Subgroups of arbitrary even ordinary depth

TL;DR

This paper constructs specific examples of groups and subgroups demonstrating that even ordinary depth can be arbitrarily large, specifically for any positive integer n, resolving an open problem in group theory.

Contribution

It provides the first known examples of subgroups with arbitrary even ordinary depth, answering a longstanding open question.

Findings

Existence of groups with subgroups of arbitrary even depth

Construction of examples for each positive integer n

Resolution of the open problem on large even ordinary depth

Abstract

We show that for each positive integer $n$, there are a group $G$ and a subgroup $H$ such that the ordinary depth is $d(H, G) = 2n$. This solves the open problem posed by Lars Kadison whether even ordinary depth larger than $6$ can occur.

Equations

$${\sigma }_n$$

$$G_n$$

$$H_n$$

$$H_n\cong N\times G^{n-1}\le G^n<G\wr C_n.$$

$$N_n=⟨N,N^{{\sigma }_n},\dots ,N^{{\sigma }_n^{n-1}}⟩=\bigcap _{i=0}^{n-1}{H}_n^{{\sigma }_n^i}$$

$$K_n=⟨G,G^{{\sigma }_n},\dots ,G^{{\sigma }_n^{n-1}}⟩\le G_n.$$

$$\begin{array}[]{l|rrrrr}&g_{1}&g_{2}&g_{3}&g_{4}\\ \hline\cr\nu_{1}&1&1&1&1\\ \nu_{2}&1&1&-1&-1\\ \nu_{3}&1&-1&1&-1\\ \nu_{4}&1&-1&-1&1\end{array}\ \ \ \ \begin{array}[]{l|rrrrr}&g_{1}&g_{2}&g_{3}^{\prime}&g_{4}^{\prime}&g_{5}\\ \hline\cr\chi_{1}&1&1&1&1&1\\ \chi_{2}&1&1&1&-1&-1\\ \chi_{3}&2&2&-1&0&0\\ \chi_{4}&3&-1&0&1&-1\\ \chi_{5}&3&-1&0&-1&1\end{array}\ \ \ \ \begin{array}[]{rcl}\chi_{1}|_{N}&=&\nu_{1}\\ \chi_{2}|_{N}&=&\nu_{1}\\ \chi_{3}|_{N}&=&2\nu_{1}\\ \chi_{4}|_{N}&=&\nu_{2}+\nu_{3}+\nu_{4}\\ \chi_{5}|_{N}&=&\nu_{2}+\nu_{3}+\nu_{4}\end{array}$$

$$X_n=\left\{{\chi }_{i_1}\times {\chi }_{i_2}\times \cdots \times {\chi }_{i_n}\in \text{mbox}\mathrm{Irr}(K_n);i_1\in \{4,5\},i_j\in \{1,2,3\}\text{mbox}for2\le j\le n\right\}$$

$$Y_n=\left\{{\chi }^{G_n};\chi \in X_n\right\}.$$

$$\left\{(i_1,i_2,\dots ,i_n);i_1\in \{4,5\},i_j\in \{1,2,3\}\text{mbox}for2\le j\le n\right\},$$

Full text

Mathematics Subject Classification (2010): primary 20C15; secondary 20B35, 05C12, 05C76.

Keywords: ordinary depth of a subgroup, distance of characters, Cartesian product of graphs, wreath product.

1 Introduction

The notion of depth was originally defined for von-Neumann algebras, see [8]. Later it was also defined for Hopf algebras, see [19]. For some recent results in this direction, see [10, 16, 9]. In [18] and later in [3], the depth of semisimple algebra inclusions was studied, by Burciu, Kadison and Külshammer. First results were considering the depth $2$ case, later it was generalized for arbitrary $n$. In the case of group algebra inclusion ${{\mathbb{C}}}G\subseteq{{\mathbb{C}}}H$ it was shown that the depth is at most $2$ if and only if $H$ is normal in $G$, see [18].

