Proof. Let $\mathcal {A}=\{\chi \in {{\operatorname {Irr}}}(G)\mid \chi \ \text {vanishes at some class in}\ \mathcal {C}\}$ . We want to see that if $\alpha ,\beta \in \mathcal {A}$ , then there exists a path in $\Gamma _v(G)$ joining $\alpha $ and $\beta $ . Let $C,D\in \mathcal {C}$ such that $\alpha (C)=\beta (D)=0$ . Since $\mathcal {C}$ is a connected component in $\Delta _v(G)$ , there exists a path

$$ \begin{align*} C=C_0\leftrightarrow C_1\leftrightarrow\cdots\leftrightarrow C_{d-1}\leftrightarrow C_d=D \end{align*} $$

joining C and D. By the definition of the graph $\Delta _v(G)$ , this means that there exist $\chi _i\in {{\operatorname {Irr}}}(G)$ such that $\chi _i(C_{i-1})=\chi _i(C_i)=0$ for $i=1,\ldots ,d$ . Therefore,

$$ \begin{align*} \alpha\leftrightarrow\chi_1\leftrightarrow\cdots\leftrightarrow\chi_d\leftrightarrow\beta \end{align*} $$

is a path in $\Gamma _v(G)$ joining $\alpha $ and $\beta $ . This proves that $\mathcal {A}$ is contained in a connected component of $\Gamma _v(G)$ .

Similarly, we can prove that if $\mathcal {A}$ is the set of characters in a connected component in $\Gamma _v(G)$ , then $\mathcal {D}=\{K\in {{\operatorname {cl}}}(G)\mid \,\chi (K)=0\ \text {for some}\ \chi \in \mathcal {A}\}$ is contained in a connected component in $\Delta _v(G)$ .

Now, suppose that $\mathcal {A}\subsetneq \mathcal {B}$ , where $\mathcal {B}$ is a connected component of $\Gamma _v(G)$ . Let $\gamma \in \mathcal {B}-\mathcal {A}$ . Therefore, $\gamma (K)\neq 0$ for every $K\in \mathcal {C}$ . Since $\alpha ,\gamma \in \mathcal {A}$ , there exists a path in $\Gamma _v(G)$ joining $\alpha $ and $\gamma $ . This means that there exist $\eta ,\mu $ in this path that are linked; one of them belongs to $\mathcal {A}$ and the other does not. Suppose that $\eta \in \mathcal {A}$ and $\mu \not \in \mathcal {A}$ . Then, there exists $D\not \in \mathcal {C}$ such that $\eta (D)=\mu (D)=0$ . Since $\eta \in \mathcal {A}$ , there exists $C\in \mathcal {C}$ such that $\eta (C)=0$ . By the definition of $\Delta _v(G)$ , C and D are linked by means of $\eta $ . This is a contradiction.

We omit the proofs of the remaining parts, which are similar.

Proof. Let $\theta \in {{\operatorname {Irr}}}(N)$ lie under $\chi $ and let $T=I_G(\theta )$ . By Clifford’s correspondence, there exists $\psi \in {{\operatorname {Irr}}}(T)$ such that $\psi ^G=\chi $ . By the formula for the induced character, $\chi $ vanishes on $G-\bigcup _{g\in G}T^g$ . Since a group is never the union of the conjugates of a proper subgroup, this implies that $\theta $ is G-invariant. Thus, $\chi _N=e\theta $ for some positive integer e and we want to see that $e=1$ . Arguing by contradiction, assume that $e>1$ .

Let $(G^*,N^*,\theta ^*)$ be a character triple isomorphic to $(G,N,\theta )$ with $N^*\leq {\mathbf Z}(G^*)$ and $\theta ^*$ linear and faithful. (We refer the reader to Section 5.4 and in particular Corollary 5.9 of [Reference Navarro33] for background on character triples.) Let $\chi ^\ast \in {{\operatorname {Irr}}}(G^\ast \mid \theta ^\ast )$ correspond to $\chi $ under the isomorphism. Note that $\chi ^*$ is not linear and $\chi ^*_{N^*}=e\theta ^*$ . Since $\theta ^*$ is linear, ${{\operatorname {Van}}}(\chi ^*)\cap N^*=\emptyset $ . Therefore, there exists $x^*\in G^*-N^*$ such that $\chi ^*(x^*)=0$ . Let $x\in G$ such that

$$ \begin{align*}(Nx)^\ast=N^\ast x^\ast,\end{align*}$$

where $^\ast :G/N\longrightarrow G^\ast /N^\ast $ is the associated group isomorphism. Note that $x\not \in N$ . By [Reference Navarro33, Lemma 5.17(a)], there exists an algebraic integer $\alpha $ such that

$$ \begin{align*}\chi(x)=\alpha\chi^\ast(x^\ast)=0,\end{align*} $$

contradicting the hypothesis that ${{\operatorname {Van}}}(\chi )\subseteq N$ .

Proof. This is [Reference Isaacs, Navarro and Wolf13, Theorem B].

Proof. If $\psi (\sigma )\not \in \{\chi (\sigma ),\chi (\sigma )/2\}$ , $\chi $ is labeled by the partition $\lambda $ , and $\mu $ is the cycle partition of $\sigma $ , then by [Reference James and Kerber14, Theorems 2.5.7 and 2.5.13] $\lambda $ is self-conjugated and $\mu $ is the partition consisting of the diagonal hook-lengths of $\lambda $ . However, in this case, $\chi (\sigma )\not =0$ by [Reference James and Kerber14, Corollary 2.4.8] and $\psi (\sigma )\not =0$ by [Reference James and Kerber14, Theorem 2.5.13].

Proof. If $\chi =\chi ^{(n-2,2)}$ and $\psi =\chi ^{(n-2,1^2)}$ (up to exchange or multiplying with $\mathrm {sgn}$ ), then by [Reference James15, page 93], we have that $\psi \equiv \chi +1\text { }(\mathrm {mod}\ 2)\,$ . So in this case, $\chi $ and $\psi $ cannot have common zeros. So we may now assume that this is not the case.

The cases $n=8$ and $9$ can be checked by looking at character tables. So assume now that $n\geq 10$ .

Let $\lambda $ and $\mu $ be the partitions labeling $\chi $ and $\psi $ , respectively, so that $\chi =\chi ^\lambda $ and $\psi =\chi ^\mu $ . Note that $\lambda ,\mu \not \in \{(n),(1^n)\}$ since $\chi $ and $\psi $ have degree larger than 1. Further, for any partition $\gamma \vdash n$ , let $\tau _\gamma \in \mathsf {S}_n$ have cycle partition $\gamma $ .

