# Schur ring and Codes for $S$-subgroups over $\mathbb{Z}_{2}^{n}$

Ronald Orozco López

###### Abstract

In this paper the relationship between $S$-subgroups in $\mathbb{Z}_{2}^{n}$
and binary codes is shown. If the codes used are both $P(T)$-codes and
$G$-codes, then the $S$-subgroup is free. The codes constructed are cyclic,
decimated or symmetric and the $S$-subgroups obtained are free under the
action the cyclic permutation subgroup, invariants under the action the
decimated permutation subgroup and symmetric under the action of symmetric
permutation subgroup, respectively. Also it is shows that there is no codes
generating whole $\mathbb{Z}_{2}^{n}$ in any $\mathcal{G}_{n}(a)$-complete
$S$-set of the $S$-ring $\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$.

Keywords: Schur ring, codes, binary sequences, decimation, autocorrelation
Mathematics Subject Classification: 05E15,94B60,05B20

##  1 Introduction

Let $G$ be a finite group with identity element $e$ and $\mathbb{C}[G]$ the
group algebra of all formal sums $\sum_{g\in G}a_{g}g$, $a_{g}\in\mathbb{C}$,
$g\in G$. For $T\subset G$, the element $\sum_{g\in T}g$ will be denoted by
$\overline{T}$. Such an element is also called a simple quantity. The
transpose of $\overline{T}=\sum_{g\in G}a_{g}g$ is defined as
$\overline{T}^{\top}=\sum_{g\in G}a_{g}(g^{-1})$. Let
$\\{T_{0},T_{1},...,T_{r}\\}$ be a partition of $G$ and let $S$ be the
subspace of $\mathbb{C}[G]$ spanned by
$\overline{T_{1}},\overline{T_{2}},...,\overline{T_{r}}$. We say that $S$ is a
Schur ring ($S$-ring, for short) over $G$ if:

  1. 1.

$T_{0}=\\{e\\}$,

  2. 2.

for each $i$, there is a $j$ such that
$\overline{T_{i}}^{\top}=\overline{T_{j}}$,

  3. 3.

for each $i$ and $j$, we have
$\overline{T_{i}}\cdot\overline{T_{j}}=\sum_{k=1}^{r}\lambda_{i,j,k}\overline{T_{k}}$,
for constants $\lambda_{i,j,k}\in\mathbb{C}$.

The numbers $\lambda_{i,j,k}$ are the structure constants of $S$ with respect
to the linear base
$\\{\overline{T_{0}},\overline{T_{1}},...,\overline{T_{r}}\\}$. The sets
$T_{i}$ are called the basic sets of the $S$-ring $S$. Any union of them is
called an $S$-sets. Thus, $X\subseteq G$ is an $S$-set if and only if
$\overline{X}\in S$. The set of all $S$-set is closed with respect to taking
inverse and product. Any subgroup of $G$ that is an $S$-set, is called an
$S$-subgroup of $G$ or $S$-group. A partition $\\{T_{0},...,T_{r}\\}$ of $G$
is called Schur partition or $S$-partition if $T_{0}=\\{e\\}$ and if for each
$i$ there is some $j$ such that $T_{i}^{-1}=\\{g^{-1}:g\in T_{i}\\}=T_{j}$. It
is known that there is a 1-1 correspondence between $S$-rings over $G$ and
$S$-partitions of $G$. By using this correspondence, in this paper we will
refer to an $S$-ring by mean of its $S$-partition.

The concept of $S$-ring was iniciated by I. Schur in their classical paper [1]
which was published in 1933. Later, the theory of $S$-ring was developed for
Wielandt [2]. But the main objective of theory was purely group theoretical
concept, especially in problem concerning the permutations groups. In the 80s
and 90s, the theory received a notable impulse by the study of $S$-ring over
cyclic groups and their applications to the graph theory [3],[4],[5],[6].

With the papers [8],[9] and [11] was initiated the research of $S$-ring over
the group $\mathbb{Z}_{2}^{n}$ and was shown the relationship between this and
Hadamard matrices, perfect binary sequences and periodic compatible binary
sequences. In this paper we will show the relationship between $S$-ring over
$\mathbb{Z}_{2}^{n}$ and binary codes. In particular we will use codes for to
construct $S$-subgroups over $\mathbb{Z}_{2}^{n}$. This point of view will be
shown as an alternative to the cocyclic matrices and to the difference sets
used for to research hadamard matrices, since with the schur rings and its
generator codes will be possible to understand the structure of the special
binary sequences above.

In this paper a code $\mathcal{X}_{n}$ generating whole $\mathbb{Z}_{2}^{n}$
is found. Then the others codes for $S$-subgroups are constructed by using
$\mathcal{X}_{n}$ as a base. A code $\mathcal{X}_{n}^{\prime}$ will be called
$G$-codes if there exists a permutation subgroup $G$ in
$Aut(\mathbb{Z}_{2}^{n})$ such that
$G\mathcal{X}_{n}^{\prime}=\mathcal{X}_{n}^{\prime}$. Other types of codes
studied are the $P(T)$-codes, closely related to the partition $P(T)$ of the
subset $T$ of $N=\\{0,1,...,n-1\\}$. In fact, the $S$-subgroups constructed
are both $P(T)$-codes and $G$-codes. This codes generating free $S$-subgroups
in $\mathbb{Z}_{2}^{n}$.

This paper is organized as follows. In section 2 basic concepts of theory of
codes are shown. In section 3 properties of $P(T)$-codes are stablished. Also
is shown that a $P(T)$-code generates a free subgroup in $\mathbb{Z}_{2}^{n}$
and that a $G$-code generates an $S$-subgroup in the same group. In section 4
is shown the connection between $G$-codes and $\mathcal{G}_{n}(a)$-complete
$S$-set. $S$-subgroups and its generator codes in $S$-rings induced by the
permutation subgroups $C_{n}$, $\Delta_{n}$, $H_{n}C_{n}$, $\Delta_{n}C_{n}$
and $H_{n}\Delta_{n}C_{n}$ of $Aut(\mathbb{Z}_{2}^{n})$ are studied in the
sections 5 to 10.

##  2 Some terminology of the Theory of Codes

In [11] the following terminology related to the theory of codes can be found

An arbitrary set $\mathcal{A}$ will be called an alphabet and its elements are
called letters. A finite sequence of letters written in the form
$s_{1}s_{2}\cdots s_{n}$, $n\geq 0$, with every $s_{i}$ in $\mathcal{A}$, is
called a word. Any subsequence of consecutive letters of a word is a subword.
When $n=0$ the word is the empty word and denoted by $1_{\mathcal{A}}$. Given
a word $w=s_{1}s_{2}\cdots s_{n}$, the number $n$ is called the lenght of $w$
and is denoted $l(w)$. Then the empty word $1_{\mathcal{A}}$ has lenght $0$,
i.e., $l(1_{\mathcal{A}})=0$.

Let $\mathcal{A}^{*}$ denote all finite words defined on $\mathcal{A}$ and let
$\mathcal{A}^{+}$ denote all finite nonempty words on $\mathcal{A}$.
$\mathcal{A}^{*}$ is equipped with an associative binary operation obtained by
concatenating two sequences:

$s_{1}s_{2}\cdots s_{n}\cdot t_{1}t_{2}\cdots t_{m}=s_{1}s_{2}\cdots
s_{n}t_{1}t_{2}\cdots t_{m}.$

The empty word is the unit element with respect to this operation and
consequently the sets $\mathcal{A}^{*}$ and $\mathcal{A}^{+}$ are a monoid and
a semigroup, respectivelly.

A factorization of a word $s\in\mathcal{A}^{*}$ is a sequence
$\\{s_{1},s_{2},...,s_{n}\\}$ of $n\geq 0$ words in $\Sigma^{*}$ such that
$s=s_{1}s_{2}\cdots s_{n}$. For a subset $X$ of $\mathcal{A}^{*}$, we denote
by $X^{*}$ the submonoid generated by $X$,

$X^{*}=\\{x_{1}x_{2}\cdots x_{n}|\ n\geq 0,\ x_{i}\in X\\}.$

Similarly, we denote by $X^{+}$ the subsemigroup generated by $X$,

$X^{{}^{+}}=\\{x_{1}x_{2}\cdots x_{n}|\ n\geq 1,\ x_{i}\in X\\}.$

By definition, each word $s$ in $X^{*}$ admits a least one factorization
$x_{1}x_{2}\cdots x_{n}$ with all $x_{i}$ in $X$. Such a factorization is
called an $X$-factorization.

A monoid $M$ is called free if it has a subset $B$ such that:

  1. 1.

$M=B^{*}$, and

  2. 2.

For all $n,m\geq 1$ and $x_{1},\cdots x_{n}$, $y_{1},\cdots y_{m}\in B$ we
have

$x_{1}\cdots x_{n}=y_{1}\cdots y_{m}\ \Rightarrow\ n=m\ \textsl{and}\
x_{i}=y_{i}\ \textsl{for}\ i=1,2,...,n.$ (2.1)

From condition 1., $B$ is a generating set of $M$ and condition 2. say us that
each element in $M$ has an unique representation as a product of elements of
$B$. The set $B$ satisfying 1. y 2. is called a base of $M$.

Let $M$ be a monoid and $B$ its generating set. We say that $B$ is minimal
generating set if no proper subset of $B$ is a generating set. A element $x$
of $M$ is called indecomposable or atomic if it cannot be expressed in the
form $x=yz$ with $y,z\neq 1$.

Now, by reformulating the condition (2.1) we obtain the following definition.
A subset $\mathcal{X}\subseteq\mathcal{A}^{*}$ is a code if it satisfies the
following condition: For all $n,m\geq 1$ and $x_{1},...,x_{n}$,
$y_{1},...,y_{m}\in\mathcal{X}$

$x_{1}\cdots x_{n}=y_{1}\cdots y_{m}\ \Rightarrow\ n=m\ \textsl{and}\
x_{i}=y_{i}\ \textsl{for}\ i=1,2,...,n.$ (2.2)

Note that the empty word $1_{\mathcal{A}}$ is never in a code. The following
theorem shows the equivalence between codes and free generating set

######  Theorem 1.

Let $\mathcal{X}\subseteq\mathcal{A}^{*}$. Then the following conditions are
equivalent:

  1. 1.

$\mathcal{X}$ is a code,

  2. 2.

$\mathcal{X}$ is a free generating set, or a base, of the monoid
$\mathcal{X}^{*}$,

  3. 3.

$\mathcal{X}^{*}$ is free and $\mathcal{X}$ is its minimal generating set.

On the other hand, let $G$ be a group with $\mathcal{A}\subseteq G$. Elements
of $\mathcal{A}^{*}$ represent elements of $G$ closed under concatenation and
inversion. The empty word represents $1_{G}$, the unity of $G$. Then
$\mathcal{A}^{*}$ is a subgroup of $G$. A word $w=s_{1}s_{2}\cdots s_{n}$ in
$\mathcal{A}^{*}$ is called reduced if $w$ contains no subwords $xx^{-1}$ or
$x^{-1}x$ for $x\in\mathcal{A}$. A group $G$ is called a free group if there
exists a generating set $X$ of $G$ such that every non-empty reduced group
word in $X$ defines a non-trivial element of $G$. Let $G$ be a free group on
$X$. Then the cardinality of $X$ is called the rank of $G$.