Now let us remind the reader to the notion of ordinary depth of a subgroup $H$ in the finite group $G$. We say that the * depth of the group algebra inclusion * ${{\mathbb{C}}}H\subseteq{{\mathbb{C}}}G$ is $2n$ if ${{\mathbb{C}}}G\otimes_{{{\mathbb{C}}}H}\cdots\otimes_{{{\mathbb{C}}}H}{{{\mathbb{C}}}G}$ ($n+1$-times ${{\mathbb{C}}}G$) is isomorphic to a direct summand of $\oplus_{i=1}^{a}{{\mathbb{C}}}G\otimes_{{{\mathbb{C}}}H}\cdots\otimes_{{{\mathbb{C}}}H}{{\mathbb{C}}}G$ ($n$ times ${{\mathbb{C}}}G$) as ${{\mathbb{C}}}G-{{\mathbb{C}}}H$-bimodules (or equivalently as ${{\mathbb{C}}}H-{{\mathbb{C}}}G$-bimodules) for some positive integer $a$.

Furthermore, ${{\mathbb{C}}}H$ is said to have depth $2n+1$ in ${{\mathbb{C}}}G$ if the same assertion holds for ${{\mathbb{C}}}H-{{\mathbb{C}}}H$-bimodules. Finally ${{\mathbb{C}}}H$ has depth $1$ in ${{\mathbb{C}}}G$ if ${{\mathbb{C}}}G$ is isomorphic to a direct summand of $\oplus_{i=1}^{a}{{\mathbb{C}}}H$ as ${{\mathbb{C}}}H-{{\mathbb{C}}}H$ bimodules. The minimal depth of group algebra inclusion ${{\mathbb{C}}}H\subseteq{{\mathbb{C}}}G$ is called (minimal) ordinary depth of $H$ in $G$, which we denote by $d(H,G)$. If one considers group algebras over a field of characterstic $p$ then one gets in a similar way the notion of modular depth and it does not depend on the field, only on the characteristic, as it is shown in [1, Remark 4.5] by Boltje, Danz and Külshammer.

The ordinary depth can be obtained from the so called inclusion or Frobenius matrix $M$. If $\chi_{1},\ldots,\chi_{s}$ are all irreducible characters of $G$ and $\psi_{1},\ldots,\psi_{r}$ are all irreducible characters of $H$, then $m_{i,j}:=(\psi_{i}^{G},\chi_{j})$. The ”powers” of $M$ are defined by $M^{2l}:=M^{2l-1}M^{T}$ and $M^{2l+1}:=M^{2l}M$. The ordinary depth $d(H,G)$ can be obtained as the smallest integer $n$ such that $M^{n+1}\leq aM^{n-1}$ for some positive integer $a$, where the inequality of matrices means that this inequality holds componentwise.

The results on characters in [3] help to determine $d(H,G)$. Two irreducible characters $\alpha,\beta\in\mbox{\rm Irr}(H)$ are called related, $\alpha\sim_{G}\beta$, if they are constituents of $\chi_{H}$, for some $\chi\in\mbox{\rm Irr}(G)$. The distance $d_{G}(\alpha,\beta)=m$ is the smallest integer $m$ such that there is a chain of irreducible characters of $H$ such that $\alpha=\psi_{0}\sim_{G}\psi_{1}\ldots\sim_{G}\psi_{m}=\beta$. If there is no such chain then $d_{G}(\alpha,\beta)=-\infty$ and if $\alpha=\beta$ then the distance is zero. If $X$ is the set of irreducible constituents of $\chi_{H}$ then we set $m(\chi):=\max\{\min\{d_{G}(\alpha,\psi);\psi\in X\};\alpha\in\mbox{\rm Irr}(H)\}$. We will use the following result from [3].