Case 1: n is even. We have that $\tau _{(n-k,k)}\in \mathsf {A}_n$ for any $1\leq k\leq n/2$ . For $1\leq k\leq 4$ , let $N_{n,k}:=\{\gamma \vdash n\mid \chi ^\gamma (\tau _{(n-k,k)})=0\}$ be the set of partitions labeling characters that do not vanish on $\tau _{(n-k,k)}$ . We then have that $N_{n,k}=B_{n,k}\cup C_{n,k}\cup D_{n,k}$ with

$$ \begin{align*} B_{n,1}&=\{(n),(1^n)\},\\C_{n,1}&=\emptyset,\\D_{n,1}&=\{(n-h,2,1^{h-2})\mid 2\leq h\leq n-2\},\\B_{n,2}&=\{(n),(n-1,1),(2,1^{n-2}),(1^n)\},\\C_{n,2}&=\{(n-2,2),(2^2,1^{n-4})\},\\D_{n,2}&=\{(n-h,3,1^{h-3})\mid 3\leq h\leq n-3\}\cup\{(n-h,2^2,1^{h-4})\mid 4\leq h\leq n-2\},\\B_{n,3}&=\{(n),(n-1,1),(n-2,1^2),(3,1^{n-3}),(2,1^{n-2}),(1^n)\},\\C_{n,3}&=\{(n-3,3),(n-3,2,1),(n-4,2^2),(3^2,1^{n-6}),(3,2,1^{n-5}),(2^3,1^{n-6})\},\\D_{n,3}&=\{(n-h,4,1^{h-4})\mid 4\leq h\leq n-4\}\cup\{(n-h,3,2,1^{h-5})\mid 5\leq h\leq n-3\}\\&\quad\cup\{(n-h,2^3,1^{h-6})\mid 6\leq h\leq n-2\},\\B_{n,4}&=\{(n),(n-1,1),(n-2,1^2),(n-3,1^3),(4,1^{n-4}),(3,1^{n-3}),(2,1^{n-2}),(1^n)\},\\C_{n,4}&=\{(n-4,4),(n-4,3,1),(n-4,2,1^2),(n-5,3,2),(n-5,2^2,1),(n-6,2^3),\\&\quad(4^2,1^{n-8}),(4,3,1^{n-7}),(4,2,1^{n-6}),(3^2,2,1^{n-8}),(3,2^2,1^{n-7}),(2^4,1^{n-8})\},\\D_{n,4}&=\{(n-h,5,1^{h-5})\mid 5\leq h\leq n-5\}\cup\{(n-h,4,2,1^{h-6})\mid 6\leq h\leq n-4\}\\&\quad\cup\{(n-h,3,2^2,1^{h-7})\mid 7\leq h\leq n-3\}\cup\{(n-h,2^4,1^{h-8})\mid 8\leq h\leq n-2\}. \end{align*} $$

To see this, note that in view of the Murnaghan–Nakayama formula (see [Reference James and Kerber14, Section 2.4.7]), if $\gamma \in N_{n,k}$ , then $\gamma $ is obtained by adding a $(n-k)$ -hook to a hook-partition of k and that this condition is also sufficient for $\gamma \in N_{n,k}$ since $k\leq 4<n/2$ (so that there is at most one way in which hooks of the given length can be recursively removed from any given partition of n).

So if $\chi $ and $\psi $ have no common zero in $\mathsf {A}_n$ , then $\{\lambda ,\mu \}\cap N_{n,k}\not =\emptyset $ for any $1\leq k\leq 4$ . Note that the sets $D_{n,k}$ are pairwise disjoint. The same is true for the sets $C_{n,k}$ since $n\geq 10$ . Assume first that neither $\lambda $ nor $\mu $ is in

$$ \begin{align*}\cup B_{n,k}\setminus\{(n),(1^n)\}=\{(n-1,1),(n-2,1^2),(n-3,1^3),(4,1^{n-4}),(3,1^{n-3}),(2,1^{n-2})\}.\end{align*} $$

Then $\lambda \in C_{n,k_1}\cap D_{n,k_2}$ and $\mu \in C_{n,k_3}\cap D_{n,k_4}$ for some pairwise different $1\leq k_i\leq 4$ . In particular, $\lambda ,\mu \in \bigcup _{k=1}^4 C_{n,k}$ . In each of these cases, it can be checked that $\chi ^\lambda (\tau _{(n-5,5)})=\chi ^\mu (\tau _{(n-5,5)})=0$ .

Next assume that $\lambda\hspace{-0.5pt} \in\hspace{-0.5pt} \{(n-3,1^3),(4,1^{n-4})\}$ . Then $\mu\hspace{-0.5pt} \in\hspace{-0.5pt} N_{n,1}\hspace{-0.5pt}\cap\hspace{-0.5pt} N_{n,2}\hspace{-0.5pt}\cap\hspace{-0.5pt} N_{n,3}\hspace{-0.5pt}=\hspace{-0.5pt}\{(n), (1^n)\}$ , leading to a contradiction.

If $\lambda \in \{(n-2,1^2),(3,1^{n-3})\}$ , then

$$ \begin{align*}\mu\in N_{n,1}\cap N_{n,2}=\{(n),(n-2,2),(2^2,1^{n-4}),(1^n)\},\end{align*} $$

so that $\mu \in \{(n-2,2),(2^2,1^{n-4})\}$ , which had been excluded at the beginning of the proof.

If $\lambda \in \{(n-1,1),(2,1^{n-2})\}$ , then $\mu \in N_{n,1}\setminus \{(n),(1^n)\}$ and so $\mu =(n-h,2,1^{h-2})$ for some $2\leq h\leq n-2$ . In this case, we have that $\chi ^\lambda (\tau _\gamma )=\chi ^\mu (\tau _\gamma )=0$ for some $\gamma \in \{(n-5,2^2,1), (n-6,3,2,1)\}$ .

Case 2: n is odd. In this case, $\tau _{(n)},\tau _{(n-k,k-1,1)}\in \mathsf {A}_n$ for $2\leq k\leq (n+1)/2$ . Let $N_{n,0}:=\{\gamma \vdash n\mid \chi ^\gamma (\tau _{(n)})=0\}$ and for $2\leq k\leq 4$ , let $N_{n,k}:=\{\gamma \vdash n\mid \chi ^\gamma (\tau _{(n-k,k-1,1)})=0\}$ . We can write $N_{n,k}=B_{n,k}\cup C_{n,k}\cup D_{n,k}$ with

$$ \begin{align*} B_{n,0}&=\emptyset,\\C_{n,0}&=\emptyset,\\D_{n,0}&=\{(n-h,1^h)\mid 0\leq h\leq n-1\},\\B_{n,2}&=\{(n),(n-2,2),(2^2,1^{n-4}),(1^n)\},\\C_{n,2}&=\{(n-1,1),(2,1^{n-2})\},\\D_{n,2}&=\{(n-h,3,1^{h-3})\mid 3\leq h\leq n-3\}\cup\{(n-h,2^2,1^{h-4})\mid 4\leq h\leq n-2\},\\B_{n,3}&=\{(n),(n-3,2,1),(3,2,1^{n-5}),(1^n)\},\\C_{n,3}&=\{(n-2,1^2),(n-4,2^2),(3^2,1^{n-6}),(3,1^{n-3})\},\\D_{n,3}&=\{(n-h,4,1^{h-4})\mid 4\leq h\leq n-4\}\cup\{(n-h,2^3,1^{h-6})\mid 6\leq h\leq n-2\},\\B_{n,4}&=\{(n),(n-2,2),(n-3,3),(2^3,1^{n-6}),(2^2,1^{n-4}),(1^n)\},\\C_{n,4}&=\{(n-3,1^3),(n-4,2,1^2),(n-5,2^2,1),(n-6,2^3),(4^2,1^{n-8}),(4,3,1^{n-7}),\\&\ \quad(4,2,1^{n-6}),(4,1^{n-4})\},\\D_{n,4}&=\{(n-h,5,1^{h-5})\mid 5\leq h\leq n-5\}\cup\{(n-h,3^2,1^{h-6})\mid 6\leq h\leq n-3\}\\&\quad\cup\{(n-h,2^4,1^{h-8})\mid 8\leq h\leq n-2\}. \end{align*} $$

This can be seen again by noting that if $\gamma \in N_{n,k}$ , then $\gamma $ can be obtained by adding an $(n-k)$ -hook to a partition $\overline {\gamma }\vdash k$ . Further, if $2\leq k\leq 4$ , then $\chi ^{\overline {\gamma }}(\tau _{(k-1,1)})\not =0$ . These conditions are again sufficient for $\gamma \in N_{n,k}$ since $k<n/2$ .

