On the number of non-G-equivalent minimal abelian codes

TL;DR

This paper explores the classification of minimal abelian codes in finite abelian groups, establishing equivalences between different subgroup notions and providing formulas for counting non-equivalent codes based on group structure.

Contribution

It proves the equivalence of G-isomorphism and isomorphism on certain subgroups, and derives formulas for counting non-G-equivalent minimal abelian codes.

Findings

Number of non-G-equivalent minimal abelian codes equals the number of divisors of the group's exponent under certain conditions.

Established the equivalence between G-isomorphism and subgroup isomorphism for cyclic quotient conditions.

Calculated the number of such codes for specific classes of abelian groups.

Abstract

Let $G$ be a finite abelian group. Ferraz, Guerreiro and Polcino Milies prove that the number of $G$-equivalence classes of minimal abelian codes is equal to the number of $G$-isomorphism classes of subgroups for which corresponding quotients are cyclic. In this article, we prove that the notion of $G$-isomorphism is equivalent to the notion of isomorphism on the set of all subgroups $H$ of $G$ with the property that $G/H$ is cyclic. As an application, we calculate the number of non-$G$-equivalent minimal abelian codes for some specific family of abelian groups. We also prove that the number of non-$G$-equivalent minimal abelian codes is equal to number of divisors of the exponent of $G$ if and only if for each prime $p$ dividing the order of $G$, the Sylow $p$-subgroups of $G$ are homocyclic.

Equations

$$G=⟨x⟩\times A=⟨y⟩\times B$$

$$G/⟨x⟩=⟨x⟩H/⟨x⟩\cong H/H\cap ⟨x⟩=H/⟨x^{p^m}⟩$$

$$H/⟨x^{p^m}⟩\cong K/⟨y^{p^m}⟩.$$

$$G/H=⟨xH⟩\cong C_{p^m}\cong ⟨yK⟩=G/K$$

$$⟨a⟩/⟨a^{p^t}⟩\times ⟨z⟩\times K_2\cong K/⟨y^{p^m}⟩≆H/⟨x^{p^m}⟩\cong ⟨b⟩\times H_2.$$

$$\theta ((x^{p^m}{)}^{l}h)=(y^{p^m}{)}^l{\theta }_1(h)$$

$$(⟨a⟩/⟨x^{p^m}⟩)\times ⟨b_1⟩\times H_1\cong H/⟨x^{p^m}⟩≆K/⟨y^{p^m}⟩\cong ⟨a_1⟩\times (⟨b⟩/⟨y^{p^m}⟩)\times K_1.$$

$$\theta (a^i\cdot h)=b^i\cdot {\theta }_1(h)$$

$$\phi (g_1{g}_2)=\phi ((x^{i_1+i_2-p^m})(x^{p^m}{h}_1{h}_2))=y^{i_1+i_2-p^m}\theta (x^{p^m}{h}_1{h}_2)=y^{i_1+i_2-p^m}\theta (x^{p^m})\theta (h_1)\theta (h_2)$$

$$G_i/G_i\cap L\cong G_{i}L/L\le G/L$$

$$G=(\prod _{i=1}^n{L}_i)\times (\prod _{i=1}^n⟨x_i⟩),$$

$$C_{p^m}\cong G/⟨a⟩=L⟨a⟩/⟨a⟩\cong L/L\cap ⟨a⟩$$

$$\{H_k\times K_j|H_k=⟨a^{p^k}b⟩,K_j=⟨b^{p^j}⟩,0\le k\le n-m,0\le j\le m\}.$$

Full text

1. Introduction

Due to Berman [1] and MacWilliams [3], an abelian code over a field is defined to be an ideal in a finite group algebra of an abelian group. An abelian code is said to be minimal if the corresponding ideal is minimal in the set of all ideals of the group algebra. Let $G$ be a finite abelian group and $\mathbb{F}$ a finite field of characteristic coprime to the order of $G$. Under these conditions, Maschke’s Theorem says that every abelian code is a direct sum of minimal abelian codes. Moreoever as defined in [4], two abelian codes $I$ and $J$ are called $G$-equivalent if there is a group automorphism $\varphi:G\rightarrow G$ whose linear extension to the group algebra maps $I$ onto $J$. It is easy to see that $G$-equivalent codes have the same weight distribution. However, the converse is not true (see Proposition IV.2 in [2] ). Therefore, knowing the number of $G$-equivalence classes of minimal abelian codes tells us a lot about the nature of codes that can be defined using the group algebra $\mathbb{F}G$.