Theorem 1.1**.**

[3, Thm 3.6, Thm  3.10]

• (i)

Let $m\geq 1$. Then $H$ has ordinary depth $\leq 2m+1$ in $G$ if and only if the distance between two irreducible characters of $H$ is at most $m$.

• (ii)

Let $m\geq 2$. Then $H$ has ordinary depth $\leq 2m$ in $G$ if and only if $m(\chi)\leq m-1$ for all $\chi\in\mbox{\rm Irr}(G)$.

Thus we have the following.

Corollary 1.2**.**

Let $H$ be a subgroup of a finite group $G$. The ordinary depth $d(H,G)$ is the minimal possible positive integer which can be determined from the following upper bounds:

• (i)

For $m\geq 1$, $d(H,G)\leq 2m+1$ if and only if the distance between two irreducible characters of $H$ is at most $m$.

• (ii)

For $m\geq 2$, $d(H,G)\leq 2m$ if and only if $m(\chi)\leq m-1$ for all $\chi\in\mbox{\rm Irr}(G)$.

• (iii)

$d(H,G)\leq 2$* if and only if $H$ is normal in $G$,*

• (iv)

$d(H,G)=1$* if and only if $G=HC_{G}(x)$ for all $x\in H$, see [2, Thm. 1.7 ].*

We will also use the following result from [3].

Theorem 1.3**.**

[3, Thm 6.9]** Suppose that $H$ is a subgroup of a finite group $G$ and $N={\mathrm{Core}}_{G}(H)$ is the intersection of $m$ conjugates of $H$. Then $d(H,G)\leq 2m$. If additionally $N\leq Z(G)$ holds then $d(H,G)\leq 2m-1$.

In recent publications, several authors determined the ordinary depth of subgroups in some special series of groups, e.g. $PSL(2,q)$, Suzuki groups, Ree groups, symmetric and alternating groups, see [3], [5], [6], [13], [14]. In [4], twisted group algebra inclusions for symmetric and alternating groups are studied.

It is known that odd ordinary depth of a subgroup in a finite group can be arbitrarily large: It is shown in [3] that the depth of the symmetric group $S_{n}$ in $S_{n+1}$ is $2n-1$.

Lars Kadison posed the following open problem on his homepage, see [17]: Are there subgroups of ordinary depth $2n$ where $n>3$?

If one looks at the results of the above papers or the calculations presented in  [12], one has the impression that in most cases the depth of subgroups is odd. However still one can find examples of arbitrarily large even depth. In our examples wreath products will play an important role. There is another depth concept, the combinatorial depth, which is considered in several investigations, see e.g. [1], [13], [14]. In this short note we will always consider ordinary depth, so in the following depth will always mean ordinary depth.

The main result of this paper is the following.

Theorem 1.4**.**

There exists a series of groups and subgroups $(G_{n},H_{n})$ such that $d(H_{n},G_{n})=2n$ for every positive integer $n$.

2 Constructing examples

Examples of subgroups of depth $8$ had been constructed earlier by the third author with the help of the GAP system [7], see [12]. Let us mention some of them:

• •

$d(A_{15}\cap(S_{12}\times S_{3}),A_{15})=8$,

• •

$d((2^{6}:U_{4}(2)),O_{8}^{-}(2))=8$, and

• •

$d(((((C_{2}\wr C_{2})\wr C_{2})\wr C_{2})\cap A_{8}\times A_{8}),((C_{2}\wr C_{2})\wr C_{2})\wr C_{2}))=8$.

It was shown already in [3] that $d(D_{8},S_{4})=4$. The first author found with GAP that

$d(D_{8}\times S_{4},S_{4}\wr C_{2})=8$. Continuing this process, we obtained that

$d((D_{8}\times S_{4})\times(S_{4}\wr C_{2}),(S_{4}\wr C_{2})\wr C_{2}))=16$ and in general

• •

$G_{0}:=S_{4}$, $H_{0}:=D_{8}$,

• •

$G_{n}:=G_{n-1}\wr C_{2}$, $H_{n}:=H_{n-1}\times G_{n-1}<G_{n-1}\times G_{n-1}<G_{n}$,

• •

$d(H_{n},G_{n})=2^{n+2}$.