Again the sets $D_{n,k}$ are pairwise disjoint; the same holds for the sets $C_{n,k}$ since $n\geq 11$ and we may assume that $\{\lambda ,\mu \}\cap N_{n,k}\not =\emptyset $ for $k\in \{0,2,3,4\}$ . Assume first that neither $\lambda $ nor $\mu $ is contained in $\cup B_{n,k}$ . Then, similarly to Case 1, $\lambda ,\mu \in \cup C_{n,k}$ . In particular, $\chi (\tau _{(n-5,4,1)})=0$ and $\psi (\tau _{(n-5,4,1)})=0$ . So we may now assume that $\lambda \in (\cup B_{n,k})\setminus \{(n),(1^n)\}$ .

If $\lambda \in \{(n-2,2),(2^2,1^{n-2})\}$ , then $\mu \in N_{n,0}\cap N_{n,3}$ and then $\mu \in \{(n-2,1^2),(3,1^{n-3})\}$ , which had been excluded at the beginning of the proof.

If $\lambda \in \{(n-3,3),(2^3,1^{n-6})\}$ , then again $\mu \in N_{n,0}\cap N_{n,3}$ , so $\mu \in \{(n-2,1^2),(3,1^{n-3})\}$ and then $\chi ^\lambda (\tau _{(n-5,4,1)})=\chi ^\mu (\tau _{(n-5,4,1)})=0$ .

If $\lambda \in \{(n-3,2,1),(3,2,1^{n-5})\}$ , then $\mu \in N_{n,0}\cap N_{n,2}\cap N_{n,4}=\{(n),(1^n)\}$ , leading to a contradiction.

Proof of Theorem A

We check that $\chi (1)=n(n-3)/2$ if and only if $\chi \in \{\chi ^{(n-2,2)},\chi ^{(2^2,1^{n-4})}\}$ and that $\chi (1)=(n-1)(n-2)/2$ if and only if $\chi \in \{\chi ^{(n-2,1^2)},\chi ^{(3,1^{n-3})}\}$ . The theorem then follows by Theorem 3.2.

The ‘if’ parts are easily checked using the hook formula. The ‘only if’ parts for $n=8$ are also easily checked. So assume now that $n\geq 9$ . Note that $\chi ^{(n)}(1),\chi ^{(1^n)}(1)=1$ and $\chi ^{(n-1,1)}(1),\chi ^{(2,1^n)}(1)=n-1$ . We show that $\chi ^\lambda (1)>(n-1)(n-2)/2$ for any $\lambda \vdash n$ with $\lambda _1,\lambda _1'\leq n-3$ (with $\lambda '$ the partition that is conjugated to $\lambda $ ), which concludes the proof of the ‘only if’ parts.

For $n{\kern-1pt}={\kern-1pt}9$ and $10$ , this can easily be checked by looking at character tables. So assume that $n\geq 11$ and that the claim holds for $n-1$ and $n-2$ . Then by induction, $\chi ^\mu (1)\geq (n-1)(n-4)/2$ for any $\mu \vdash n-1$ with $\mu _1,\mu _1'\leq n-3$ and $\chi ^\nu (1)>(n-3)(n-4)/2$ for any $\nu \vdash n-2$ with $\nu _1,\nu _1'\leq n-5$ . If $\lambda $ has at least two removable nodes A and B, then, by the branching rule,

$$ \begin{align*}\chi^\lambda(1)\geq\chi^{\lambda\setminus\{A\}}(1)+\chi^{\lambda\setminus\{B\}}(1)\geq (n-1)(n-4)>(n-1)(n-2)/2.\end{align*} $$

If $\lambda $ has only one removable node, then $\lambda =(a^b)$ for some $a,b$ with $ab=n$ . Further, $2\leq a,b\leq n/2<n-5$ . So

$$ \begin{align*}\chi^\lambda(1)= \chi^{(a^{b-2},(a-1)^2)}(1)+\chi^{(a^{b-1},a-2)}(1)>(n-3)(n-4)>(n-1)(n-2)/2,\end{align*} $$

which concludes the proof.

Proof. For $n\leq 9$ , the result can be proved looking at character tables. So assume from now on that $n\geq 10$ .

Let $\tau $ be the cycle partition of $\sigma $ . If $\tau _1\geq 4$ , then there exists a $\tau _1$ -core $\lambda $ of n by [Reference Granville and Ono9, Theorem 1]. In particular, $\chi ^\lambda (\sigma )=0$ in view of the Murnaghan–Nakayama formula. If $\tau _1\leq n-4$ , then neither $(n-2,2)$ nor $(2^2,1^{n-4})$ is a $\tau _1$ -core. If $\tau _1\geq n-3$ , then we can take $\lambda =(n-5,5)$ as a $\tau _1$ -core.

So we may now assume that $\tau =(3^a,2^b,1^c)$ for some $a,b,c\geq 0$ with $3a+2b+c=n$ . Taking $\lambda =(n-4,2,1^2)$ , $(n-1,1)$ , or $(n-3,1^2)$ depending on whether n is congruent to 0, 1, or 2 modulo 3, respectively, we may assume that $n-3a\geq 4$ (since for this choice of $\lambda $ , we have that $|\lambda _{(3)}|\geq 4$ , so that $\chi ^\lambda (\sigma )=0$ if $n-3a\leq 3$ ).

If b is odd, then $\chi ^\lambda (\tau )=0$ for any $\lambda =\lambda '$ , in particular for $\lambda =(n/2,2,1^{n/2-2})$ or $((n+1)/2,1^{(n-1)/2})$ depending on the parity of n, since then $\chi ^\lambda =\chi ^\lambda \cdot \mathrm {sgn}$ . So we may assume that b is even.

Using the determinantal formula, we have that

$$ \begin{align*}\chi^{(n-2,2)}=1{\uparrow}_{\mathsf{S}_{n-2,2}}^{\mathsf{S}_n}-1{\uparrow}_{\mathsf{S}_{n-1}}^{\mathsf{S}_n},\end{align*} $$

where $\mathsf {S}_{n-2,2}\cong \mathsf {S}_{n-2}\times \mathsf {S}_2$ is a maximal Young subgroup. We now show that $\chi ^{(n-2,2)}(\sigma )\not =0$ . Multiplying with $\mathrm {sgn}$ , this also gives $\chi ^{(2^2,1^{n-4})}(\sigma )\not =0$ , so that the lemma follows.

By the above formulas, we have that

$$ \begin{align*}\chi^{(n-2,2)}(\sigma)=b+\binom{c}{2}-c=b+\frac{c(c-3)}{2}.\end{align*} $$

So $\chi ^{(n-2,2)}(\sigma )\not =0$ unless $(b,c)\in \{(0,0),(0,3),(1,1),(1,2)\}$ . Each of these choices contradicts $2b+c=n-3a\geq 4$ or b even.

Proof of Theorem 3.3

For $n=7$ or $8$ , the theorem can be easily checked. So assume that $n\geq 9$ . By Theorem 3.2, we have that $\Gamma _v(\mathsf {S}_n)$ is connected with diameter 2. Further if $g,h\in \mathsf {S}_n$ are vanishing elements, then by Lemma 3.4, there are $\chi ,\psi \in {{\operatorname {Irr}}}(\mathsf {S}_n)\setminus \{\chi ^{(n-2,2)},\chi ^{(2^2,1^{n-4})}\}$ with $\chi (g)=0$ and $\psi (h)=0$ . Since $\chi $ and $\psi $ have a common zero, it follows that $\Delta _v(\mathsf {S}_n)$ is connected and that it has diameter at most 2.