##  3 Schur rings and codes for $S$-subgroups over $\mathbb{Z}_{2}^{n}$

In this paper denote by $\mathbb{Z}_{2}$ the cyclic group of order 2 with
elements $+$ and $-$(where + and $-$ mean 1 and $-1$ respectively). Let
$\mathbb{Z}_{2}^{n}=\overset{n}{\overbrace{\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}}}$.
Then all $X\in\mathbb{Z}_{2}^{n}$ are sequences of $+$ and $-$ and will be
called $\mathbb{Z}_{2}$-sequences or binary sequences. All binary sequence in
$\mathbb{Z}_{2}^{n}$ is of the form $(x_{0},x_{1},...,x_{n-1})$. Let 1 denote
the sequence $(1,1,...,1)$. As $X^{2}=\textbf{1}$ for all $X$ in
$\mathbb{Z}_{2}^{n}$, then all reduced word in $\mathbb{Z}_{2}^{n}$ contains
no the subword $XX$. Now we will find a code generating whole
$\mathbb{Z}_{2}^{n}$.

We define the following subset of $\mathbb{Z}_{2}^{n}$

$\displaystyle\mathcal{X}_{n}=\left\\{\begin{array}[]{c}X_{0}=-+\cdots++\\\
X_{1}=+-\cdots++\\\ \vdots\\\ X_{n-2}=++\cdots-+\\\
X_{n-1}=++\cdots+-\end{array}\right\\}$ (3.6)

where each $-$ is in the $i$-th position. In the following theorem we shall
show that $\mathcal{X}_{n}$ is a base for all $\mathbb{Z}_{2}^{n}$

######  Theorem 2.

$\mathcal{X}_{n}$ is a code for $\mathbb{Z}_{2}^{n}$.

###### Proof.

As $|\mathcal{X}_{n}|=n$, then all word on $\mathcal{X}_{n}$ has the form

$w=X_{0}^{\epsilon_{0}}X_{1}^{\epsilon_{1}}\cdots X_{n-1}^{\epsilon_{n-1}},$

with $\epsilon_{i}=0,1$. Thus, the number of words on $\mathcal{X}_{n}$ of
lenght $k$ is $\binom{n}{k}$, with $k$ ranging in $[1,n-1]$. As the empty word
corresponds to 1 and as $X_{0}X_{1}\cdots X_{n-1}=-\textbf{1}$, the number
total of words constructed with the codewords $X_{i}$ is $2^{n}$. Hence there
exist a 1-1 correspondence between all words on $\mathcal{X}_{n}$ and all
binary sequences in $\mathbb{Z}_{2}^{n}$. Consequently
$\mathcal{X}_{n}^{*}=\mathbb{Z}_{2}^{n}$ as we announce. ∎

Let $Aut(\mathbb{Z}_{2}^{n})$ denote the automorphism group of
$\mathbb{Z}_{2}^{n}$ and take $G$ by a subgroup of $Aut(\mathbb{Z}_{2}^{n})$.
We shall denote with $\mathfrak{S}(\mathbb{Z}_{2}^{n},G)$ an $S$-partition of
$\mathbb{Z}_{2}^{n}$ under the action of $G$. As $\mathcal{X}_{n}$ generates
whole $\mathbb{Z}_{2}^{n}$, we wish to find codes on $\mathcal{X}_{n}$, this
is, with codewords factorizable on $\mathcal{X}_{n}$, for $S$-subgroups of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},G)$. We start with the following definition

######  Definition 1.

Let $T=\\{i_{1},i_{2},...,i_{r}\\}$ be a subset of $N=\\{0,1,...,n-1\\}$ and
let $P(T)=\\{T_{1},...,T_{s}\\}$ denote a partition on $T$. A code
$\mathcal{X}_{n}^{\prime}$ on $\mathcal{X}_{n}$ is a $P(T)$-code if
$\mathcal{X}_{n}^{\prime}=\\{Y_{T_{1}},...,Y_{T_{s}}\\}$, where
$Y_{T_{j}}=X_{j_{1}}\cdots X_{j_{k}}$ and $T_{j}=\\{j_{1},...,j_{k}\\}$.

The map $T_{j}\mapsto Y_{T_{j}}$ establishes an 1-1 correspondence between the
blocks $T_{j}$ of $T$ and the codewords $Y_{T_{j}}$ of
$\mathcal{X}_{n}^{\prime}$. Then it is easily inferred that

######  Theorem 3.

Let $\mathcal{X}_{n}^{\prime}$ be a $P(T)$-code. Then

$|\mathcal{X}_{n}^{\prime*}|=2^{|\mathcal{X}_{n}^{\prime}|}.$ (3.7)

We will call to $\mathcal{X}_{n}^{\prime*}$ a $P(T)$-free group.

###### Proof.

Let $P(T)=\\{T_{1},...,T_{s}\\}$ be a partition of some subset $T$ of $N$.
Then all word on $\mathcal{X}_{n}^{\prime}$ is irreducible. Hence the number
of words on $\mathcal{X}_{n}^{\prime}$ of lenght $k$ is
$\binom{|\mathcal{X}^{\prime}|}{k}$ and

$|\mathcal{X}_{n}^{\prime*}|=1+\sum_{k=1}^{|\mathcal{X}_{n}^{\prime}|}\binom{|\mathcal{X}_{n}^{\prime}|}{k}=2^{|\mathcal{X}_{n}^{\prime}|}$

∎

######  Corollary 1.

$\mathcal{X}_{n}$ in (3.6) is the only $P(T)$-code generating whole
$\mathbb{Z}_{2}^{n}$.

###### Proof.

$\mathcal{X}_{n}$ is a $P(T)$-code with
$P(T)=\\{\\{0\\},\\{1\\},...,\\{n-1\\}\\}$. The statement is followed from
here. ∎

Now, we will find the number of $P(T)$-free subgroups in $\mathbb{Z}_{2}^{n}$

######  Theorem 4.

The number of $P(T)$-free subgroup in $\mathbb{Z}_{2}^{n}$ is $B_{|T|+1}$
where $B_{|T|}$ are the Bell numbers.

###### Proof.

By the correspondence is clear that the number of $P(T)$-free subgroups for
any subset $T$ is $B_{|T|}$. As $\binom{n}{|T|}$ indicates the number of
$|T|$-element subsets of an $n$-element set, then $\binom{n}{|T|}B_{|T|}$
indicates the number of $P(T)$-free subgroups with fixed size. Assuming that
$B_{0}$ is the number of empty words we obtain
$\sum_{|T|=0}^{n}\binom{n}{|T|}B_{|T|}=B_{|T|+1}$. ∎

For example, the following are all $P(T)$-free subgroup of
$\mathbb{Z}_{2}^{3}$

$\displaystyle\\{X_{0},X_{1},X_{2}\\}^{*},\\{X_{0}X_{1},X_{2}\\}^{*},\\{X_{0}X_{2},X_{1}\\}^{*},\\{X_{1}X_{2},X_{0}\\}^{*}$
$\displaystyle\\{X_{0}X_{1}X_{2}\\}^{*}$
$\displaystyle\\{X_{0},X_{1}\\}^{*},\\{X_{0}X_{1}\\}^{*},\\{X_{0},X_{2}\\}^{*},\\{X_{0}X_{2}\\}^{*},\\{X_{1},X_{2}\\}^{*},\\{X_{1}X_{2}\\}^{*}$
$\displaystyle\\{X_{0}\\}^{*},\\{X_{1}\\}^{*},\\{X_{2}\\}^{*},$
$\displaystyle\\{1\\}^{*}$

The reason for deal with $P(T)$-code will be showed now. Let

$\mathcal{X}_{7}=\left\\{\begin{array}[]{c}X_{0}=-++++++\\\ X_{1}=+-+++++\\\
X_{2}=++-++++\\\ X_{3}=+++-+++\\\ X_{4}=++++-++\\\ X_{5}=+++++-+\\\
X_{6}=++++++-\\\ \end{array}\right\\}$ (3.8)

a code for $\mathbb{Z}_{2}^{7}$ and define the set

$\mathcal{X}_{7}^{\prime}=\\{X_{3}X_{5}X_{6},X_{2}X_{4}X_{5},X_{1}X_{3}X_{4},X_{0}X_{2}X_{3},X_{6}X_{1}X_{2},X_{5}X_{0}X_{1},X_{4}X_{6}X_{0}\\}.$
(3.9)

on $\mathcal{X}_{7}$. It is easy to show that $\mathcal{X}_{7}^{\prime}$ is
not a code. For example, $X_{0}X_{2}X_{5}X_{6}$ has at least two factorization
in $\mathcal{X}^{\prime}$, namely $X_{0}X_{2}X_{3}\cdot X_{3}X_{5}X_{6}$ and
$X_{2}X_{4}X_{5}\cdot X_{4}X_{6}X_{0}$. Also, $|\mathcal{X}_{7}^{\prime*}|=16$
and not $128$ as desirable. Hence $\mathcal{X}_{7}^{\prime*}$ is not free
group. However, from theorem 3 a $P(T)$-code always generates a free group.

The following theorem say us as to obtain new $P(T)$-free subgroup from old.

######  Theorem 5.

Let $\mathcal{X}_{i}$ be $P(T_{i})$-codes, $1\leq i\leq r$, in
$\mathbb{Z}_{2}^{n}$ such that $T_{i}\cap T_{j}=\emptyset$, $i\neq j$, and
$T_{i}\subset N$. Then

$\left(\bigcup_{i=1}^{r}\mathcal{X}_{i}\right)^{*}=\prod_{i=1}^{r}\mathcal{X}_{i}^{*}.$
(3.10)

and

$|\prod_{i=1}^{r}\mathcal{X}_{i}^{*}|=2^{\sum_{i=1}^{r}|\mathcal{X}_{i}|}$
(3.11)

###### Proof.

Follows by induction on number of $T_{i}$-codes $\mathcal{X}_{i}$. ∎

Now we will obtain a necessary condition for the existence of an $S$-subgroup

######  Theorem 6.

Let $G$ be a permutation automorphic subgroup of $Aut(\mathbb{Z}_{2}^{n})$
acting on some set $\mathcal{X}$ in $\mathbb{Z}_{2}^{n}$. Then
$\mathcal{X}^{*}$ is an $S$-subgroup in $\mathfrak{S}(\mathbb{Z}_{2}^{n},G)$.

###### Proof.

We take a word $Y_{i_{1}}Y_{i_{2}}\cdots Y_{i_{r}}$ in $\mathcal{X}^{\prime*}$
and a $g$ in $G$. Then

$\displaystyle g(Y_{i_{1}}Y_{i_{2}}\cdots Y_{i_{r}})$ $\displaystyle=$
$\displaystyle g(Y_{i_{1}})g(Y_{i_{2}})\cdots g(Y_{i_{r}})$ $\displaystyle=$
$\displaystyle Y_{j_{1}}Y_{j_{2}}\cdots Y_{j_{r}}.$

As $g$ is arbritary, then $g(Y_{i_{1}}Y_{i_{2}}\cdots Y_{i_{r}})$ is in
$\mathcal{X}^{\prime*}$ for all $g$ in $G$. Hence $G$ defines a partition on
$\mathcal{X}^{\prime*}$ and $\mathcal{X}^{\prime*}$ is an $S$-subgroup of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},G)$. ∎

Not all $S$-subgroup is free. $\mathcal{X}_{7}^{\prime*}$ in (3.9) is an
$S$-subgroup of $\mathfrak{S}(\mathbb{Z}_{2}^{7},C_{7})$, where
$C_{7}=\left\langle C\right\rangle$ is the cyclic permutation automorphic
subgroup of $Aut(\mathbb{Z}_{2}^{7})$ of order 7 with $C$ the cyclic
permutation acting on all component of some $Y$ in $\mathbb{Z}_{2}^{7}$. But
$\mathcal{X}_{7}^{\prime*}$ is not a free subgroup of $\mathbb{Z}_{2}^{7}$.