The idea of the proof is to use Thm. 1.3 to prove that $d(H_{n},G_{n})\leq 2^{n+2}$. Then we show that the depth cannot be at most $2^{n+2}-1=2(2^{n+1}-1)+1$, since by Cor. 1.2 then the distance of any two characters of $H_{n}$ would be at most $2^{n+1}-1$, however there are irreducible characters of $H_{n}$ of distance exactly $2^{n+1}$. The proof is a rather complicated induction, see [15].

We wanted to simplify the construction. Our aim was also to construct as depth more even numbers. We can generalize the first two steps of the former construction in another way as follows:

• •

$d(D_{8},S_{4})=4,$

• •

$d(D_{8}\times S_{4},S_{4}\wr C_{2})=8,$

• •

$d(D_{8}\times S_{4}\times S_{4},S_{4}\wr C_{3})=12.$

In general, we have that

• •

$G_{1}:=S_{4}$, $H_{1}:=D_{8}$,

• •

$G_{n}:=G_{1}\wr C_{n}$,

• •

$H_{n}:=H_{1}\times G_{1}^{n-1}<G_{1}^{n}<G_{n}$.

Then we have that $d(H_{n},G_{n})=4n$. The proof is again using Thm. 1.3 to prove that $d(H_{n},G_{n})\leq 4n$. If $d(H_{n},G_{n})\leq 4n-1=2(2n-1)+1$, then by Cor. 1.2 any two irreducible characters of $H_{n}$ have distance at most $2n-1$. However, one can show that there exist irreducible characters of $H_{n}$ of distance $2n$.

If we want to get every even number then we can use a modified construction. We may take the Klein four group $V_{4}\triangleleft S_{4}$ instead of $D_{8}$ and get:

• •

$d(V_{4},S_{4})=2,$

• •

$d(V_{4}\times S_{4},S_{4}\wr C_{2})=4,$

• •

$d(V_{4}\times S_{4}\times S_{4},S_{4}\wr C_{3})=6.$

In general, we have a series of groups and subgroups such that $d(H_{n},G_{n})=2n$ holds. The idea of the proof will be the same as before, for the inequality we will use again Thm. 1.3, and to prove that it cannot be a strict inequality, we find two irreducible characters of distance $n$ in $H_{n}$. For that, we consider suitable characters of the base group of the wreath product and define a Cartesian product of graphs that encodes the relation $\sim$.

3 Proof of Theorem 1.4

Let $G$ be the symmetric group on four points, and $N$ be its normal Klein four subgroup. Set $G_{1}=G$, $H_{1}=N$. Then $d(H_{1},G_{1})=2$, by Cor. 1.2. Define for $n\geq 2$

[TABLE]

Let $C_{n}$ denote the cyclic group of order $n$. Then $H_{n}<G_{n}\cong G\wr C_{n}$ and

[TABLE]

Let $N_{n}={\mathrm{Core}}_{G_{n}}(H_{n})$, the largest normal subgroup of $G_{n}$ that is contained in $H_{n}$. Then $N_{1}=N$, and

[TABLE]

is an intersection of $n$ conjugates of $H_{n}$, and Thm. 1.3 yields $d(H_{n},G_{n})\leq 2n$. Set

[TABLE]

Then $H_{n}\leq K_{n}\cong G^{n}$.

The character tables of $N$ and $G$ are as follows, where the columns are indexed by the conjugacy classes of the elements $g_{1}=()$, $g_{2}=(1,3)(2,4)$, $g_{3}=(1,2)(3,4)$, $g_{3}^{\prime}=(1,2,3)$, $g_{4}=(1,4)(2,3)$, $g_{4}^{\prime}=(1,3)$, $g_{5}=(1,2,3,4)$.