Let now $\chi ,\psi \in {{\operatorname {Irr}}}(\mathsf {A}_n)$ have degrees larger than 1. Then $\chi $ and $\psi $ have a common zero by Lemma 3.1 and Theorem 3.2 if and only if $\{\chi ,\psi \}\not =\{(\chi ^{(n-2,2)})_{\mathsf {A}_n},(\chi ^{(n-2,1^2)})_{\mathsf {A}_n}\}$ (note that $(\chi ^{(n-2,2)})_{\mathsf {A}_n}$ and $(\chi ^{(n-2,1^2)})_{\mathsf {A}_n}$ are both irreducible by [Reference James and Kerber14, Theorem 2.5.7]). Further, any vanishing element $g\in \mathsf {A}_n$ is a zero of some irreducible character $\not =(\chi ^{(n-2,2)})_{\mathsf {A}_n}$ by Lemmas 3.1 and 3.4. We can then conclude as in the $\mathsf {S}_n$ case.

Proof. Let $\chi ,\psi \in {{\operatorname {Irr}}}(G)$ be nonlinear. Since nilpotent groups are monomial and maximal subgroups of nilpotent groups are normal, we have that there exist normal maximal subgroups M and N of G such that $\chi $ is induced from M and $\psi $ is induced from N. By the character-induction formula, both characters vanish outside $M\cup N$ . Since $G\neq M\cup N$ by comparing orders, the result follows.

Proof. This follows from the proof of [Reference Manz and Wolf26, Lemma 18.1].

Proof. By Lemma 4.2, all the classes in $G-{\mathbf F}(G)$ are linked. Let $\Delta _1$ be the connected component containing these classes. Suppose first that ${\mathbf F}(G)$ is abelian. Then Lemma 2.2 implies that for any $\chi \in {{\operatorname {Irr}}}(G)$ nonlinear, ${{\operatorname {Van}}}(\chi )\not \subseteq {\mathbf F}(G)$ . Therefore, all the vanishing classes are linked to some class in $G-{\mathbf F}(G)$ , which implies that $\Delta _v(G)$ is connected.

Suppose now that ${\mathbf F}(G)$ is not abelian and that $\Delta _v(G)$ is not connected. Let $\Delta _1,\ldots ,\Delta _t$ be the connected components of the graph, where $\Delta _1$ contains all classes in $G-{\mathbf F}(G)$ . So all the classes in $\Delta _2,\ldots ,\Delta _t$ are contained in ${\mathbf F}(G)$ . Let y be a representative of a class in one of these components. Then, y is a vanishing element and for every $\chi \in {{\operatorname {Irr}}}(G)$ satisfying $\chi (y)=0$ ,

$$ \begin{align*}\chi_{{\mathbf F}(G)}\in{{\operatorname{Irr}}}({\mathbf F}(G)),\end{align*} $$

by Lemma 2.2. In particular, y is a vanishing element of ${\mathbf F}(G)$ , that is, all the classes in $\Delta _2,\ldots ,\Delta _t$ are contained in ${\mathbf F}(G)-{\mathbf Z}({\mathbf F}(G))$ . Suppose that $y_1,y_2\in {\mathbf F}(G)-{\mathbf Z}({\mathbf F}(G))$ lie in vanishing classes. Let $\chi _i\in {{\operatorname {Irr}}}(G)$ such that $\chi _i(y_i)=0$ . Then,

$$ \begin{align*}\varphi_i:=(\chi_i)_{{\mathbf F}(G)}\in{{\operatorname{Irr}}}({\mathbf F}(G)).\end{align*} $$

By Theorem 4.1, there exists $t\in {\mathbf F}(G)$ such that $\varphi _i(t)=0$ for $i=1,2$ . However, then $y_1$ and t are linked by means of $\chi _1$ , and t and $y_2$ are linked by means of $\chi _2$ . Thus, $y_1$ and $y_2$ belong to the same connected component. It follows that all the vanishing classes in ${\mathbf F}(G)$ belong to the same connected component. The result follows.

Proof. We may assume that $n>1$ . Set $F:={\mathbf F}_{n-1}(G)$ and let $Z/{\mathbf F}_{n-1}(G)={\mathbf Z}(G/{\mathbf F}_{n-1}(G))$ . By Lemma 2.3 and Theorem 4.1, we know that all the classes in $G-Z$ belong to the same connected component $\Delta _1$ . Since $Z/F$ is abelian, Lemma 4.2 applied to $Z/{\mathbf F}_{n-2}(G)$ implies that there exists $\varphi \in {{\operatorname {Irr}}}(Z)$ that vanishes on all the classes in $Z-F$ . Therefore, the same holds for any $\chi \in {{\operatorname {Irr}}}(G|\varphi )$ . In particular, all the classes in $Z-F$ belong to the same connected component $\Delta _2$ (possibly $\Delta _1=\Delta _2$ ). We claim that any vanishing element $x\in F$ belongs to either $\Delta _1$ or $\Delta _2$ . Let $\chi \in {{\operatorname {Irr}}}(G)$ such that $\chi (x)=0$ . We may assume that ${{\operatorname {Van}}}(\chi )\subseteq F$ . By Lemma 2.2, $\chi _F\in {{\operatorname {Irr}}}(F)$ . Let $\psi \in {{\operatorname {Irr}}}(G/F)$ be nonlinear. By Gallagher’s theorem [Reference Isaacs12, Corollary 6.17], $\chi \psi \in {{\operatorname {Irr}}}(G)$ vanishes at both x and some element in $G-F$ . This proves the claim. Thus, $\Delta _1$ and $\Delta _2$ are all the connected components.

Proof. By Lemma 2.1, it suffices to show that $\Delta _v(G)$ has at most two connected components. Let n be the Fitting height of G. By Theorem 4.1, we may assume that $n>1$ . By Lemma 4.4, we may assume that $G/{\mathbf F}_{n-1}(G)$ is abelian. By Lemma 4.2, all the classes in $G-{\mathbf F}_{n-1}(G)$ belong to the same connected component, say $\Delta _1$ . By Lemma 4.3, we may assume that $n\geq 3$ .

Suppose first that ${\mathbf F}_{n-1}(G)/{\mathbf F}_{n-2}(G)$ is abelian. Using Lemma 4.2 again, the classes in ${\mathbf F}_{n-1}(G)-{\mathbf F}_{n-2}(G)$ belong to the same connected component, say $\Delta _2$ (possibly $\Delta _2=\Delta _1$ ). Suppose that $x\in {\mathbf F}_{n-2}(G)$ is a vanishing element. We want to see that the class of x belongs to $\Delta _1$ or $\Delta _2$ . Suppose not. Then, if $\chi \in {{\operatorname {Irr}}}(G)$ is such that $\chi (x)=0$ , we have that $\chi $ does not have any zeros in $G-{\mathbf F}_{n-2}(G)$ . By Lemma 2.2, it follows that $\chi _{{\mathbf F}_{n-2}(G)}\in {{\operatorname {Irr}}}({\mathbf F}_{n-2}(G))$ . Using Gallagher’s theorem, we deduce that $\chi \psi \in {{\operatorname {Irr}}}(G)$ for every $\psi \in {{\operatorname {Irr}}}(G/{\mathbf F}_{n-2}(G))$ . We can assume that $\psi $ is nonlinear. Since this character has zeros on $G-{\mathbf F}_{n-2}(G)$ , we have a contradiction.