Again let $G$ be a permutation automorphic subgroup of
$Aut(\mathbb{Z}_{2}^{n})$ and let $Y_{G}$ denote the orbit of some $Y$ in
$\mathbb{Z}_{2}^{n}$ under the action of $G$. From previous theorem
$Y_{G}^{*}$ is an $S$-subgroup. We will called to $Y_{G}^{*}$ a basic
$S$-subgroup of $\mathfrak{S}(\mathbb{Z}_{2}^{n},G)$. If for a code
$\mathcal{X}$ it is true that $\mathcal{X}=Y_{G}$ for some $Y$ in
$\mathbb{Z}_{2}^{n}$, then we will say that $\mathcal{X}$ is a $G$-code. In
the following sections we construct $S$-subgroups by using $G$-codes.

##  4 Schur ring $\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$

Let $\omega(X)$ denote the Hamming weight of $X\in\mathbb{Z}_{2}^{n}$. Thus,
$\omega(X)$ is the number of $+$ in any $\mathbb{Z}_{2}-$sequences $X$ of
$\mathbb{Z}_{2}^{n}$. Now let $\mathcal{G}_{n}(k)$ be the subset of
$\mathbb{Z}_{2}^{n}$ such that $\omega(X)=k$ for all $X\in\mathcal{G}_{n}(k)$,
where $0\leq k\leq n$.

We let $T_{i}=\mathcal{G}_{n}(n-i)$. It is straightforward to prove that the
partition
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})=\\{\mathcal{G}_{n}(0),...,\mathcal{G}_{n}(n)\\}$
induces an $S$-partition over $\mathbb{Z}_{2}^{n}$, where $S_{n}\leq
Aut(\mathbb{Z}_{2}^{n})$ is the permutation group on $n$ objects. From [7] it
is know that the constant structure $\lambda_{i,j,k}$ is equal to

$\lambda_{i,j,k}=\begin{cases}0&\mbox{if }i+j-k\ \mbox{is an odd number}\\\
\binom{k}{(j-i+k)/2}\binom{n-k}{(j+i-k)/2}&\mbox{if }i+j-k\ \mbox{is an even
number}\end{cases}$ (4.1)

From (4.1) follows that

$\mathcal{G}_{n}(a)\mathcal{G}_{n}(b)=\begin{cases}\bigcup\limits_{i=0}^{a}\mathcal{G}_{n}(n-a-b+2i),&0\leq
a\leq\left[\dfrac{n}{2}\right],a\leq b\leq n-a,\\\
\bigcup\limits_{i=0}^{n-a}\mathcal{G}_{n}(a+b-n+2i),&\left[\dfrac{n}{2}\right]+1\leq
a\leq n,n-a\leq b\leq a.\end{cases}$ (4.2)

From (3.6) we know that $\mathcal{X}_{n}=\mathcal{G}_{n}(n-1)$. Then
$\mathcal{X}_{n}$ is an $S_{n}$-code for
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$. In the following corollary we found
another $S_{n}$-code for $\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$

######  Corollary 2.

$\mathcal{G}_{n}(1)$ is an $S_{n}$-code for
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$.

###### Proof.

It is enough to take into account that
$-\mathcal{G}_{n}(n-1)=\mathcal{G}_{n}(1)$. ∎

We prefer to use the $S_{n}$-code $\mathcal{G}_{n}(n-1)$ and not
$\mathcal{G}_{n}(1)$ because in $\mathcal{G}_{n}(n-1)$ the positions of the
negative components are easily obtained. Indeed, $+++-\cdot+-++=+-+-$ in
$\mathcal{G}_{4}(3)$ but in $\mathcal{G}_{4}(1)$ we have $---+\cdot-+--=+-+-$.

Next we will see that a $G$-code is contained in any $S$-set of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$

######  Proposition 1.

A $G$-code $\mathcal{X}$ is contained in $\mathcal{G}_{n}(a)$ for some $a$
ranging in $[1,n-1]$.

###### Proof.

Take $X$ in $\mathcal{X}$. It is easy to note that $\omega(gX)=\omega(X)$ for
all $g\in G$. Hence $\mathcal{X}\subseteq\mathcal{G}_{n}(a)$ for some $a$ in
$[1,n-1]$. ∎

On the other hand, it is follows directly from (4.1) that
$\lambda_{i,j,2k+1}=0$ if $i+j$ is even and $\lambda_{i,j,2k}=0$ if $i+j$ is
odd. The union of all basic sets $\mathcal{G}_{n}(2a)$ in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$ will be denoted by $\mathcal{E}_{n}$
and the union of all basic sets $\mathcal{G}_{n}(2a+1)$ in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$ will be denoted $\mathcal{O}_{n}$.
The sets $\mathcal{E}_{2n}$ and $\mathcal{O}_{2n+1}$ are subgroups of order
$2^{2n-1}$ and $2^{2n}$, respectively. Then

$\mathfrak{S}(\mathcal{E}_{2n},S_{n})=\\{\mathcal{G}_{2n}(0),\mathcal{G}_{2n}(2),...,\mathcal{G}_{2n}(2n)\\}$

and

$\mathfrak{S}(\mathcal{O}_{2n+1},S_{n})=\\{\mathcal{G}_{2n+1}(1),\mathcal{G}_{2n+1}(3),...,\mathcal{G}_{2n+1}(2n+1)\\}$

are $S$-subgroups of $\mathfrak{S}(\mathbb{Z}_{2}^{2n},S_{2n})$ and
$\mathfrak{S}(\mathbb{Z}_{2}^{2n+1},S_{2n+1})$, respectively.

From (4.2),
$\mathcal{G}_{2n}(n)^{2}=\bigcup_{i=0}^{n}\mathcal{G}_{2n}(2i)=\mathcal{E}_{2n}$
and
$\mathcal{G}_{2n+1}(n)^{2}=\bigcup_{i=0}^{n}\mathcal{G}_{2n+1}(2i+1)=\mathcal{O}_{2n+1}$.
Therefore, neither $\mathcal{G}_{4n}(2n)$ nor $\mathcal{G}_{4n+3}(2n+1)$
contains some code $\mathcal{X}$ generating whole $\mathbb{Z}_{2}^{4n}$ and
$\mathbb{Z}_{2}^{2n+1}$, respectively. This remark is generalized below.

From [7] is obtained the following definition

######  Definition 2.

Take $\mathcal{G}_{n}(a)$ in $\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$. Let
$\mathfrak{S}^{\prime}\subset\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$ be a set
of basic sets. We will call $\mathfrak{S}^{\prime}$ a
$\mathcal{G}_{n}(a)$-complete $S$-set if it holds

  1. 1.

$\mathcal{G}_{n}(i)\mathcal{G}_{n}(j)\supset\mathcal{G}_{n}(a)$ for all
$\mathcal{G}_{n}(i),\mathcal{G}_{n}(j)\in\mathfrak{S}^{\prime}$,

  2. 2.

There is no $\mathcal{G}_{n}(b)\in\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$ such
that $\mathcal{G}_{n}(b)^{2}\supset\mathcal{G}_{n}(a)$ and
$\mathcal{G}_{n}(b)\mathcal{G}_{n}(k)\supset\mathcal{G}_{n}(a)$ for all
$\mathcal{G}_{n}(k)\in\mathfrak{S}^{\prime}$.

A important result obtained is that there is no $\mathcal{G}_{n}(a)$-complete
for all $n$ and all $a$

######  Theorem 7.

  1. 1.

There is no $\mathcal{G}_{2n}(2a+1)$-complete $S$-sets in
$\mathfrak{S}(\mathbb{Z}_{2}^{2n},S_{n})$.

  2. 2.

There is no $\mathcal{G}_{2n+1}(2a)$-complete $S$-sets in
$\mathfrak{S}(\mathbb{Z}_{2}^{2n+1},S_{n})$.

In the following theorem is shown the relationship between codes generating
whole $\mathbb{Z}_{2}^{n}$ and non $\mathcal{G}_{n}(a)$-complete $S$-sets in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$

######  Theorem 8.

There is no a code $\mathcal{X}$ generating whole $\mathbb{Z}_{2}^{n}$ in a
$\mathcal{G}_{n}(a)$-complete $S$-set.

###### Proof.

Let $\mathfrak{S}^{\prime}$ denote a $\mathcal{G}_{2n}(2a)$-complete $S$-set.
From (4.2)

$\mathcal{G}_{2n}(2b)^{2}=\bigcup_{i=0}^{2b}\mathcal{G}_{2n}(2n-4b+2i)$

for all $\mathcal{G}_{2n}(2b)$ in $\mathfrak{S}^{\prime}$. Then all powers of
$\mathcal{G}_{2n}(2b)$ will contain basic sets $\mathcal{G}_{2n}(2k)$ only.
Therefore the basic sets in a $\mathcal{G}_{2n}(2a)$-complete can generate the
$S$-subgroup $\mathcal{E}_{2n}$ at the most. With a similar argument is shown
for basic sets in $\mathcal{G}_{2n+1}(2a+1)$-complete $S$-sets. ∎

We finish this section showing some basic sets of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$ that can to contain $G$-codes
generating all $\mathbb{Z}_{2}^{n}$.

######  Proposition 2.

  1. 1.

$\mathcal{G}_{4n}(2n-1)^{3}=\mathcal{O}_{4n}$,
$\mathcal{G}_{4n}(2n-1)^{4}=\mathcal{E}_{4n}$.

  2. 2.

$\mathcal{G}_{4n+2}(2n-1)^{3}=\mathcal{O}_{4n+2}$,
$\mathcal{G}_{4n}(2n-1)^{4}=\mathcal{E}_{4n+2}$.

  3. 3.

$\mathcal{G}_{4n+1}(2n-2)^{3}=\mathcal{E}_{4n+1}$,
$\mathcal{G}_{4n+1}(2n-2)^{4}=\mathcal{O}_{4n+1}$.

  4. 4.

$\mathcal{G}_{4n+3}(2n)^{3}=\mathcal{E}_{4n+3}$,
$\mathcal{G}_{4n+3}(2n)^{4}=\mathcal{O}_{4n+3}$.

###### Proof.

By using (4.2) we have

$\displaystyle\mathcal{G}_{4n}(2n-1)^{2}$ $\displaystyle=$
$\displaystyle\mathcal{G}_{4n}(2)\cup\cdots\cup\mathcal{G}_{4n}(4n)$
$\displaystyle\supset$ $\displaystyle\mathcal{G}_{4n}(2n).$

As

$\mathcal{G}_{4n}(2n-1)\mathcal{G}_{4n}(2n)=\mathcal{G}_{4n}(1)\cup\cdots\cup\mathcal{G}_{4n}(4n-1)$
(4.4)

then $\mathcal{G}_{4n}(2n-1)^{3}=\mathcal{O}_{4n}$. From
$\mathcal{G}_{4n}(2n-1)\mathcal{G}_{4n}(2n+1)\supset\mathcal{G}_{4n}(0)$ and
from (4) is followed that $\mathcal{G}_{4n}(2n-1)^{4}=\mathcal{E}_{4n}$. ∎

As $S_{n}$ induces a $S$-partition on $\mathbb{Z}_{2}^{n}$ is straightforward
to prove that $G$ induces a $S$-partition on $\mathbb{Z}_{2}^{n}$ for all
$G\leq S_{n}\leq Aut(\mathbb{Z}_{2}^{n})$. In the following sections we will
construct $S$-subgroups by using $G$-codes in $S$-ring
$\mathfrak{S}(\mathbb{Z}_{2}^{n},G)$.