[TABLE]

Set

[TABLE]

and

[TABLE]

Let $\Gamma_{1}$ be the undirected graph with vertex set $\{4,5\}$ and edge set $\{\{4,5\}\}$, $\Gamma_{0}$ be the undirected graph with vertex set $\{1,2,3\}$ and edge set $\{\{1,3\},\{2,3\},\{1,2\}\}$. For $n\geq 2$, let $\Gamma_{n}$ be the Cartesian product of $\Gamma_{1}$ and $n-1$ copies of $\Gamma_{0}$, that is, $\Gamma_{n}$ has vertex set

[TABLE]

and there is an edge between $(i_{1},i_{2},\ldots,i_{n})$ and $(i_{1}^{\prime},i_{2}^{\prime},\ldots,i_{n}^{\prime})$ if and only if there is a (unique) $j$ such that $i_{k}=i_{k}^{\prime}$ for $k\not=j$ and $i_{j}\not=i_{j}^{\prime}$ and either $\{i_{j},i_{j}^{\prime}\}=\{4,5\}$ or $\{i_{j},i_{j}^{\prime}\}\subset\{1,2,3\}$.

Lemma 3.1**.**

• (i)

$Y_{n}\subseteq\mbox{\rm Irr}(G_{n})$, and mapping $\chi$ to $\chi^{G_{n}}$ defines a bijection from $X_{n}$ to $Y_{n}$.

• (ii)

For $\psi\in Y_{n}$ and $\psi^{\prime}\in\mbox{\rm Irr}(G_{n})$, if $\psi|_{H_{n}}$ and $\psi^{\prime}|_{H_{n}}$ have a common constituent then $\psi^{\prime}\in Y_{n}$.

• (iii)

Let $\psi=\chi^{G_{n}}$, $\psi^{\prime}=(\chi^{\prime})^{G_{n}}$ for $\chi,\chi^{\prime}\in X_{n}$, with $\psi\not=\psi^{\prime}$. Then $\psi|_{H_{n}}$ and $\psi^{\prime}|_{H_{n}}$ have a common constituent if and only if there is an edge between $(i_{1},i_{2},\ldots,i_{n})$ and $(i_{1}^{\prime},i_{2}^{\prime},\ldots,i_{n}^{\prime})$ in $\Gamma_{n}$, where $\chi=\chi_{i_{1}}\times\chi_{i_{2}}\times\cdots\times\chi_{i_{n}}$ and $\chi^{\prime}=\chi_{i_{1}^{\prime}}\times\chi_{i_{2}^{\prime}}\times\cdots\times\chi_{i_{n}^{\prime}}$.

• (iv)

The distance of the vertices $(4,1,1,\ldots,1)$ and $(4,2,2,\ldots,2)$ of $\Gamma_{n}$ is $n-1$.

• (v)

The distance $d_{G_{n}}(\alpha_{n},\omega_{n})$ of the characters $\alpha_{n}:=\nu_{2}\times\chi_{1}\times\cdots\times\chi_{1}$ and

$\omega_{n}:=\nu_{2}\times\chi_{2}\times\cdots\times\chi_{2}$* of $H_{n}$ is $n$.*

• Proof.

Let $\psi=\chi^{G_{n}}$, where $\chi=\chi_{i_{1}}\times\chi_{i_{2}}\times\cdots\times\chi_{i_{n}}\in X_{n}$, that is, $\chi_{i_{1}}$ is faithful and the other $\chi_{i_{j}}$ are not.