Finally, we may assume that ${\mathbf F}_{n-1}(G)/{\mathbf F}_{n-2}(G)$ is not abelian. In this case, we prove that $\Gamma _v(G)$ has at most two connected components. Let $\Phi $ be the (normal) subgroup of G such that

$$ \begin{align*}\Phi/{\mathbf F}_{n-2}(G)=\Phi(G/{\mathbf F}_{n-2}(G)).\end{align*} $$

By Lemma 4.2 applied to $G/\Phi $ , there exists $\lambda \in {{\operatorname {Irr}}}({\mathbf F}_{n-1}(G)/\Phi )$ such that

$$ \begin{align*}\psi:=\lambda^G\in{{\operatorname{Irr}}}(G).\end{align*} $$

Clearly, by Lemma 2.2, all the irreducible characters of G whose restriction to ${\mathbf F}_{n-1}(G)$ is not irreducible belong to the connected component of $\psi $ . Now we claim that all the characters in

$$ \begin{align*}\mathcal{A}= \{\chi\in{{\operatorname{Irr}}}(G)\mid\chi_{\Phi}\in{{\operatorname{Irr}}}(\Phi),\chi(1)>1\}\end{align*} $$

belong to the connected component of $\psi $ and that all the characters in

$$ \begin{align*}\mathcal{B}=\{\chi\in{{\operatorname{Irr}}}(G)\mid\chi_{{\mathbf F}_{n-1}}(G)\in{{\operatorname{Irr}}}({\mathbf F}_{n-1}(G)), \chi_{\Phi}\not\in{{\operatorname{Irr}}}(\Phi)\}\end{align*} $$

belong to the same connected component. The result follows.

Let $\chi \in \mathcal {A}$ . By the definition of $\mathcal {A}$ , $\chi _{\Phi }\in {{\operatorname {Irr}}}(\Phi )$ . By Gallagher again, $\chi \psi \in {{\operatorname {Irr}}}(G)$ , which implies that $\chi $ and $\chi \psi $ are linked in $\Gamma _v(G)$ . Since $\chi \psi $ vanishes on $G-{\mathbf F}_{n-1}(G)$ , we deduce that $\chi \psi $ is linked to $\psi $ . Therefore, there is a path of length $2$ joining $\chi $ and $\psi $ . We have thus seen that the characters in $\mathcal {A}$ and $\psi $ are in the same connected component.

It remains to prove that the characters in $\mathcal {B}$ belong to the same connected component. First, we see that for any $\chi \in \mathcal {B}$ , there exists $U_{\chi }\trianglelefteq G$ such that $\Phi \leq U_{\chi }<{\mathbf F}_{n-1}(G)$ and $\chi $ vanishes on ${\mathbf F}_{n-1}(G)-U_{\chi }$ . Let $\varphi \in {{\operatorname {Irr}}}(\Phi )$ lie under $\chi $ . Suppose that $\varphi $ is ${\mathbf F}_{n-1}(G)$ -invariant, so that $({\mathbf F}_{n-1}(G),\Phi ,\varphi )$ is a character triple with ${\mathbf F}_{n-1}(G)/\Phi ={\mathbf F}(G/\Phi )$ abelian by Gaschütz’s theorem [Reference Manz and Wolf26, Theorem 1.12]. The existence of $U_{\chi }$ follows from [Reference Wolf37, Lemma 2.2]. Now, we assume that $\varphi $ is not ${\mathbf F}_{n-1}(G)$ -invariant. Let $T:=I_G(\varphi )$ and note that $G=T{\mathbf F}_{n-1}(G)$ (because $\chi $ restricts irreducibly to ${\mathbf F}_{n-1}(G)$ ). In particular, $T\cap {\mathbf F}_{n-1}(G)\trianglelefteq G$ (because $T\cap {\mathbf F}_{n-1}(G)\trianglelefteq T$ , ${\mathbf F}_{n-1}(G)/\Phi $ is abelian and T contains $\Phi $ , so $T\cap {\mathbf F}_{n-1}(G)$ is also normal in ${\mathbf F}_{n-1}(G)$ ). Set $U_{\chi }=T\cap {\mathbf F}_{n-1}(G)$ and note that it satisfies the properties that we want.

Therefore, if $\chi ,\psi \in \mathcal {B}$ , then both characters vanish on ${\mathbf F}_{n-1}(G)-(U_{\chi }\cup U_{\psi })$ . Since

$$ \begin{align*}U_{\chi}\cup U_{\psi}\subsetneq{\mathbf F}_{n-1}(G),\end{align*} $$

we conclude that all the characters in $\mathcal {B}$ are linked. This completes the proof.

Proof. By [Reference Yang38, Theorem 5.2], there exists $\mu \in {{\operatorname {Irr}}}({\mathbf F}_8(G))$ such that $\chi =\mu ^G\in {{\operatorname {Irr}}}(G)$ . In particular, $\chi $ vanishes on $G-{\mathbf F}_8(G)$ . Thus, all classes in $G-{\mathbf F}_8(G)$ belong to the same connected component $\Delta _1$ . Furthermore, they are linked. Let $x\in {{\operatorname {Van}}}(G)$ and assume that x is not linked to any class in $G-{\mathbf F}_8(G)$ . Let $\psi \in {{\operatorname {Irr}}}(G)$ such that $\psi (x)=0$ . Therefore, ${{\operatorname {Van}}}(\psi )\subseteq {\mathbf F}_8(G)$ . Lemma 2.2 implies that $\psi _{{\mathbf F}_8(G)}\in {{\operatorname {Irr}}}({\mathbf F}_8(G))$ . Thus, $\psi \gamma \in {{\operatorname {Irr}}}(G)$ for every nonlinear $\gamma \in {{\operatorname {Irr}}}(G/{\mathbf F}_8(G))$ . This character vanishes both at x and at some element in $G-{\mathbf F}_8(G)$ . This is a contradiction.

Suppose now that $|G|$ is odd. By [Reference Moretó and Wolf30, Theorem D], there exists $\chi \in {{\operatorname {Irr}}}(G)$ such that $\chi $ vanishes on $G-{\mathbf F}_3(G)$ . The result follows by similar arguments as above.

Proof. As mentioned above, we may assume that $G=\mathrm {P}\Omega _{2n}^+(q)$ with $n\geq 4$ , and we aim to prove that $\Gamma _v(G)$ is connected. Let $G_{sc}$ be the corresponding finite reductive group of simply connected type, so that $G_{sc}=\mathrm {Spin}_{2n}(q)$ is the full covering group of G and $G=G_{sc}/{\mathbf Z}(G_{sc})$ .

Maximal tori and their orders of finite reductive groups are well known, see for example [Reference Malle21, Section 3A]. Here a maximal torus of $G_{sc}$ is defined to be the F-fixed points of a maximal torus of the ambient algebraic group $\mathbf G$ under a suitable Frobenius map $F:\mathbf G\rightarrow \mathbf G$ such that $G_{sc}=\mathbf G^F$ . Specifically, the $G_{sc}$ -conjugacy classes of F-stable maximal tori of $\mathbf G$ are parameterized by pairs of partitions $(\lambda ,\mu )$ of n (that is, $\lambda =(\lambda _1,\lambda _2,\ldots )$ and $\mu =(\mu _1,\mu _2,\ldots )$ with $\sum _i \lambda _i+\sum _j \mu _j=n$ ) such that the number of parts of $\mu $ is even. The order of the corresponding (conjugate) maximal tori of $G_{sc}$ is

$$ \begin{align*} \prod_{\lambda_i}(q^{\lambda_i}-1)\prod_{\mu_j}(q^{\mu_j}+1). \end{align*} $$

Consider three tori $\mathcal {T}_i$ ( $1\leq i\leq 3$ ) of $G_{sc}$ of orders

$$ \begin{align*} |\mathcal{T}_1|=(q^{n-1}+1)(q+1), |\mathcal{T}_2|=(q^{n-2}+1)(q^2+1),\end{align*} $$

and

$$ \begin{align*} |\mathcal{T}_3| = \begin{cases} q^{n}-1 &\text{if } n \text{ is odd},\\ (q^{n-1}-1)(q-1) &\text{if } n \text{ is even}. \end{cases} \end{align*} $$

Assume for a moment that $n\geq 5$ and $(n,q)\neq (5,2)$ . In particular, primitive prime divisors

$$ \begin{align*}\ell_1:=\ell(2n-2), \ell_2:=\ell(2n-4),\end{align*} $$

and

$$ \begin{align*} \ell_3:= \begin{cases} \ell(n) &\text{if } n \text{ is odd},\\ \ell(n-1) &\text{if } n \text{ is even}, \end{cases} \end{align*} $$

exist. Furthermore, $\ell _i$ divides $|\mathcal {T}_i|$ and the $\ell _i$ -Sylow subgroups of $G_{sc}$ are cyclic (see [Reference Malle and Testerman24, Theorem 25.14]). Let $g_i\in \mathcal {T}_i$ be of order $\ell _i$ . We use the same notation $g_i$ and $\mathcal {T}_i$ for their images under the natural projection from $G_{sc}$ to G.