A one-to-one correspondence between $G$-equivalence classes of minimal abelian codes and $G$-isomorphism classes of cocyclic subgroups of $G$ is established by Ferraz, Guerreiro and Polcino Milies. (For the details see Proposition III.2, Proposition III.7 and Proposition III.8 in [2] ). According to [2], two subgroups $H$ and $K$ of $G$ are called $G$-isomorphic if there is an automorphism $\varphi$ of $G$ which maps $H$ onto $K$. A subgroup $L\leq G$ is called a cocyclic subgroup of $G$ if $G/L$ is cyclic. Note that this definition is not the same definition as in [2]. We take into account $G$ itself also as a cocyclic subgroup to count the minimal abelian code which corresponds to the subgroup $G$ of $G$. From the definition, it is clear that if two subgroups of $G$ are $G$-isomorphic, then they are isomorphic. However, the converse of this statement is not true for arbitrary subgroups of $G$. We observe that the notion of $G$-isomorphism is equivalent to the notion of isomorphism on the set of cocyclic subgroups of $G$ as follows.

Proposition 1.1**.**

Let $G$ be a finite abelian group and let $H$, $K$ be cocyclic subgroups of $G$. Then $H$ and $K$ are $G$-isomorphic if and only if they are isomorphic.

This proposition, together with Proposition III.2, Proposition III.7 and Proposition III.8 in [2] leads us to write the following theorem.

Theorem 1.2**.**

Let $G$ be a finite abelian group. The number of non-$G$-equivalent minimal abelian codes over $\mathbb{F}$ is equal to the number of isomorphism classes of cocyclic subgroups of $G$.

Let $\eta({\mathbb{F}G})$ denote the number of non-$G$-equivalent minimal abelian codes over $\mathbb{F}$. As an application of Theorem 1.2, we prove the following results. Among these, the first result is the following.

Theorem 1.3**.**

Let $H$ be a finite abelian group and let $G$ be a direct product of finite number of copies of $H$. Then we have that $\eta({\mathbb{F}G})$=$\eta({\mathbb{F}H})$.

We observe that under an assumption on the exponent of the direct factors, multiplying a finite abelian group by a homocyclic group does not change the number $\eta({\mathbb{F}G}).$ Here a homocyclic group is a direct product of pairwise isomorphic cyclic groups.

Theorem 1.4**.**

Let $K$ be a finite homocyclic group and $H$ a finite abelian group such that $\exp(K)=\exp(H)$. If $G=K\times H$, then we have that $\eta({\mathbb{F}G})$=$\eta({\mathbb{F}H})$.

As emphasized in [4], the codes arising from the group algebra $\mathbb{F}_{2}(C_{m}\times C_{n})$, where $m$ and $n$ are positive odd integers, are referred as two-dimensional linear recurring arrays, linear recurring planes or two-dimensional cyclic codes in the works [5] and [6]. These codes are related to the problem of constructing perfect maps and have applications to $x$-ray photography. In [4, Theorem 3.6], it is stated that the number of non-equivalent minimal codes of $\mathbb{F}_{2}(C_{m}\times C_{n})$ is equal to the number of divisors of the exponent of the corresponding group. Ferraz, Guerreiro and Polcino Milies point out that this result is not true by calculating the number of non-equivalent minimal codes of $\mathbb{F}_{2}(C_{p^{n}}\times C_{p})$ as $2n$ where $p$ is an odd prime. (see [2, Propostion IV.3]). The following theorem generalizes this result.