For part (i), $\chi$ has inertia subgroup $K_{n}$ inside $G_{n}$. Hence by Clifford’s Theorem, see [11, Thm. 6.11], $\chi^{G_{n}}$ is irreducible. The irreducible constituents of the restriction $\psi|_{K_{n}}$ are the $n$ conjugates of $\chi$ by $\sigma_{n}$, i. e., those characters where the $n$ components of $\chi$ are cyclically permuted. Thus each constituent has exactly one faithful component. Hence $\chi$ is the only constituent of $\psi|_{K_{n}}$ that lies in $X_{n}$. Thus we get an inverse to the map $\chi\mapsto\chi^{G_{n}}$.

For part (ii), consider the restriction of the constituents of $\psi|_{K_{n}}$ to $H_{n}$. We get irreducible constituents where the first component is a nontrivial character of $N$ and all other components are non-faithful characters of $G$, and irreducible constituents where the first component is the trivial character of $N$ and exactly one other component is faithful. Let $\psi^{\prime}\in\mbox{\rm Irr}(G_{n})$ have the property that $\psi^{\prime}|_{H_{n}}$ and $\psi|_{H_{n}}$ have a common irreducible constituent, which means that $0\neq(\psi|_{H_{n}},\psi^{\prime}|_{H_{n}})=((\psi|_{H_{n}})^{G_{n}},\psi^{\prime})$. If this constituent is of the first kind then inducing it to $K_{n}$ yields a character with first component $\chi_{4}+\chi_{5}$ and all other components non-faithful. If the common constituent is of the second kind then inducing it to $K_{n}$ yields a character with first component $\chi_{1}+\chi_{2}+2\chi_{3}$ and exactly one other component faithful. (Here we used that $(\mu\times\theta_{2}\times\cdots\times\theta_{n})^{K_{n}}=(\mu^{G}\times\theta_{2}\times\cdots\times\theta_{n})$, where $\theta_{i}\in\mbox{\rm Irr}(G)$, for $i=2\ldots n$, $\mu\in\mbox{\rm Irr}(N)$.)

In both cases, the irreducible constituents are cyclic shifts of characters in $X_{n}$, thus inducing further from $K_{n}$ to $G_{n}$ yields characters all whose irreducible constituents lie in $Y_{n}$. Now note that $\psi^{\prime}$ is one of them.

For part (iii), note that there is an edge between $(i_{1},i_{2},\ldots,i_{n})$ and $(i_{1}^{\prime},i_{2}^{\prime},\ldots,i_{n}^{\prime})$ in $\Gamma_{n}$ if and only if $\chi:=\chi_{i_{1}}\times\chi_{i_{2}}\times\cdots\times\chi_{i_{n}}$ and $\chi^{\prime}:=\chi_{i_{1}^{\prime}}\times\chi_{i_{2}^{\prime}}\times\cdots\times\chi_{i_{n}^{\prime}}$ differ in exactly one component $\chi_{i_{j}}$, $\chi_{i_{j}^{\prime}}$, such that $\chi_{i_{j}}|_{N}$ and $\chi_{i_{j}^{\prime}}|_{N}$ have a common constituent. Let $\psi:=(\chi_{i_{1}}\times\chi_{i_{2}}\times\cdots\times\chi_{i_{n}})^{G_{n}}$, and $\psi^{\prime}:=(\chi_{i_{1}^{\prime}}\times\chi_{i_{2}^{\prime}}\times\cdots\times\chi_{i_{n}^{\prime}})^{G_{n}}$. Then $\psi|_{K_{n}}$ contains as a constituent $\chi_{i_{1}}\times\chi_{i_{2}}\times\cdots\times\chi_{i_{n}}$ and all its cyclic shifts, $\psi^{\prime}_{K_{n}}$ contains as a constituent $\chi_{i_{1}^{\prime}}\times\chi_{i_{2}^{\prime}}\times\cdots\times\chi_{i_{n}^{\prime}}$ and all its cyclic shifts. When restricted further to $H_{n}$ the scalar product can be nonzero if and only if some of cyclic shifts of $\chi$ and some of cyclic shifts of $\chi^{\prime}$ have in the first component a restriction that have a common component and all other components are equal. But then they must be shifted in the same way, since otherwise the faithful components were in different place. Thus $\chi_{i_{1}}\times\chi_{i_{2}}\times\cdots\times\chi_{i_{n}}$ and $\chi_{i_{1}^{\prime}}\times\chi_{i_{2}^{\prime}}\times\cdots\times\chi_{i_{n}^{\prime}}$ differ in exactly one component $\chi_{i_{j}}$, $\chi_{i_{j}^{\prime}}$, such that $\chi_{i_{j}}|_{N}$ and $\chi_{i_{j}^{\prime}}|_{N}$ have a common constituent.