We first argue that all the nonunipotent characters of G (as well as of $G_{sc}$ ) are contained in one connected component. (Here, a character of G is called nonunipotent if its lift to $G_{sc}$ is nonunipotent. See [Reference Digne and Michel5, Definition 13.19] for the definition of unipotent characters of finite reductive groups.) For this, it is sufficient to show that any such character of $G_{sc}$ vanishes on at least two $g_i$ . This can be argued similarly as in [Reference Malle21, Section 3B]. Assume otherwise. Then there is $\chi $ belonging to the Lusztig series $\mathcal E(G_{sc},s)$ for some nontrivial semisimple element

$$ \begin{align*}s\in G_{ad}:=\mathrm{P}(\mathrm{CO}_{2n}(q)^0)\end{align*} $$

such that $\chi $ is nonzero on at least two $g_i$ .

Suppose that $\chi (g_1)\neq 0$ . Then $\ell _1$ does not divide $\chi (1)$ (otherwise, since every character degree is a product of a power of q and some $\Phi _k$ up to a constant, we have that $\Phi _{2n-2}$ appears in the product for $\chi (1)$ ; which would imply that $\chi $ is of $\ell _1$ -defect zero, and hence vanishes at $g_1$ , which is a contradiction). It follows from the character-degree formula in Lusztig’s parameterization [Reference Digne and Michel5, Remark 13.24] that $|{\mathbf C}_{G_{ad}}(s)|$ is divisible by $\ell _1$ and moreover ${\mathbf C}_{G_{ad}}(s)$ contains a conjugate of $\mathcal {T}_1^\ast $ , where $\mathcal {T}_i^\ast $ is the torus of $G_{ad}$ dual to $\mathcal {T}_i$ . What we have shown also applies to the cases $\chi (g_2)\neq 0$ and $\chi (g_3)\neq 0$ . Since $\chi $ is nonzero on at least two $g_i$ , we deduce that ${\mathbf C}_{G_{ad}}(s)$ contains certain conjugates of at least two $\mathcal {T}_i^\ast $ . Using the known structure of centralizers of semisimple elements in finite reductive groups (see [Reference Nguyen34, Lemmas 2.3 and 2.5], for instance, for the case of split orthogonal groups), we see that s must be trivial, violating the nonunipotent assumption on $\chi $ .

Lusztig’s classification of ordinary irreducible characters of finite reductive groups, together with the aforementioned character-degree formula and the known centralizers of semisimple elements also show that, for each i, G possesses a nonunipotent character of $\ell _i$ -defect $0$ , which thus vanishes on $g_i$ . For instance, for $i=1$ , we choose a semisimple element $s\in G =[G_{ad},G_{ad}]$ so that $\Phi _{2n-2}$ , a polynomial in q, is not a factor of $|\mathbf {C}_{G_{ad}}(s)|$ . Every character in the Lusztig series $\mathcal E(G_{sc},s)$ then has degree divisible by $\Phi _{2n-2}$ and $\ell _1$ -defect $0$ . Moreover, these characters (of $G_{sc}$ ) restrict trivially to ${\mathbf Z}(G_{sc})$ (see [Reference Hung11, Lemma 5.8], for instance), and therefore they are lifts of characters of G.

We now turn to (nontrivial) unipotent characters. First assume that n is odd. As mentioned in [Reference Malle21, Section 3G], all these characters except the Steinberg one have degrees divisible by either $\ell _1$ or $\ell _3$ , and therefore have either $\ell _1$ -defect or $\ell _3$ -defect zero, and thus vanish on either $g_1$ or $g_3$ . We now know that all members of ${{\operatorname {Irr}}}(G)\backslash \{\mathbf {1}_G,{\mathsf {St}}_G\}$ are contained in just one connected component of $\Gamma _v(G)$ . Thus, we would be done if ${\mathsf {St}}_G$ has a common zero with any other irreducible character of the group. This is not difficult to see. Let r be the defining characteristic of the group. Consider $g\in G$ that is an r-singular element but not an r-element and $p\neq r$ that is a prime divisor of $|g|$ . Then g is a vanishing element for both ${\mathsf {St}}_G$ and any p-defect zero characters.

Now assume that n is even. According to [Reference Malle21, Section 3G], if a nontrivial unipotent character of G has degree not divisible by either $\ell _1$ or $\ell _3$ , it must be either the Steinberg character or one of the two others labeled by the symbols

$$ \begin{align*}{n-1 \choose 1} \quad\text{and}\quad {0\hspace{6pt}\cdots \hspace{6pt} n-3\hspace{6pt}n-1 \choose 1\hspace{6pt}\cdots\hspace{6pt}n-2\hspace{6pt} n-1}.\end{align*} $$

(We refer the reader to [Reference Carter2, Section 13.8] for the labeling and degree formulas of unipotent characters of classical groups.) The degrees of these two characters, however, are divisible by $\Phi _{2n-4}$ . They are therefore of $\ell _2$ -defect zero, and thus vanish at $g_2$ , proving that they are in the same connected component with nonunipotent characters. As with the case of odd n, $\Gamma _v(G)$ is therefore connected.

Consider $G=\mathrm {P}\Omega ^+_8(q)$ . As the case $(n,q)=(4,2)$ can be checked using [8], we assume that $q>2$ , so that the primitive prime divisors $\ell _1,\ell _2,\ell _3$ still exist. (Note that $|G|$ is now divisible by $\Phi _4^2$ and the condition $\chi (g_2)\neq 0$ does not imply that ${\mathbf C}_{G_{ad}}(s)$ contains a conjugate of $\mathcal {T}_2^\ast $ , as we had earlier.) We still have that every nonunipotent character of G vanishes at either $g_1$ or $g_3$ . However, the (two) unipotent characters of G labeled by the symbol ${2 \choose 2}$ have degree $q^2\Phi _3\Phi _6$ , and hence are of both $\ell _1$ - and $\ell _3$ -defect zero. They therefore vanish on both $g_1$ and $g_3$ , and it follows that all the nonunipotent characters of G are contained in one connected component of $\Gamma _v(G)$ . As mentioned above, this connected component also contains all (nontrivial) unipotent characters except possibly the Steinberg one or the two characters labeled by ${3 \choose 1}$ and ${0\hspace {6pt}1 \hspace {6pt} 3 \choose 1\hspace {6pt}2\hspace {6pt}3}$ , which are of degrees $q\Phi _4^2$ and $q^7\Phi _4^2$ , respectively. These characters are of $\ell _2$ -defect zero and vanish on every $\ell _2$ -singular element. However, G has another $\ell _2$ -defect zero unipotent character, namely the one labeled by ${1\hspace {6pt}2 \choose 0\hspace {6pt}3}$ , of degree $q^3\Phi _3\Phi _4^2/2$ , and thus the two exceptional characters are in the same connected component with the nonunipotent characters. Finally, the Steinberg character is handled as above, and $\Gamma _v(G)$ is connected.