##  5 Schur ring $\mathfrak{S}(\mathbb{Z}_{2}^{n},C_{n})$

Let $C$ denote the cyclic permutation on the components $+$ and $-$ of $X$ in
$\mathbb{Z}_{2}^{n}$ such that

$C(X)=C\left(x_{0},x_{1},...,x_{n-2},x_{n-1}\right)=\left(x_{1},x_{2},x_{3},...,x_{0}\right),$
(5.1)

that is, $C(x_{i})=x_{(i+1)modn}$. The permutation $C$ is a generator of
cyclic group $C_{n}=\left\langle C\right\rangle$ of order $n$. Let
$X_{C}=Orb_{C_{n}}X=\\{C^{i}(X):C^{i}\in C_{n}\\}$. Therefore, $C_{n}$ defines
a partition in equivalent class on $\mathbb{Z}_{2}^{n}$ which is an
$S$-partition and this we shall denote by
$\mathbb{Z}_{2C}^{n}=\mathfrak{S}(\mathbb{Z}_{2}^{n},C_{n})$. It is worth
mentioning that this Schur ring corresponds to the orbit Schur ring induced by
the cyclic permutation automorphic subgroup $C_{n}\leq S_{n}\leq
Aut(\mathbb{Z}_{2}^{n})$.

On the other hand, let $X=\\{x_{i}\\}$ and $Y=\\{y_{i}\\}$ be two complex-
valued sequences of period $n$. The periodic correlation of $X$ and $Y$ at
shift $k$ is the product defined by:

$\mathsf{P}_{X,Y}(k)=\sum\limits_{i=0}^{n-1}x_{i}\overline{y}_{i+k},\
k=0,1,...,n-1,$ (5.2)

where $\overline{a}$ denotes the complex conjugation of $a$ and $i+k$ is
calculated modulo $n$. If $Y=X$, the correlation $\mathsf{P}_{X,Y}(k)$ is
denoted by $\mathsf{P}_{X}(k)$ and is the autocorrelation of $X$. Obviously,

$\displaystyle\mathsf{P}_{X}(k)$ $\displaystyle=$
$\displaystyle\overline{\mathsf{P}_{X}(n-k)},$ (5.3)
$\displaystyle\mathsf{P}_{-X}(k)$ $\displaystyle=$
$\displaystyle\mathsf{P}_{X}(k),$ (5.4) $\displaystyle\mathsf{P}_{C^{i}X}(k)$
$\displaystyle=$ $\displaystyle\mathsf{P}_{X}(k),$ (5.5)

for all $0\leq i\leq n-1$ and for all $X$ in $\mathbb{Z}_{2}^{n}$.

If $X$ is a $\mathbb{Z}_{2}$-sequence of length $n$,
$\mathsf{P}_{X}(k)=2\omega\left\\{Y_{k}\right\\}-n$, where $Y_{k}=XC^{k}X$.
Also by (4.2), if $X\in\mathcal{G}_{n}(a)$, then

$\mathsf{P}_{X}(k)=n-4a+4i_{k},$ (5.6)

for some $0\leq i_{k}\leq a$ and $n-\mathsf{P}_{X}(k)$ is divisible by 4 for
all $k$.

We know from theorem 2 that $\mathcal{G}_{n}(n-1)$ is a code for
$\mathfrak{S}(\mathbb{Z}_{2}^{n},S_{n})$ for all $n$. As
$\mathcal{G}_{n}(n-1)=\\{X,CX,C^{2}X,...,C^{n-1}\\}=\mathcal{X}_{C}$, then
$\mathcal{X}_{C}$ is a $C_{n}$-code for
$\mathfrak{S}(\mathbb{Z}_{2}^{n},C_{n})$. Then for to obtain information from
each basic set $Y_{C}$ in $\mathfrak{S}(\mathbb{Z}_{2}^{n},C_{n})$ we must do
it through of its $\mathcal{X}_{C}$-factorization with the $C_{n}$-code
$\mathcal{X}_{C}$. The advantage of using this code lies in its simplicity,
since each $C^{i}X$ has exactly a $-$ as its component and thereby it is
possible to know exactly the Hamming weight of each word writing with this
basis.

All word $Y$ in $\mathcal{G}_{n}(a)$ has the form $C^{i_{1}}XC^{i_{2}}X\cdots
C^{i_{r}}X$ with length $|Y|=r$ and with $a=n-r$. Then, every basic set in
$\mathcal{G}_{nC}(a)$ has form

$Y_{C}=\bigcup_{k=0}^{n-1}C^{i_{1}+k}XC^{i_{2}+k}X\cdots C^{i_{r}+k}X,$ (5.7)

and in this way if $Z=C^{j_{1}}XC^{j_{2}}X\cdots C^{j_{s}}X$, we have

$\displaystyle Y_{C}Z_{C}$ $\displaystyle=$
$\displaystyle\bigcup_{k=0}^{n-1}(YC^{k}Z)_{C}$ $\displaystyle=$
$\displaystyle\bigcup_{k=0}^{n-1}(C^{i_{1}}X\cdots
C^{i_{r}}XC^{j_{1}+k}X\cdots C^{j_{s}+k}X)_{C}.$

Each word $YC^{k}Z=C^{i_{1}}X\cdots C^{i_{r}}XC^{j_{1}+k}X\cdots C^{j_{s}+k}X$
can be reduced if exist two equal letters. Thereupon $YC^{k}Z$ decreases its
length an even number. Therefore $YC^{k}Z$ belong to $\mathcal{G}_{n}(b)$ with
$b=n-(r+s)+2w$ where $2w$ is the number of canceled letters. If both $Y$ and
$Z$ belongs to $\mathcal{G}_{n}(a)$, then $b=n-2r+2w$ and
$\mathsf{P}_{Y,Z}(k)=n-4r+4w_{k}$.

Next we will obtain the algebraic version of (5.3), (5.4) and (5.5)

######  Proposition 3.

Let $Y$ denote the binary sequence $C^{i_{1}}XC^{i_{2}}X\cdots C^{i_{r}}X$. If
$YC^{k}Y\in\mathcal{G}_{n}(a)$, then

  1. 1.

$YC^{n-k}Y$ and $(C^{j}X)C^{k}(C^{j}X)$ are in $\mathcal{G}_{n}(a)$ too.

  2. 2.

$(-Y)C^{k}(-Y)\in\mathcal{G}_{n}(a)$.

###### Proof.

1\. It is clear that

$YC^{k}Y=C^{i_{1}}XC^{i_{2}}X\cdots C^{i_{r}}XC^{i_{1}+k}XC^{i_{2}+k}X\cdots
C^{i_{r}+k}X\in\mathcal{G}_{n}(a)$

with $a=n-2r+2w$, where $2w$ are the number of canceled letters. We wish to
show that the cancellation numbers of $YC^{k}Y$ and $YC^{n-k}Y$ coincide.
Suppose that $i_{j}=i_{1}+k$ for some $j$ and some $k$. Then this implies that
$n-k+i_{j}=i_{1}$ reduced module $n$. Therefore

$YC^{n-k}Y=C^{i_{1}}X\cdots C^{i_{r}}XC^{n-k+i_{1}}X\cdots C^{n-k+i_{r}}X$

has the same number of cancellations as $YC^{k}Y$. Equally is proved for
$(C^{j}X)C^{k}(C^{j}X)$. 2\. As $C^{k}(-Y)=-C^{k}Y$, then
$(-Y)C^{k}(-Y)=YC^{k}Y\in\mathcal{G}_{n}(a)$. ∎

Now we will show other advantage of to use the $C_{n}$-code $\mathcal{X}_{C}$

######  Proposition 4.

For all $n\leq 2$ we have

$\overline{\mathcal{G}_{n}(n-2)}^{2}=n+2\overline{\mathcal{G}_{n}(n-2)}+(n-3)\overline{\mathcal{G}_{n}(n-4)}.$
(5.8)

###### Proof.

All word $Y$ in $\mathcal{G}_{n}(n-2)$ has the form $C^{i}XC^{j}X$ with $i<j$.
Then $YC^{k}Y=C^{i}XC^{j}XC^{i+k}XC^{j+k}X$ and there exist a $k$ such that
either $i+k=j$ or $j+k=i$ and for all the remaining values of $k$ we have that
$C^{i}XC^{j}XC^{i+k}XC^{j+k}X$ is a reduced word. As $YC^{k}Y$ and $YC^{n-k}Y$
are in $\mathcal{G}_{n}(a)$ for some $a$, then $Y_{C}^{2}$ contains $2$ words
in $\mathcal{G}_{n}(n-2)$, $n-3$ words in $\mathcal{G}_{n}(n-4)$ and the
trivial word in $\mathcal{G}_{n}(n)$. ∎

On the other hand, let

$F_{d}(\mathbb{Z}_{2}^{n})=\bigcup_{|X|=d}X.$ (5.9)

Clearly $d$ divides to $n$ and the $X\in F_{d}(\mathbb{Z}_{2}^{n})$ have the
form $X=(Y,Y,...,Y)$, with $Y\in\mathbb{Z}_{2}^{d}$. Then
$F_{d}(\mathbb{Z}_{2C}^{n})=\bigcup_{|X_{C}|=d}X_{C}$ is an $S$-set of
$\mathbb{Z}_{2C}^{n}$, for each $d|n$. When $d=n$, we will to say that $C_{n}$
acts freely on $X_{C}$ and we denote $F_{n}(\mathbb{Z}_{2C}^{n})$ as
$F(\mathbb{Z}_{2C}^{n})$. When $d<n$, we will to say that $C_{n}$ don’t act
freely on $X_{C}$ and let $\widehat{F}(\mathbb{Z}_{2C}^{n})$ denote the set of
the $X_{C}$ which are not frees under the action of $C_{n}$, namely

$\widehat{F}(\mathbb{Z}_{2C}^{n})=\bigcup_{d\mid
n,d<n}F_{d}(\mathbb{Z}_{2C}^{n}).$ (5.10)

Therefore,

$\displaystyle\mathbb{Z}_{2C}^{n}$ $\displaystyle=$ $\displaystyle
F(\mathbb{Z}_{2C}^{n})\cup\widehat{F}(\mathbb{Z}_{2C}^{n})$ (5.11)
$\displaystyle=$ $\displaystyle\bigcup_{d\mid n}F_{d}(\mathbb{Z}_{2C}^{n}).$

The set $F_{d}(\mathbb{Z}_{2}^{n})$ is constructed with codewords in

$\mathcal{X}_{F,d}=\\{A_{0,d}X,A_{1,d}X,\cdots,A_{d-1,d}X\\}$ (5.12)

where $A_{i,d}X=C^{i}XC^{i+d}X\cdots C^{i+kd}X$ for $i=0,1,2,...,d-1$ with
$k=\frac{n}{d}-1$, $X\in\mathcal{G}_{n}(n-1)$ and

$\\{i,i+d,i+2d,\cdots,i+kd\\}$

is an arithmetic progression. Then $\mathcal{X}_{F,d}$ is a $P(T)$-code with

$P(T)=\\{\\{i,i+d,i+2d,...,i+kd\\}:\ i=0,1,...,d-1\\}$

and any word in $\mathcal{X}_{F,d}^{*}$ has the form

$Y=A_{0,d}^{\epsilon_{0}}XA_{1,d}^{\epsilon_{1}}X\cdots
A_{d-1,d}^{\epsilon_{d-1}}X$ (5.13)

$\epsilon_{i}=0,1$. It is clear that $C^{i}A_{j,d}X=A_{i+j,d}X$. Hence
$\mathcal{X}_{F,d}$ is a $C_{n}$-code.