For part (iv), observe that any shortest path from $(4,1,\ldots,1)$ to $(4,2,\ldots,2)$ in $\Gamma_{n}$ replaces in each step exactly one $1$ by a $2$.

For part (v), fix $n$ and let $\alpha_{n}\sim_{G_{n}}\psi_{1}\sim_{G_{n}}\psi_{2}\sim_{G_{n}}\cdots\sim_{G_{n}}\psi_{m}\sim_{G_{n}}\omega_{n}$ be a shortest path of related characters in $\mbox{\rm Irr}(H_{n})$, of length $m+1$. This means that there are irreducible characters $\Phi_{1},\Phi_{2},\ldots,\Phi_{m+1}$ of $G_{n}$ such that $\alpha_{n}$ and $\psi_{1}$ are constituents of $\Phi_{1}|_{H_{n}}$, $\psi_{i}$ and $\psi_{i+1}$ are constituents of $\Phi_{i+1}|_{H_{n}}$, for $1\leq i\leq m-1$, and $\psi_{m}$ and $\omega_{n}$ are constituents of $\Phi_{m+1}|_{H_{n}}$. By Frobenius reciprocity we have that $(\alpha_{n}^{G_{n}},\Phi_{1})\neq 0$. Since $\alpha_{n}^{K_{n}}=(\chi_{4}+\chi_{5})\times\chi_{1}\times\cdots\times\chi_{1}$ is a sum of characters in $X_{n}$, we know that $\Phi_{1}\in Y_{n}$, and part (ii) implies that $\Phi_{i}\in Y_{n}$ for all $i\in\{1,2,\ldots,m+1\}$. Let $\Theta_{i}$ be the unique character in $X_{n}$ with the property $\Phi_{i}=\Theta_{i}^{G_{n}}$, for $1\leq i\leq m+1$. By part (iii), $\Theta_{i}$ and $\Theta_{i+1}$ differ in at most one component. Now $\Theta_{1}$ has $n-1$ components $\chi_{1}$, and $\Theta_{m+1}$ has $n-1$ components $\chi_{2}$, thus $m\geq n-1$ holds. Conversely, any path of length $n-1$ between $(4,1,1,\ldots,1)$ and $(4,2,2,\ldots,2)$ in $\Gamma_{n}$ yields a path of related characters from $\alpha_{n}$ to $\omega_{n}$, of length $n$, hence $m+1=n$. $\square$

In order to prove that $d(H_{n},G_{n})=2n$, it remains to show that $d(H_{n},G_{n})\geq 2n$ holds. If $d(H_{n},G_{n})\leq 2n-1=2(n-1)+1$, then by Cor. 1.2 (i) we have that every two irreducible characters of $H_{n}$ have distance at most $n-1$. However, the characters $\alpha_{n}$ and $\omega_{n}$ constructed in Lemma 3.1 have distance $n$, which is a contradiction. So we are done.

Acknowledgments

The first author was supported by the Stipendium Hungaricum PhD fellowship at the Budapest University of Technology and Economics. The second author gratefully acknowledges support by the German Research Foundation (DFG) within the SFB-TRR 195 Symbolic Tools in Mathematics and their Applications. The third author was supported by the NKFI-Grants No. 115288 and 115799.