When $(n,q)=(5,2)$ , the above arguments still go through with $\mathcal {T}_2$ replaced by a maximal torus of order $(q^3-1)(q^2-1)$ and $\ell _2$ being $\ell (3)$ (and keep $\mathcal {T}_1$ and $\mathcal {T}_3$ ). This concludes the proof.

Proof. This can be checked using GAP [8].

Proof. We provide arguments only for $\lambda =(n-2,2)$ and $\mu =(n-2,1^2)$ with $n\geq 4$ . The other cases are similar. Let $\sigma _1:=(1\, \cdots \,n-3)$ and $\sigma _2:=\sigma _1(n-2 \,\,n-1)$ . It is easy to see that $\chi ^\lambda $ vanishes on both $\sigma _1$ and $\sigma _2$ , while $\chi ^\mu $ takes values 1 on $\sigma _1$ and $-1$ on $\sigma _2$ . This shows that the partitions determined by $\chi ^\lambda $ and $\chi ^\mu $ are different, as desired.

Proof. Suppose that $\lambda $ is the partition of n corresponding to $\chi $ . The assumption on $\chi $ implies that $\lambda $ is not self-conjugate, or equivalently, the Young diagram $[\lambda ]$ of $\lambda $ is not symmetric. Following [Reference James and Kerber14], we use $R_{ij}$ for the part of the rim of $[\lambda ]$ corresponding to the hook at $(i,j)$ . The length of the hook at $(i,j)$ is denoted by $h_{ij}$ .

If $\lambda \in \{(n),(1^n)\}$ , then $\chi $ is the trivial or sign character, so the result is clear ( $n\geq 2$ as $\lambda $ is not self-conjugate). So we may now assume that $(1,2)$ and $(2,1)$ are both nodes of $\lambda $ .

First we consider the case where $[\lambda ]\backslash R_{11}$ is not symmetric (in particular, $[\lambda ]\backslash R_{11}$ is nonempty). Let $\overline {\lambda }$ be the partition with Young diagram $[\lambda ]\backslash R_{11}$ . By induction, there exist $\overline {\pi }$ and $\overline {\sigma }$ in $\mathsf {S}_{n-h_{11}}$ of different signature such that

$$ \begin{align*} \chi^{\overline{\lambda}}(\overline{\pi})=\pm \chi^{\overline{\lambda}}(\overline{\sigma})\neq 0. \end{align*} $$

Let $\tau $ be a cycle of length $n-h_{11}$ in $\mathsf {S}_{n-h_{11}}$ , and set

$$ \begin{align*} \pi:=\tau\overline{\pi} \quad\text{and}\quad \sigma:=\tau\overline{\sigma}. \end{align*} $$

By the Murnaghan–Nakayama formula, we have

$$ \begin{align*} \chi^\lambda(\pi)=\pm \chi^{\overline{\lambda}}(\overline{\pi}) \quad\text{and}\quad \chi^\lambda(\sigma)=\pm \chi^{\overline{\lambda}}(\overline{\sigma}), \end{align*} $$

which implies what we wanted.

It remains to consider the case where $[\lambda ]\backslash R_{11}$ is symmetric. Since $[\lambda ]$ itself is not symmetric but $[\lambda ]\backslash R_{11}$ is, it follows that $h_{12}\neq h_{21}$ . Without loss of generality, assume that $h_{12}>h_{21}$ , so that $h_{12}$ is the second largest hook length in $[\lambda ]$ and it occurs with multiplicity 1. Note that $[\lambda ]\backslash R_{12}$ can be obtained from $[\lambda ]\backslash R_{11}$ by adding some nodes to the first column and that $[\lambda ]\backslash R_{12}$ has at least two nodes in the first column. It follows that $[\lambda ]\backslash R_{12}$ is not symmetric. Repeating the above arguments, we arrive at the same conclusion.

Proof. Let $\lambda ,\mu $ be the partitions of n with $\chi =\chi ^\lambda $ and $\psi =\chi ^\mu $ . We may assume that $\mu \not \in \{\lambda ,\lambda '\}$ (with $\lambda $ and $\lambda '$ being conjugated partitions).

Then by [Reference Morotti31, Theorem 2], $H(\lambda )\not = H(\mu )$ or $H([\lambda ]\backslash R_{11})\not =H([\mu ]\backslash R_{11})$ . The lemma then follows by the proofs of [Reference Morotti32, Propositions 3.3.8 and 3.3.9].

Proof of Theorem 6.1

The result can be checked for $n\leq 7$ from the known character tables, so we suppose that $n\geq 8$ . Assume that $\chi ,\psi \in {{\operatorname {Irr}}}(\mathsf {S}_n)\backslash \{\textbf {1}_{\mathsf {S}_n},\mathrm {sgn}\}$ such that $\mathcal {P}(\mathbf {d}_\chi )=\mathcal {P}(\mathbf {d}_\psi )$ , and let $\lambda $ and $\mu $ be the partitions of n corresponding to $\chi $ and $\mu $ , respectively.

We know that $\{\lambda ,\mu \}$ is not one of the pairs considered in Lemma 6.2. It follows that, by Theorem A, $\chi $ and $\psi $ have a common zero. As $\mathcal {P}(\mathbf {d}_\chi )=\mathcal {P}(\mathbf {d}_\psi )$ , we deduce that ${{\operatorname {Van}}}(\chi )={{\operatorname {Van}}}(\psi )$ , and thus, by Lemma 6.4,

$$ \begin{align*} \psi\in\{\chi, \mathrm{sgn}\cdot\chi\}. \end{align*} $$

We therefore would be done if $\chi $ and $\mathrm {sgn}\cdot \chi $ produce different partitions on $\mathsf {S}_n$ ; that is, $\mathcal {P}(\chi )\neq \mathcal {P}(\mathrm {sgn}\cdot \chi )$ . For this, it is sufficient to show that there exist permutations $\pi $ and $\sigma $ of different signature such that

$$ \begin{align*}\chi(\pi)=\pm\chi(\sigma)\neq 0.\end{align*} $$

This is done in Lemma 6.3, and the proof is complete.

Question 7.1. Let G be a finite group. Is it true that if $\Gamma (G)$ is connected, then $\Gamma _v(G)$ is connected?

Proof. By symmetry, it suffices to prove that ${{\operatorname {Van}}}(\chi )\not \subseteq {{\operatorname {Van}}}(\psi )$ . Let $\chi (1)$ be a $\pi $ -number for some set of primes $\pi $ , so that $\psi (1)$ is a $\pi '$ -number. By way of contradiction, suppose that ${{\operatorname {Van}}}(\chi )\subseteq {{\operatorname {Van}}}(\psi )$ . By [Reference Malle, Navarro and Olsson22], there exists $p\in \pi $ and $x\in G$ of p-power order such that $\chi (x)=0$ . It follows that $\psi (x)=0$ . Note that $\psi $ has degree not divisible by p. Therefore, [Reference Navarro33, Corollary 4.20] implies that $\psi (x)\neq 0$ , which is a contradiction.

Proof. This is stated in [Reference Bubboloni, Dolfi and Spiga1, Example 1, Section 5], but there is an error in the proof appearing after [Reference Bubboloni, Dolfi and Spiga1, Proposition 5.1] of why $\chi ^G$ does not vanish on Q, so we give a new proof of this fact here. Further, it is not stated there that $\chi ^G$ vanishes outside Q, but this follows from Q being normal.

Let $H=\{(a_i)\in Q\mid a_i\in {\mathbb {F}}_q\}$ and $H_k:=\{(a_i)\in H\mid a_i=0\text { for }i<k\}$ for $1\leq k\leq s$ .