If $d=1$, then $k=n-1$, $i=0$ and

$\mathcal{X}_{F,1}=\\{A_{0,1}X\\}=\\{XCXC^{2}X\cdots C^{n-1}X\\}$ (5.14)

is a code with a codeword. As $XCXC^{2}X\cdots C^{n-1}X=-\textbf{1}$, then
$\mathcal{X}_{F,1}^{*}=\\{\textbf{1},-\textbf{1}\\}$ for all $n$.

If $d=n$, then $k=0$, $i=0,1,\cdots,n-1$ and

$\mathcal{X}_{F,n}=\\{A_{0,n}X,A_{1,n}X,\cdots,A_{n-1,n}X\\}=\\{X,CX,C^{2}X,\cdots,C^{n-1}X\\}$
(5.15)

is an code with exactly $n$ codewords and $\mathcal{X}_{F,d}=\mathcal{X}_{C}$.
Let $|\mathcal{X}_{F,d}|$ be the rank of $\mathcal{X}_{F,d}$. We then note
that $1<|\mathcal{X}_{F,d}|<n$ for $1<d<n$.

Now we will see the relationship between free subgroup $\mathcal{X}_{F,d}^{*}$
and the sets $F_{d}(\mathbb{Z}_{2}^{n})$

######  Theorem 9.

$\mathcal{X}_{F,d}^{*}$ is a subgroup of $\mathbb{Z}_{2}^{n}$ of order $2^{d}$
with $\mathcal{X}_{F,d}^{*}=\bigcup_{r|d}F_{d}(\mathbb{Z}_{2}^{n})$. We will
donote this subgroup with $\mathbb{G}_{d}(n)$.

###### Proof.

As $\mathcal{X}_{F,d}$ is a $P(T)$-code, then $\mathbb{G}_{d}(n)$ is a
subgroup of $\mathbb{Z}_{2}^{n}$ of order $2^{d}$ for all divisor $d$ of $n$.
Then we only will show that $\mathbb{G}_{d}(n)$ has the desired structure. If
$d$ is a prime divisor of $n$, then all words of $\mathbb{G}_{d}(n)$ are in
$F_{d}(\mathbb{Z}_{2}^{n})$, except for 1 and $-\textbf{1}$. Hence

$\mathbb{G}_{d}(n)=F_{1}(\mathbb{Z}_{2}^{n})\cup F_{d}(\mathbb{Z}_{2}^{n})$
(5.16)

with

$\displaystyle-\textbf{1}$ $\displaystyle=$
$\displaystyle\prod_{i=0}^{d-1}A_{i,d}X$ 1 $\displaystyle=$
$\displaystyle(A_{i,d}X)^{2}\ for\ all\ i.$

Suppose that $d$ is no prime. Then

$A_{i,d}XA_{i+r,d}X\cdots A_{i+(\frac{d}{r}-1)r,d}X$ (5.17)

is contained in $F_{r}(\mathbb{Z}_{2}^{n})$, $r|d$ and $i=0,1,...,r-1$.
Therefore $\mathbb{G}_{d}(n)=\bigcup_{r|d}F_{r}(\mathbb{Z}_{2}^{n})$. ∎

######  Corollary 3.

$\mathbb{G}_{dC}(n)$ is an $S$-subgroup of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},C_{n})$.

###### Proof.

The group $\left\langle C_{n}\right\rangle$ defines a partition on each
$F_{r}(\mathbb{Z}_{2}^{n})$ in $\mathbb{G}_{d}(n)$ and hence we obtain the
desired statement. ∎

######  Example 1.

The subgroup $\mathbb{G}_{3}(9)$ of $\mathbb{Z}_{2}^{9}$ is given by

$\displaystyle\mathbb{G}_{3}(9)$ $\displaystyle=$ $\displaystyle
F_{1}(\mathbb{Z}_{2}^{9})\cup F_{3}(\mathbb{Z}_{2}^{9})$ $\displaystyle=$
$\displaystyle\\{-\textbf{1},\textbf{1}\\}\cup\\{XC^{3}XC^{6}X,\
CXC^{4}XC^{7}X,\ C^{2}XC^{5}XC^{8}X,$ $\displaystyle
XCXC^{3}XC^{4}XC^{6}XC^{7}X,\ XC^{2}XC^{3}XC^{5}XC^{6}XC^{8}X,$ $\displaystyle
CXC^{2}XC^{4}XC^{5}XC^{7}XC^{8}X\\}$ $\displaystyle=$
$\displaystyle\\{-\textbf{1},\textbf{1},-++-++-++,++-++-++-,$
$\displaystyle+-++-++-+,-+--+--+-,--+--+--+,$ $\displaystyle+--+--+--\\}.$

And the $S$-subgroup $\mathbb{G}_{3C}(9)$ of
$\mathfrak{S}(\mathbb{Z}_{2}^{9},C_{9})$ is given by

$\displaystyle\mathbb{G}_{3C}(9)$ $\displaystyle=$
$\displaystyle\\{\textbf{1},-\textbf{1},(-++-++-++)_{C},(-+--+--+-)_{C}\\}.$

######  Theorem 10.

$\mathbb{G}_{d}(n)\subseteq\bigcup_{a=0}^{d}\mathcal{G}_{n}\left(\frac{na}{d}\right)$
with equality only for $d=1,n$.

###### Proof.

Follows from
$F_{d}(\mathbb{Z}_{2}^{n})\subseteq\mathcal{G}_{n}\left(\frac{n}{d}\right)\cup\mathcal{G}_{n}\left(n-\frac{n}{d}\right)$.
∎

Now we show the lattice of $S$-subgroups $\mathbb{G}_{dC}(60)$ of
$\mathfrak{S}(\mathbb{Z}_{2}^{60},C_{60})$ ordered by inclusion.

$\mathbb{G}_{30C}(60)$$\mathbb{G}_{15C}(60)$$\mathbb{G}_{10C}(60)$$\mathbb{G}_{6C}(60)$$\mathbb{G}_{5C}(60)$$\mathbb{G}_{3C}(60)$$\mathbb{G}_{2C}(60)$$\mathbb{G}_{1C}(60)=\\{\textbf{1},-\textbf{1}\\}$$\mathbb{G}_{60C}(60)=\mathbb{Z}_{2C}^{60}$$\mathbb{G}_{20C}(60)$$\mathbb{G}_{12C}(60)$$\mathbb{G}_{4C}(60)$

We finish this section we provide other proof to the Theorems 6 and 7 in [7].
We start with the lemma

######  Lemma 1.

$XC^{n}X\in F_{n}(\mathbb{Z}_{2}^{2n})$ (5.18)

for all $X$ in $\mathcal{X}_{C}=\mathcal{G}_{2n}(2n-1)$

###### Proof.

Clearly $XC^{n}X$ is in $F_{n}(\mathbb{Z}_{2}^{2n})$ when $X=-+++\cdots+++$.
As

$(C^{i}X)C^{n}(C^{i}X)=C^{i}(XC^{n}X)$

for any other codeword $C^{i}X$ in the code $\mathcal{X}_{C}$, then
$C^{i}(XC^{n}X)\in F_{n}(\mathbb{Z}_{2}^{2n})$ for all $1\leq i\leq 2n-1$,
since $\left\langle C_{n}\right\rangle$ defines a partition on
$F_{n}(\mathbb{Z}_{2}^{2n})$. ∎

######  Theorem 11.

If $Y_{C}\in F(\mathbb{Z}_{2C}^{2n})$, then
$Y_{C}^{2}\setminus\\{\textbf{1}\\}\not\in F(\mathbb{Z}_{2C}^{2n})$.

###### Proof.

Take $Y=C^{i_{1}}X\cdots C^{i_{r}}X$ in $F(\mathbb{Z}_{2}^{2n})$. Then

$\displaystyle YC^{n}Y$ $\displaystyle=$ $\displaystyle C^{i_{1}}X\cdots
C^{i_{r}}XC^{i_{1}+n}X\cdots C^{i_{r}+n}X$ $\displaystyle=$ $\displaystyle
C^{i_{1}}(XC^{n}X)\cdots C^{i_{r}}(XC^{n}X).$

From the above lemma $XC^{n}X$ is in $F_{n}(\mathbb{Z}_{2}^{2n})$ for all $X$
in $\mathcal{X}_{C}$. Then $YC^{n}Y\in F_{n}(\mathbb{Z}_{2}^{2n})$ for all $Y$
in $F(\mathbb{Z}_{2}^{n})$ and

$Y_{C}^{2}\setminus\\{\textbf{1}\\}\notin F(\mathbb{Z}_{2C}^{2n})$

as we promised to show. ∎

######  Lemma 2.

$XC^{k}X\notin F_{d}(\mathbb{Z}_{2}^{2n+1})$ (5.19)

for no $X$ in $\mathcal{X}_{C}=\mathcal{G}_{2n+1}(2n)$ and for no $k$ ranging
in $[1,2n]$, $d<2n+1$.

###### Proof.

Take $X=-++\cdots++$ in $\mathcal{X}_{C}$. Then
$XC^{k}X=-\overset{2n-k}{\overbrace{+\cdots+}}-\overset{k-1}{\overbrace{+\cdots+}}$.
If we want $k-1=2n-k$, then $k=\frac{2n+1}{2}$, which is not possible. Hence
$XC^{k}X$ is no contained in $F_{d}(\mathbb{Z}_{2}^{2n+1})$, $d<2n+1$. As
$C^{i}(XC^{k}X)=(C^{i}X)C^{k}(C^{i}X)$, it is followed the statement. ∎

######  Theorem 12.

If $X_{C}\in F(\mathbb{Z}_{2C}^{2n+1})$, then
$X_{C}^{2}\setminus\\{\textbf{1}\\}\in F(\mathbb{Z}_{2C}^{2n+1})$.

###### Proof.

Let $Y=C^{i_{1}}XC^{i_{2}}X\cdots C^{i_{r}}X$ such that all the $i_{j}$ are
not in arithmetic progression. If
$YC^{k}Y=C^{i_{1}}(XC^{k}X)C^{i_{2}}(XC^{k}X)\cdots C^{i_{r}}(XC^{k}X)$ is
contained in some $F_{d}(\mathbb{Z}_{2}^{2n+1})$, $d|(2n+1)$, $d<2n+1$, then
$XC^{k}X$ must be contained in $F_{d}(\mathbb{Z}_{2}^{2n+1})$, but is not
possible by the previous lemma. ∎

##  6 Schur ring $\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n})$

Let $\delta_{a}\in S_{n}$ act on $X\in\mathbb{Z}_{2}^{n}$ by decimation, that
is, $\delta_{a}(x_{i})=x_{ai(\mod n)}$ for all $x_{i}$ in $X$, $(a,n)=1$ and
let $\Delta_{n}$ denote the set of this $\delta_{a}$. The set $\Delta_{n}$ is
a group of order $\phi(n)$ isomorphic to $\mathbb{Z}_{n}^{*}$, the group the
units of $\mathbb{Z}_{n}$, where $\phi$ is called the Euler totient function.
Clearly $\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n})$ is an $S$-partition of
$\mathbb{Z}_{2}^{n}$. In this section, we will construct $\Delta_{n}$-codes
for $S$-subgroups of $\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n})$.

