤O_p(G) = F and thus P = F is abelian. Therefore, G is an A-group as wanted. □

Corollary 4.3

Let G be a finite group and let x ∈ G be a root of unity. Then $\mathbf {Z}(G)=\mathbf {Z}_{\infty }(G)$, that is, G/Z(G) has a trivial center, and $G^{\prime }\cap \mathbf {Z}(G)=1$.

Proof

By Proposition 4.2, G is a finite solvable A-group. The result now follows from (3.8) and Theorem 4.1 in [10]. □

Proof Proof of Theorem B

Let G be a finite group, let Z := Z(G) and F := F(G). Suppose first that x ∈ G is a root of unity element. If G is abelian, then we are done. Assume that G is non-abelian. By Theorem A, x ∈ F, F and G/F are abelian and G is metabelian. By Corollary 4.3, $\mathbf {Z}_{\infty }(G)=\mathbf {Z}(G)$, G/Z has trivial center and $G^{\prime }\cap \mathbf {Z}(G)=1$. Now, the conclusion follows from Lemma 4.1 (1).

For the converse, assume that G is nonabelian. So $F=G^{\prime }Z$ is abelian, $G^{\prime }\cap Z=1$ and $G/Z\cong {\prod }_{i=1}^{m}{\varGamma }_{q_{i}}$ for some integer m ≥ 1 and prime powers q_i > 2. In particular, G is a metabelian group. Write $\overline {G}=G/Z$ and use the ‘bar’ notation. By Lemma 2.5, $\overline {G}$ has a root of unity element $\overline {x}$ for some x ∈ G. Note that the hypothesis above implies that $F=G^{\prime }\times Z$.

We claim that x is also a root of unity element of G, that is, |χ(x)| = 1 for all χ ∈Irr(G). As $|\chi (x)|=|\chi (\overline {x})|=1$ for every χ ∈Irr(G/Z), it suffices to show that if 1≠λ ∈Irr(Z) and χ ∈Irr(G) lying over λ, then |χ(x)| = 1.

Let 1≠λ ∈Irr(Z). Since $F=Z\times G^{\prime }$, $\theta =\lambda \times 1_{G^{\prime }}\in \text {Irr}(F)$ is an extension of λ and $G^{\prime }\leq \text {Ker}(\theta )$. So 𝜃 can be considered an irreducible character of $F/G^{\prime }$ and thus 𝜃 extends to $\phi \in \text {Irr}(G/G^{\prime })$. Thus λ extends to ϕ ∈Irr(G). By Gallagher’s theorem, every χ ∈Irr(G) lying above λ has the form ϕμ for some irreducible character $\mu \in \text {Irr}(\overline {G})$. Since ϕ is linear, we have |ϕ(x)| = 1. We also have $|\mu (x)|=|\mu (\overline {x})|=1$ as $\mu \in \text {Irr}(\overline {G})$ and $\overline {x}$ is a root of unity in $\overline {G}$. Therefore

$$ |\chi(x)|=|\phi(x)\mu(x)|=|\phi(x)|\cdot|\mu(x)|=1, $$

hence x is a root of unity element of G. □