Though it is not stated in [Reference Bubboloni, Dolfi and Spiga1, Proposition 5.1], the characters $\chi _\alpha ^c$ appearing there are pairwise distinct, see the proof of [Reference Hanaki and Okuyama10, Theorem 4.8], so that indeed, for some $\alpha \not =1$ , $\chi ^G=\chi _\alpha ^G$ is irreducible. By [Reference Hanaki and Okuyama10, Theorem 4.1], any element in Q is G-conjugate to some element in H. So it is enough to prove that $\chi _\alpha ^G$ does not vanish on H.

Fix any $1\not =\alpha \in {{\operatorname {Irr}}}(H)$ and any $h\in H$ . We may assume that $h\not =1$ . Let $1\leq k,j<~s$ with $\alpha \in {{\operatorname {Irr}}}(H/H_{k+1})\setminus {{\operatorname {Irr}}}(H/H_k)$ and $h\in H_j\setminus H_{j+1}$ . For any $g\in G$ , we have that if $h^g=(a_i)$ , then $a_i=0$ for $i<j$ and $a_j\not =0$ . If further $g\in Q$ and $c\in C$ , then the j th components of $h^c$ and $(h^c)^g$ coincide.

If $j>k$ , then $h^c\in Q_{k+1}$ for every $c\in C$ , so in this case, the argument from [Reference Bubboloni, Dolfi and Spiga1, Section 5] works.

Assume next that $j<k$ . Since $C=\langle x\rangle $ with $x\in {\mathbb {F}}_{q^s}^*$ of order $p=(q^s-1)/(q-1)$ , then the j th component of $h^c$ is not in ${\mathbb {F}}_q$ and so by the above paragraph, it follows that $h^c$ is not Q-conjugate to any element of $HQ_k$ for $1\not =c\in C$ . So in this case, $\chi _\alpha ^G(h)\not =0$ by the argument in [Reference Bubboloni, Dolfi and Spiga1, Section 5].

Consider now $j=k$ . Then $h^c\in Q_k\subseteq HQ_k$ for every $c\in C$ . By [Reference Hanaki and Okuyama10, Lemma 3.3, Proposition 4.7], we have that $\{(a_i)\in Q_k\mid \text {Tr}(a_k)=0\}\subseteq \operatorname {Ker}(\chi _\alpha )$ (apply [Reference Hanaki and Okuyama10, Lemma 3.3] with $m=k$ to $Q/Q_{k+1}$ ). In particular, if $h_c=(0^{k-1},\text {Tr}((h^c)_k),0^{s-k-1})$ , then $h^c=h_ck$ for some $k\in \operatorname {Ker}(\chi _\alpha )$ . In view of [Reference Hanaki and Okuyama10, Proposition 4.7], we then have that

$$ \begin{align*}\chi_\alpha^G(h)=q^{(s-1)(k-1)/2}(\alpha(h_1)+\cdots+\alpha(h_p))\end{align*} $$

with $h_1,\ldots ,h_p\in H_k$ . Note that H is abelian (see [Reference Bubboloni, Dolfi and Spiga1, Example 1, Section 5]). As H is a q-group with q a prime and $\alpha $ is a irreducible character of H, $\alpha (h_1)+\cdots +\alpha (h_p)$ is a sum of p not necessarily primitive $q^a$  th roots of unity for some $a\geq 1$ . As p is also prime and by definition, $p\not =q$ , it follows that also in this case, $\chi _\alpha ^G(h)\not =0$ (this is similar to the proof of [Reference Navarro33, Lemma 4.19]).

Question 7.5. Let G be a finite p-group. Suppose that $\chi ,\psi \in {{\operatorname {Irr}}}(G)$ and ${{\operatorname {Van}}}(\chi )={{\operatorname {Van}}}(\psi )$ . Is it true that $\chi (1)=\psi (1)$ ?

Proof. The result is clear if $\chi $ is induced from N. Now, assume that $\chi (x)=0$ for every $x\in G-N$ . We want to see that $\chi $ is induced from N. Assume not. Then $\chi _N\in {{\operatorname {Irr}}}(N)$ by [Reference Isaacs12, Corollary 6.19]. Hence,

$$ \begin{align*} 1&=[\chi,\chi]=\frac{1}{|G|}\sum_{g\in G}|\chi(g)|^2= \frac{1}{|G|}\sum_{g\in N}|\chi(g)|^2\\ &=(1/p)\frac{1}{|N|}\sum_{g\in N}|\chi(g)|^2=(1/p)[\chi_N,\chi_N]=1/p, \end{align*} $$

which is a contradiction.

Proof. Without loss of generality, we may assume that $\chi $ is faithful. Note that ${\mathbf Z}(G)<N$ . Therefore, $N/{\mathbf Z}(G)\cap {\mathbf Z}_2(G)/{\mathbf Z}(G)>1$ . Thus, there exists a noncentral element $x\in N\cap {\mathbf Z}_2(G)$ . Now, a similar argument as at the end of the proof of [Reference Moretó and Sangroniz29, Theorem C] shows that $\chi (x)=0$ .

Proof. We argue by induction on $|G|$ . Suppose first that there exists M maximal in G such that $\chi _M,\psi _M\in {{\operatorname {Irr}}}(M)$ . Then the result follows from the inductive hypothesis.

Now let M be a maximal subgroup of G. The hypothesis ${{\operatorname {Van}}}(\chi )={{\operatorname {Van}}}(\psi )$ and Lemma 7.6 imply that one of the characters is induced from M if and only if the other character is also induced from M. Therefore, we may assume that for any M maximal in G, both $\chi $ and $\psi $ are induced from M.

Let ${\mathbf Z}(G)\leq N\trianglelefteq G$ such that $G/N$ is elementary abelian of order $p^2$ . Let $\nu \in {{\operatorname {Irr}}}(N)$ lie under $\chi $ . Let $T:=I_G(\nu )$ . Since $\chi (1)=p$ , Clifford’s correspondence implies that $N<T$ . Therefore, if $T<G$ , then T is a maximal subgroup of G. Write $G/N=T/N\times U/N$ for some U maximal in G. Since $I_U(\nu )=N$ , $\nu $ induces irreducibly to $\mu \in {{\operatorname {Irr}}}(U)$ . Note that $\nu $ lies under $\chi $ , so by comparing degrees, we have $\chi _U=\mu $ , and this is a contradiction. It follows that $\nu $ is G-invariant. The previous paragraph also implies that $G-N\subseteq {{\operatorname {Van}}}(\chi )={{\operatorname {Van}}}(\psi )$ . Now, by [Reference Isaacs12, Problem 6.3], $\chi $ is fully ramified with respect to $G/N$ . In particular, since $\nu $ is linear, we conclude that $\chi $ has no zeros in N, whence

$$ \begin{align*}G-N={{\operatorname{Van}}}(\chi)={{\operatorname{Van}}}(\psi).\end{align*} $$

Now let $\delta \in {{\operatorname {Irr}}}(N)$ lie under $\psi $ . Suppose first that $L:=I_G(\delta )$ is maximal in G. It follows from the Clifford theory that $\psi (1)/\delta (1)=p$ . Write $G/N=L/N\times V/N$ for some V maximal in G, so that $\eta =\delta ^V\in {{\operatorname {Irr}}}(V)$ lies under $\psi $ . We conclude that $\chi _V=\eta $ . By Lemma 7.6, $G-V\not \subseteq {{\operatorname {Van}}}(\psi )$ , which is a contradiction.

Now assume that $L=G$ . Using [Reference Isaacs12, Problem 6.3] again, we see that $\psi $ is fully ramified with respect to $G/N$ . Therefore, $\psi _N=p\delta $ . Since ${{\operatorname {Van}}}(\psi )\cap N=\emptyset $ , it follows that $\delta $ is linear, by Burnside’s theorem. The result follows in this case too.

Finally, assume that $L=N$ , so that $\psi =\delta ^G$ . By Lemma 7.7, $\psi $ has some zero in N. This is the final contradiction.