We will use the commutation relation $C^{i}\delta_{a}=\delta_{a}C^{ia}$ for to
prove all of results in this section. We begin for show that
$\mathcal{G}_{n}(n-1)$ is partitioned in three equivalence class

######  Proposition 5.

$\Delta_{n}\mathcal{G}_{n}(n-1)=\\{X\\}\cup\mathcal{X}_{\mathbb{Z}_{n}^{*}}\cup\mathcal{X}_{\mathbb{Z}_{n}\setminus\mathbb{Z}_{n}^{*}}$
(6.1)

where $X=-++\cdots++$ and

$\displaystyle\mathcal{X}_{\mathbb{Z}_{n}^{*}}$ $\displaystyle=$
$\displaystyle\\{C^{a_{1}}X,C^{a_{2}}X,...,C^{a_{\phi(n)}}X:\ (a_{i},n)=1\\}$
(6.2) $\displaystyle\mathcal{X}_{\mathbb{Z}_{n}\setminus\mathbb{Z}_{n}^{*}}$
$\displaystyle=$ $\displaystyle\\{C^{d_{1}}X,C^{d_{2}}X,...,C^{d_{r}}X:\
(d_{i},n)\neq 1\\}$ (6.3)

$r=n-\phi(n)-1$.

###### Proof.

It is very easy to see that $X=-++\cdots++$ is fixed under the action of
$\Delta_{n}$. Also, as $\delta_{a}C^{i}X=C^{a^{-1}i}\delta_{a}X=C^{a^{-1}i}X$,
then
$\Delta_{n}\mathcal{X}_{\mathbb{Z}_{n}^{*}}=\mathcal{X}_{\mathbb{Z}_{n}^{*}}$
and
$\Delta_{n}\mathcal{X}_{\mathbb{Z}_{n}\setminus\mathbb{Z}_{n}^{*}}=\mathcal{X}_{\mathbb{Z}_{n}\setminus\mathbb{Z}_{n}^{*}}$.
∎

As each $C^{i}X$ in $\mathcal{X}_{\mathbb{Z}_{n}^{*}}$ or in
$\mathcal{X}_{\mathbb{Z}_{n}\setminus\mathbb{Z}_{n}^{*}}$ is atomic, then
$\mathcal{X}_{\mathbb{Z}_{n}^{*}}$ and
$\mathcal{X}_{\mathbb{Z}_{n}\setminus\mathbb{Z}_{n}^{*}}$ are
$\Delta_{n}$-code and hence $\mathcal{X}_{\mathbb{Z}_{n}^{*}}^{*}$ and
$\mathcal{X}_{\mathbb{Z}_{n}\setminus\mathbb{Z}_{n}^{*}}^{*}$ are
$S$-subgroups in $\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n})$ with
$|\mathcal{X}_{\mathbb{Z}_{n}^{*}}^{*}|=2^{\phi(n)}$ and
$|\mathcal{X}_{\mathbb{Z}_{n}\setminus\mathbb{Z}_{n}^{*}}^{*}|=2^{n-\phi(n)-1}$.

On the other hand, let

$(\mathsf{P}_{Y}(0),\mathsf{P}_{Y}(1),...,\mathsf{P}_{Y}(n-1))$

denote the autocorrelation vector of $Y$ in $\mathbb{Z}_{2}^{n}$ and let
$\mathfrak{A}(\mathbb{Z}_{2}^{n})$ denote the set of all this. Let
$X_{1}+X_{2}+\cdots+X_{n}=a$ denote the plane in $\mathbb{Z}^{n}$ in the
indeterminates $X_{i}$, $i=1,2,...,n$ and let
$\theta:\mathbb{Z}_{2}^{n}\rightarrow\mathfrak{A}(\mathbb{Z}_{2}^{n})$ be the
map defined by
$\theta(Y)=(\mathsf{P}_{Y}(0),\mathsf{P}_{Y}(1),\dots,\mathsf{P}_{Y}(n-1))$.

The decimation group $\Delta_{n}$ do not alter the set of values which
$\mathsf{P}_{X}(k)$ takes on, but merely the order in which they appear, i.e.,
if $Y=\delta_{a}X$ then $\mathsf{P}_{Y}(k)=\mathsf{P}_{X}(ka)$. Therefore, we
have the commutative diagram

$\textstyle{\mathbb{Z}_{2}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta}$$\scriptstyle{\delta_{r}}$$\textstyle{\mathbb{Z}_{2}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta}$$\textstyle{\mathfrak{A}(\mathbb{Z}_{2}^{n})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta_{r}}$$\textstyle{\mathfrak{A}(\mathbb{Z}_{2}^{n})}$
(6.4)

and $\theta\circ\delta_{r}=\delta_{r}\circ\theta.$

Let $Y\in\mathbb{Z}_{2}^{n}$ such that $\theta(Y)=(n,d,d,...,d)$. Such a
binary sequence is known as binary sequence with $2$-levels autocorrelation
value and are important by its applications on telecommunication. We want to
construct a $\Delta_{n}$-code for some $S$-subgroup $H$ of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n})$ containing such $Y$. From (6.4)
is followed that $\theta(Y)=\delta_{a}\theta(Y)=\theta(\delta_{a}Y)$, for all
$\delta_{a}\in\Delta_{n}$. Hence $Y$ and $\delta_{a}Y$ have the same
autocorrelation vector. For $Y$ fullfilling $\delta_{a}Y=Y$ for some
$\delta_{a}$ in $\Delta_{n}$ we have the following definition

######  Definition 3.

Let $a$ be a unit in $\mathbb{Z}_{n}^{*}$. A word $Y$ in $\mathbb{Z}_{2}^{n}$
is $\delta_{a}$-invariant if $\delta_{a}Y=Y$. Denote by $\mathbb{I}_{n}(a)$
the set of these $Y$.

If $Y$ is in $\mathbb{I}_{n}(a)$, then $\delta_{r}Y$ is in
$\mathbb{I}_{n}(a)$, too. Also $\delta_{a}(YZ)=\delta_{a}Y\delta_{a}Z=YZ$ for
all $Y,Z$ in $\mathbb{I}_{n}(a)$. Then $\mathbb{I}_{n}(a)$ is an $S$-subgroup
of $\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n})$. Now, we shall see that all
factorization of words in $\mathbb{I}_{n}(a)$ is relationated with cyclotomic
coset of $a$ module $n$. First, we have the following definition

######  Definition 4.

Let $a$ relative prime to $n$. The cyclotomic coset of $a$ module $n$ is
defined by

$\mathsf{C}_{s}=\\{s,sa,sa^{2},\cdots,sa^{t-1}\\}.$

where $sa^{t}\equiv s\mod n$. A subset $\\{s_{1},s_{2},\dots,s_{r}\\}$ of
$\mathbb{Z}_{n}$ is called complete set of representatives of cyclotomic coset
of $a$ modulo $n$ if $\mathsf{C}_{i_{1}}$,$\mathsf{C}_{i_{2}}$,…,
$\mathsf{C}_{i_{r}}$ are distinct and are a partition of $\mathbb{Z}_{n}$.

Take $Y=C^{i_{1}}XC^{i_{2}}X\cdots C^{i_{r}}X$ in $\mathbb{I}_{n}(a)$ with
$X=-++\cdots++$. We want $\delta_{a}Y=Y$. Then

$\displaystyle\delta_{a}Y$ $\displaystyle=$
$\displaystyle\delta_{a}C^{i_{1}}X\delta_{a}C^{i_{2}}X\cdots\delta_{a}C^{i_{r}}X$
$\displaystyle=$ $\displaystyle
C^{i_{1}a^{-1}}\delta_{a}XC^{i_{2}a^{-1}}\delta_{a}X\cdots
C^{i_{r}a^{-1}}\delta_{a}X$ $\displaystyle=$ $\displaystyle
C^{i_{1}a^{-1}}XC^{i_{2}a^{-1}}X\cdots C^{i_{r}a^{-1}}X$

since $\delta_{a}X=X$. As must be $\delta_{a}Y=Y$, then $i_{k}=a^{-1}i_{j}$ or
$i_{j}=ai_{k}$ for $1\leq k,j\leq r$. Let $\mathsf{C}_{s}X$ denote the word
$C^{s}XC^{sa}X\cdots C^{sa^{t_{s}-1}}X$. Then all $Y$ in $\mathbb{I}_{n}(a)$
has the form
$Y=\mathsf{C}_{s_{1}}^{\epsilon_{1}}X\mathsf{C}_{s_{2}}^{\epsilon_{r}}X\cdots\mathsf{C}_{s_{r}}^{\epsilon}X$,
with $\epsilon_{i}=0,1$. As $\mathbb{I}_{n}(a)$ is an $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n})$,
$\delta_{r}\mathsf{C}_{s_{i}}X=\mathsf{C}_{s_{j}}X$ and

$\mathcal{X}_{\mathbb{I}(a)}=\\{X,\mathsf{C}_{s_{1}}X,\
\mathsf{C}_{s_{2}}X,...,\ \mathsf{C}_{s_{r}}X\\}$ (6.5)

is a $\Delta_{n}$-code for $\mathbb{I}_{n}(a)$. Also
$\mathcal{X}_{\mathbb{I}(a)}$ is a $P(T)$-code with

$P(T)=\\{\\{0\\},\mathsf{C}_{s_{1}},\mathsf{C}_{s_{2}},...,\mathsf{C}_{s_{r}}\\}$

and $\\{s_{1},s_{2},...,s_{r}\\}$ a complete set of representatives. Hence
$\mathcal{X}_{\mathbb{I}(a)}^{*}$ has order $2^{r+1}$, where $r$ is the number
of cyclotomic cosets of $a$ module $n$

In the table 1, binary sequences with $2$-level autocorrelation values with
their respective $\delta_{a}$-invariants $S$-subgroups are shown

Table 1: Binary sequences with $2$-level autocorrelation values Sequences | Autocorrelation vector |  $S$-subgroup
---|---|---
$\mathsf{C}_{0}X=X\in\mathcal{G}_{n}(n-1)$ | $(n,n-4,...,n-4)$ |  $\mathbb{I}_{n}(a)$ for all $a\in\mathbb{Z}_{n}^{*}$
$\mathsf{C}_{2}X=CXC^{2}XC^{4}X$ | $(7,-1,...,-1)$ | $\mathbb{I}_{7}(3)$
$\mathsf{C}_{3}X=CXC^{3}XC^{4}XC^{5}XC^{9}X$ | $(11,-1,...,-1)$ | $\mathbb{I}_{11}(3)$
$\mathsf{C}_{0}X\mathsf{C}_{3}X=XCXC^{3}XC^{9}X$ | $(13,1,...,1)$ | $\mathbb{I}_{13}(3)$
$\mathsf{C}_{0}X\mathsf{C}_{5}X\mathsf{C}_{7}X\mathsf{C}_{10}X\mathsf{C}_{11}X$ | $(15,-1,...,-1)$ | $\mathbb{I}_{15}(4)$

On the other hand, we have the following theorem

######  Theorem 13.

If $\left\langle b\right\rangle$ is a subgroup of $\left\langle
a\right\rangle$, then $\mathbb{I}_{n}(a)\leq\mathbb{I}_{n}(b)$.

###### Proof.

Let $\mathsf{C}_{1}^{a}$ and $\mathsf{C}_{1}^{b}$ denote the classes
$\\{1,a,a^{2},...,a^{t-1}\\}$ and $\\{1,b,b^{2},...,b^{s-1}\\}$. By hypothesis
$\left\langle b\right\rangle\leq\left\langle a\right\rangle$, then
$\mathsf{C}_{1}^{b}\subseteq\mathsf{C}_{1}^{a}$. Hence there exists $y_{i}$ in
$\left\langle a\right\rangle$ such that

$\mathsf{C}_{1}^{a}=\mathsf{C}_{1}^{b}\cup
y_{1}\mathsf{C}_{1}^{b}\cup\cdots\cup y_{k}\mathsf{C}_{1}^{b},$

and $k=[\left\langle a\right\rangle:\left\langle b\right\rangle]$. Then is
follows that

$\mathsf{C}_{s}^{a}=\mathsf{C}_{s}^{b}\cup
y_{1}\mathsf{C}_{s}^{b}\cup\cdots\cup y_{k}\mathsf{C}_{s}^{b}.$

Therefore
$|\mathcal{X}_{\mathbb{I}_{n}(a)}|\leq|\mathcal{X}_{\mathbb{I}_{n}(b)}|$ and
$\mathbb{I}_{n}(a)\leq\mathbb{I}_{n}(b)$. ∎

We finish this section constructing some $\delta_{a}$-invariants $S$-subgroups

######  Proposition 6.

  1. 1.

$\mathbb{I}_{2n+1}(2n)=\\{X,\ \mathsf{C}_{q}X:\ q\in\\{1,2,...,n\\}\\}^{*}$,
$\mathsf{C}_{q}=\\{q,2n+1-q\\}$

  2. 2.

$\mathbb{I}_{2n}(2n-1)=\\{X,\ C^{n}X,\ \mathsf{C}_{q}X:\
q\in\\{1,2,...,n-1\\}\\}^{*}$, $\mathsf{C}_{q}=\\{q,2n-q\\}$.

###### Proof.

We proof $1$. The proof of $2$ it is analogous. We note that

$(2n)^{2}=(2n+1-1)^{2}=(2n+1)^{2}-2(2n+1)+1\equiv 1\mod(2n+1),$

then $\mathsf{C}_{1}=\\{1,2n\\}$. As $2n+1\nmid 2n-1$ and $q<2n+1$, then
$2nq\not\equiv q\mod(2n+1)$ and the $\mathsf{C}_{q}=\\{q,2nq\\}$ are
cyclotomic cosets of $2n$ module $2n+1$. Finally, it is easy to note that
$2nq$ is congruent to $2n+1-q$ module $2n+1$,

$2nq-2n-1+q=(2n+1)(q-1)\equiv 0\mod(2n+1).$

∎

######  Proposition 7.

Let $2p+1$ be an prime number with $p$ an odd prime number. The $S$-subgroups
invariants in $\mathbb{Z}_{2}^{2p+1}$ are $\mathbb{I}_{2p+1}(x)$,
$\mathbb{I}_{2p+1}(y)$ and $\mathbb{I}_{2p+1}(2p)$, where $x$ is a primitive
root module $2p+1$ and $y$ is not neither primitive root module $2p+1$ nor
$2p$.

###### Proof.

Let $P=\\{x_{1},x_{2},...,x_{t}\\}$ denote the set of primitive roots module
$2p+1$. Then

$\mathbb{I}_{2p+1}(x_{1})=\mathbb{I}_{2p+1}(x_{2})=\cdots=\mathbb{I}_{2p+1}(x_{t}).$

As $|\left\langle x_{i}\right\rangle|=2p$ for any $x_{i}\in P$, then
$\left\langle x_{i}\right\rangle$ has exactly a subgroup of order $2$ and a
subgroup of order $p$. Therefore there exist $S$-subgroups invariants
$\mathbb{I}_{2p+1}(y)$ and $\mathbb{I}_{2p+1}(2p)$ where $y\in
P^{c}\setminus\\{2p\\}$, with $P^{c}$ the complement of $P$ in
$\mathbb{Z}_{2p+1}^{*}$. ∎

##  7 Schur ring $\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n})$

We note by $RY$ the reversed sequence $RY=(y_{n-1},...,y_{1},y_{0})$ and let
$H_{n}$ denote the permutation automorphic subgroup $H_{n}=\\{1,R\\}\leq
S_{n}\leq Aut(Z_{2}^{n})$. Hence $H_{n}$ defines a partition on
$\mathbb{Z}_{2}^{n}$ and $\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n})$ is a schur
ring.

######  Definition 5.

Let $Y\in\mathbb{Z}_{2}^{n}$. We shall call $Y$ symmetric if $RY=Y$ and
otherwise we say it is non symmetric. We make $Sym(\mathbb{Z}_{2}^{n})$ the
set of all $Y$ symmetric and $\widehat{Sym}(\mathbb{Z}_{2}^{n})$ the set of
all $Y$ nonsymmetric.

Take $Y\in\mathbb{Z}_{2}^{n}$ such that $Y$ is of the form
$C^{i_{1}}XC^{i_{2}}X\cdots C^{i_{r}}X$. We want to understand the structure
of the words in $Sym(\mathbb{Z}_{2}^{n})$. As it must be fulfilled that
$RY=Y$, then taking $X=+\cdots+-+\cdots+$ in $\mathcal{G}_{n}(n-1)$ with $n$
an odd number we have

$\displaystyle RY$ $\displaystyle=$ $\displaystyle
RC^{i_{1}}XRC^{i_{2}}X\cdots RC^{i_{r}}X$ $\displaystyle=$ $\displaystyle
C^{-i_{1}}RXC^{-i_{2}}RX\cdots C^{-i_{r}}RX$ $\displaystyle=$ $\displaystyle
C^{n-i_{1}}XC^{n-i_{2}}X\cdots C^{n-i_{r}}X$

where we have used that $RX=X$. Hence if $Y$ is symmetric, then must be
$n-i_{j}=i_{k}$ for $j\neq k$ ranging in $[1,r]$. Thereby $Y$ has the form

$Y_{0}^{\epsilon_{0}}Y_{1}^{\epsilon_{1}}\cdots
Y_{(n-1)/2}^{\epsilon_{(n-1)/2}}$ (7.1)

for $\epsilon_{i}=0,1$ and $Y_{0}=X$ and $Y_{i}=C^{i}XC^{n-i}X$. As
$C^{i}XC^{n-i}X$ is in $Sym(\mathbb{Z}_{2}^{n})$ for all $i$, then

$\mathcal{X}_{Sym^{O}}=\\{X,\ CXC^{n-1}X,\ C^{2}XC^{n-2}X,...,\
C^{(n-1)/2}XC^{(n+1)/2}X\\}$ (7.2)

is a $H_{n}$-code for $Sym(\mathbb{Z}_{2}^{n})$. For the case $n$ an even
number it is easily followed that

$\mathcal{X}_{Sym^{E}}=\\{XC^{n-1}X,\ CXC^{n-2}X,\ C^{2}XC^{n-3}X,...,\
C^{(n-2)/2}XC^{n/2}X\\}$ (7.3)

is a $H_{n}$-code for $Sym(\mathbb{Z}_{2}^{n})$, where $X=+++\cdots++-$. Also
it is clear that $\mathcal{X}_{Sym^{E}}$ and $\mathcal{X}_{Sym^{O}}$ are
$P(T)$-codes, therefore $Sym(\mathbb{Z}_{2}^{n})$ is an free $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n})$ and $Sym(\mathbb{Z}_{2}^{2n+1})$ and
$Sym(\mathbb{Z}_{2}^{2n})$ have order $2^{n+1}$ and $2^{n}$, respectively.

Finally, the relationship between the symmetric subgroup
$Sym(\mathbb{Z}_{2}^{{}^{2n+1}})$ and the $\delta_{2n}$-invariant $S$-subgroup
$\mathbb{I}_{2n+1}(2n)$ is shown

######  Theorem 14.

$Sym(\mathbb{Z}_{2}^{{}^{2n+1}})=\mathbb{I}_{2n+1}(2n)$.

###### Proof.

By proposition 6 the codewords in $\mathcal{X}_{\mathbb{I}_{2n+1}(2n)}$ are
$X$ and $C^{q}XC^{2n+1-q}X$, $1\leq q\leq n$, with $X=+\cdots+-+\cdots+$. We
want to show that all of codewords in $\mathcal{X}_{\mathbb{I}_{2n+1}(2n)}$ is
symmetric. For this we note that $RX=X$ and

$\displaystyle R(C^{q}XC^{2n+1-q}X)$ $\displaystyle=$ $\displaystyle
RC^{q}XRC^{2n+1-q}X$ $\displaystyle=$ $\displaystyle C^{2n+1-q}RXC^{q}RX$
$\displaystyle=$ $\displaystyle C^{2n+1-q}XC^{q}X.$

∎

##  8 Schur ring $\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}C_{n})$

In this section we will use the commutation relation

$RC^{i}=C^{n-i}R$ (8.1)

to show that the $S$-subgroups $\mathbb{G}_{d}(n)$ and
$Sym(\mathbb{Z}_{2}^{n})$ are $S$-subgroups in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}C_{n})$.

######  Theorem 15.

$\mathbb{G}_{dC}(n)$ is an $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}C_{n})$

###### Proof.

By having in mind the commutation relation (8.1), we can to show that
$\mathcal{X}_{F,d}$ is an $H_{n}$-code. In this way we have

$\displaystyle RA_{i,d}X$ $\displaystyle=$ $\displaystyle
R(C^{i}XC^{i+d}X\cdots C^{i+n-d}X)$ $\displaystyle=$ $\displaystyle
RC^{i}XRC^{i+d}X\cdots RC^{i+n-d}X$ $\displaystyle=$ $\displaystyle
C^{n-i}RXC^{n-i-d}RX\cdots C^{d-i}RX$

As $X=-++\cdots++$, then $RX=CX$. Hence

$\displaystyle RA_{i,d}X$ $\displaystyle=$ $\displaystyle
C^{n-i}CXC^{n-i-d}CX\cdots C^{d-i}CX$ $\displaystyle=$ $\displaystyle
C(C^{n-i}XC^{n-i-d}X\cdots C^{d-i}X)$

Finally, reordering and rewriting

$\displaystyle RA_{i,d}X$ $\displaystyle=$ $\displaystyle C(C^{d-i}X\cdots
C^{d-i+(n-2d)}XC^{d-i+(n-d)}X)$ $\displaystyle=$ $\displaystyle CA_{d-i,d}X$
$\displaystyle=$ $\displaystyle A_{d-i+1,d}X.$

∎

######  Definition 6.

A $S$-set $Y_{C}$ in $\mathbb{Z}_{2C}^{n}$ is symmetric if $R\cdot
Y_{C}=Y_{C}$, where $R\cdot Y_{C}$ means the action of $R$ on the elements of
$Y_{C}$. The set of all symmetric $S$-sets will be denoted by
$Sym(\mathbb{Z}_{2C}^{n})$ and the set of all non-symmetric $S$-sets will be
denoted by $\widehat{Sym}(\mathbb{Z}_{2C}^{n})$.

It is easy to note that $R\cdot Y_{C}=(RY)_{C}$. Therefore $Y_{C}$ is a
symmetric $S$-set if and only if contains some $C^{i}Y$ symmetric.

######  Theorem 16.

$Sym(\mathbb{Z}_{2C}^{n})$ is an $S$-subgroup of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}C_{n})$.

###### Proof.

Take $Y_{C},Z_{C}$ in $Sym(\mathbb{Z}_{2C}^{n})$ and suppose that $Y,Z\in
Sym(\mathbb{Z}_{2}^{n})$. As

$R(YC^{k}Z)_{C}=(RYRC^{k}Z)_{C}=(YC^{n-k}Z)_{C},$

then $R(Y_{C}Z_{C})=R(Y_{C})R(Z_{C})=Y_{C}Z_{C}$. ∎

##  9 Schur ring $\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n}C_{n})$

In this section we will use the commutation relation

$\delta_{a}C^{i}=C^{ia^{-1}}\delta_{a}$ (9.1)

to show that the $S$-subgroups $\mathbb{G}_{d}(n)$ and $\mathbb{I}_{n}(a)$ are
$S$-subgroups in $\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n}C_{n})$.

######  Theorem 17.

$\mathbb{G}_{dC}(n)$ is an $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n}C_{n})$.

###### Proof.

We want to show that $\mathcal{X}_{F,d}$ is a $\Delta_{n}$-code. Take
$A_{i,d}X$ in $\mathcal{X}_{F,d}$. Then if $k=\frac{n}{d}-1$ we have

$\displaystyle\delta_{a}A_{i,d}X$ $\displaystyle=$
$\displaystyle\delta_{a}(C^{i}XC^{i+d}X\cdots C^{i+kd}X)$ $\displaystyle=$
$\displaystyle\delta_{a}C^{i}X\delta_{a}C^{i+d}X\cdots\delta_{a}C^{i+kd}X)$
$\displaystyle=$ $\displaystyle
C^{ia^{-1}}\delta_{a}XC^{(i+d)a^{-1}}\delta_{a}X\cdots
C^{(i+kd)a^{-1}}\delta_{a}X$ $\displaystyle=$ $\displaystyle
C^{ia^{-1}}XC^{(i+d)a^{-1}}X\cdots C^{(i+kd)a^{-1}}X.$

Define the map $\vartheta:\mathcal{X}_{F,d}\rightarrow\\{ni/d:\
i=1,2,...,d-1\\}$ by

$\displaystyle\vartheta(A_{i,d}X)$ $\displaystyle=$
$\displaystyle\vartheta(C^{i}XC^{i+d}X\cdots C^{i+kd}X)$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{k}(i+jd)$ $\displaystyle=$ $\displaystyle
i(k+1)+\frac{k(k+1)}{2}d$ $\displaystyle=$
$\displaystyle\frac{ni}{d}+\left(\frac{n}{2d}-\frac{1}{2}\right)n$
$\displaystyle\equiv$ $\displaystyle\frac{ni}{d}\mod n$

Then $\vartheta$ is a biyection. As
$\vartheta(\delta_{a}A_{i,d}X)\equiv\frac{a^{-1}ni}{d}\mod n$ and
$\vartheta(C^{l}\delta_{a}A_{i,d}X)\equiv(a^{-1}+l)\frac{ni}{d}\mod n$, is
followed that $\delta_{a}A_{i,d}X\in\mathcal{X}_{F,d}$ and therefore
$\mathcal{X}_{F,d}$ is a $\Delta_{n}$-code. ∎

######  Definition 7.

Let $a$ be a unit in $\mathbb{Z}_{n}^{*}$. A basic set $Y_{C}$ in
$\mathbb{Z}_{2C}^{n}$ is $\delta_{a}$-invariant if $\delta_{a}\cdot
Y_{C}=Y_{C}$. Denote by $\mathbb{I}_{nC}(a)$ the set of these $Y$.

By having in mind the basic set $Y_{C}$, we can note that $\delta_{a}\cdot
Y_{C}=(\delta_{a}Y)_{C}$. Hence $Y_{C}$ is $\delta_{a}$-invariant in
$\mathbb{Z}_{2C}^{n}$ if and only if contains some $C^{i}Y$
$\delta_{a}$-invariant in $\mathbb{Z}_{2}^{n}$.

######  Theorem 18.

$\mathbb{I}_{nC}(a)$ is an $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n}C_{n})$.

###### Proof.

Take $Y_{C},Z_{C}$ in $\mathbb{I}_{nC}(a)$ and suppose that
$Y,Z\in\mathbb{I}_{n}(a)$. As
$\delta_{a}(YC^{k}Z)_{C}=(\delta_{a}Y\delta_{a}C^{k}Z)_{C}=(YC^{ka^{-1}}Z)_{C}$,
then $\delta_{a}(Y_{C}Z_{C})=\delta_{a}(Y_{C})\delta_{a}(Z_{C})=Y_{C}Z_{C}$. ∎

##  10 Schur ring $\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}\Delta_{n}C_{n})$

Finally we show that the $S$-subgroups $\mathbb{G}_{d}(n)$,
$Sym(\mathbb{Z}_{2}^{n})$ and $\mathbb{I}_{n}(a)$ are $S$-subgroups in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}\Delta_{n}C_{n})$. Let $Y_{C}$ be any
basic set in $\mathbb{Z}_{2C}^{n}$. It is a very easy to notice that
$Y_{C}=(C^{k}Y)_{C}$ for all $k$. We will use this fact for to prove the
following lemma

######  Lemma 3.

  1. 1.

$\delta_{a}Y_{iC}=Y_{a^{-1}iC}$ for
$Y_{i}=C^{i}XC^{2n+1-i}X\in\mathcal{X}_{Sym^{O}}$.

  2. 2.

$\delta_{a}Y_{iC}=Y_{\left(a^{-1}i+\frac{a^{-1}-1}{2}\right)C}$ for
$Y_{i}=C^{i}XC^{2n-1-i}X\in\mathcal{X}_{Sym^{E}}$.

###### Proof.

$1.$ Take $\delta_{a}$ in the group $\Delta_{n}$. It is clear that
$\delta_{a}X=C^{k_{a}}X$ for some $k_{a}$ depending on $a$, where
$X=+\cdots+-+\cdots+$ is the word in $\mathcal{X}_{C}$ used to construct all
codewords in $\mathcal{X}_{Sym^{O}}$. Then

$\displaystyle\delta_{a}Y_{iC}$ $\displaystyle=$
$\displaystyle(\delta_{a}C^{i}X\delta_{a}C^{2n+1-i}X)_{C}$ $\displaystyle=$
$\displaystyle(C^{a^{-1}i}\delta_{a}XC^{a^{-1}(2n+1-i)}\delta_{a}X)_{C}$
$\displaystyle=$ $\displaystyle(C^{k_{a}}(C^{a^{-1}i}XC^{2n+1-a^{-1}i}X))_{C}$
$\displaystyle=$ $\displaystyle(C^{a^{-1}i}XC^{2n+1-a^{-1}i}X)_{C}$
$\displaystyle=$ $\displaystyle Y_{a^{-1}iC}.$

$2.$ Equally, $\delta_{a}X=C^{k_{a}}X$ for some $k_{a}$ depending on $a$,
where $X=++\cdots++-$ in $\mathcal{X}_{C}$ is used to construct all codewords
in $\mathcal{X}_{Sym^{E}}$. Then

$\displaystyle\delta_{a}Y_{iC}$ $\displaystyle=$
$\displaystyle(\delta_{a}C^{i}X\delta_{a}C^{2n-1-i}X)_{C}$ $\displaystyle=$
$\displaystyle(C^{a^{-1}i}\delta_{a}XC^{a^{-1}(2n-1-i)}\delta_{a}X)_{C}$
$\displaystyle=$
$\displaystyle(C^{k_{a}}(C^{a^{-1}i}XC^{2n-a^{-1}(i+1)}X))_{C}$
$\displaystyle=$
$\displaystyle\left(C^{\frac{a^{-1}-1}{2}}(C^{a^{-1}i}XC^{2n-a^{-1}(i+1)}X)\right)_{C}$
$\displaystyle=$
$\displaystyle\left(C^{a^{-1}i+\frac{a^{-1}-1}{2}}XC^{2n-a^{-1}(i+1)+\frac{a^{-1}-1}{2}}X\right)_{C}$
$\displaystyle=$ $\displaystyle Y_{\left(a^{-1}i+\frac{a^{-1}-1}{2}\right)C}.$

∎

We will use this lemma for to show that the symmetric binary sequences form an
$S$-subgroup in $\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}\Delta_{n}C_{n})$

######  Theorem 19.

$Sym(\mathbb{Z}_{2C}^{n})$ is an $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}\Delta_{n}C_{n})$.

###### Proof.

Clearly $Sym(\mathbb{Z}_{2C}^{n})$ is an $S$-subgroup of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}C_{n})$. From the previous lemma is
followed that $\Delta_{n}$ defines a partition on $\mathcal{X}_{Sym^{E}}$ and
$\mathcal{X}_{Sym^{O}}$. Hence they are $\Delta_{n}$-codes and
$Sym(\mathbb{Z}_{2C}^{n})$ is an $S$-subgroup of
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}\Delta_{n}C_{n})$. ∎

######  Theorem 20.

$\mathbb{G}_{dC}(n)$ is an $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}\Delta_{n}C_{n})$.

###### Proof.

Follows from theorems 15 and 17. ∎

######  Theorem 21.

$\mathbb{I}_{nC}(a)$ is an $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},H_{n}\Delta_{n}C_{n})$.

###### Proof.

From theorem 19 the case $a=n-1$ is excluded. In the previous section already
was proved that $\mathbb{I}_{nC}(a)$ is an $S$-subgroup in
$\mathfrak{S}(\mathbb{Z}_{2}^{n},\Delta_{n}C_{n})$. Now, we wish to show that
$H_{n}$ defines a partition on $\mathbb{I}_{nC}(a)$ by using (8.1). Take the
codeword $\mathsf{C}_{s}X$ in $\mathcal{X}_{\mathbb{I}_{n}(a)}$. We have then

$\displaystyle R(\mathsf{C}_{s}X)_{C}$ $\displaystyle=$ $\displaystyle
R(C^{s}XC^{sa}X\cdots C^{sa^{t_{s}-1}}X)_{C}$ $\displaystyle=$
$\displaystyle(RC^{s}XRC^{sa}X\cdots RC^{sa^{t_{s}-1}}X)_{C}$ $\displaystyle=$
$\displaystyle(C^{n-s}CXC^{n-sa}CX\cdots C^{n-sa^{t_{s}-1}}CX)_{C}$
$\displaystyle=$ $\displaystyle(C(C^{n-s}XC^{n-sa}X\cdots
C^{n-sa^{t_{s}-1}}X))_{C}$ $\displaystyle=$
$\displaystyle(C^{n-s}XC^{(n-s)a}X\cdots C^{(n-s)a^{t_{s}-1}}X))_{C}$
$\displaystyle=$ $\displaystyle(\mathsf{C}_{n-s}X)_{C}.$

∎

## References

  * [1] I. Schur. Zur Theorie der einfach transitiven Permutationsgruppen, Sitzungsber. Preuss. Akad. Wiss., Phys.-Math. Kl., 598–623, 1993.
  * [2] H. Wielandt. Finite Permutation Groups, Academic Press, New York-London, 1964.
  * [3] M. Klin, R. Poschel. The konig problem, the isomorphism problem for cyclic graphs and the method of schur rings, Algebraic Methods in Graph Theory, 1, 2, 1978.
  * [4] S. L. Ma. On association schemes, schur rings, strongly regular graphs and partial difference sets, Ars Combin., 21:211-220, 1989.
  * [5] A. Heinze. Applications of Schur rings in algebraic combinatorics: graphs, partial difference sets and cyclotomic schemes, PhD thesis, Universitat Oldenburg, 2001.
  * [6] M. Muzychuk, M. Klin, R. Poschel. The isomorphism problem for circulant graphs via Schur ring theory, Dis. Math. The. Com. Sci. 56, 241-264, 2001.
  * [7] M. E. Muzichuk, The Subschemes of the Hamming Scheme, Investigations in Algebraic Theory of Combinatorial Objects Volume 84 of the series Mathematics and Its Applications pp 187-208, 1992.
  * [8] R. Orozco, Schur Ring over Group $\mathbb{Z}_{2}^{n}$, Circulant $S-$Sets Invariant by Decimation and Hadamard Matrices, 2018, arXiv:1802.05788.
  * [9] R. Orozco, An Approximation to Proof of the Circulant Hadamard Conjecture, arXiv:1804.05007, 2018.
  * [10] R. Orozco, Schur Ring, Run Structure and Periodic Compatible Binary Sequences, arXiv:1807.10849, 2018.
  * [11] J. Berstel, D. Perrin, Theory of Codes, 2002, Universite de Marne-la-Vallee

E-mail address<EMAIL_ADDRESS>