Theorem 1.5**.**

If $G=C_{p^{n}}\times C_{p^{m}}$ and $n>m$, then $\eta({\mathbb{F}G})=(n-m+1)(m+1)$.

As a corollary we obtain the following result.

Corollary 1.6**.**

Let $n$ be a positive integer such that $n={p_{1}}^{k_{1}}{p_{2}}^{k_{2}}\dots p_{t}^{k_{t}}$ where $p_{i}$’s are distinct prime numbers and $k_{i}$’s are positive integers. Then for $G=C_{n^{l}}\times C_{n^{s}}$ where $n,l,s$ are positive integers and $l>s$ we have that $\eta(\mathbb{F}G)=\prod_{i=1}^{t}(k_{i}l-k_{i}s+1)(k_{i}s+1)$.

In [4], for an abelian group $G$ of odd order, it is proved that the number of non-$G$-equivalent minimal abelian codes over $\mathbb{F}_{2}$ is equal to the number of divisors of the exponent of $G$. In [2], it is shown that this statement is not true and moreover it is shown that if $G$ is homocyclic, the number of non-$G$-equivalent minimal abelian codes over $\mathbb{F}$ is equal to the number of divisors of exponent of $G$ (see Theorem V.6 in [2]). In the following theorem, we extend this result and give a characterization of an abelian group whose number of non-equivalent minimal codes is equal to the number of divisors of its exponent.

Theorem 1.7**.**

Let $G$ be a finite abelian group and $\mathbb{F}$ a finite field of characteristic coprime to order of $G$. The number of non-$G$-equivalent minimal abelian codes over $\mathbb{F}$ is equal to the number of divisors of exponent of $G$ if and only if for each prime $p$ dividing the order of $G$, the Sylow $p$-subgroups of $G$ are homocyclic.

Note that Theorem V.6 in [2] now follows from the Theorem 1.7 as a corollary.

The structure of the paper is as follows. In section 2, we establish some resullts on isomorphic cocyclic subgroups of a finite abelian group and give the proofs of Proposition 1.1 and Theorem 1.2, Theorem 1.3 and Theorem 1.4 . We also present some important examples related to Theorem 1.4. In section 3, we prove Theorem 1.5. In section 4, we present the proof of Theorem 1.7.

2. Proof of Proposition 1.1 and its consequences

It is not very easy to determine whether two subgroups of a given group $G$ are $G$-isomorphic or not. We begin by showing that isomorphisms between cocyclic subgroups of $G$ can be extended to an automorphism of $G$. Then, we continue to prove some other facts to give a proof for Proposition 1.1.

Lemma 2.1**.**

Let $G$ be an abelian $p$-group of exponent $p^{n}$ and $H$ a cocyclic subgroup of $G$. If the exponent of $H$ is strictly less than $p^{n}$, then for any $z\in G$ of order $p^{n}$, $zH$ generates the cyclic group $G/H$.

Proof.

Let $z$ be any element in $G$ of order $p^{n}$. As ${\rm exp}(H)<p^{n}$, we have that $z\notin H$, so $zH$ is a non-trivial element in $G/H$. Since $H$ is a cocyclic subgroup of $G$, $G/H\cong C_{p^{m}}$ for some $1\leq m\leq n$. Then there exists some $b\in G$ such that $G/H=\langle bH\rangle$. Since $zH$ is a non-trivial element in $G/H$, we have that $zH=(bH)^{p^{i}}$ for some $i$ where $0\leq i<m$. This implies that $(b^{p^{i}})^{-1}z\in H$. As ${\rm exp}(H)<p^{n}$, this is possible only if $z=b^{p^{i}}$. Since $z$ has order $p^{n}$, $i$ should be equal to zero so that $z=b$ and hence $G/H=\langle zH\rangle$. ∎

Lemma 2.2**.**

Let $G$ be an abelian $p$-group of exponent $p^{n}$ and $H<G$ a cocyclic subgroup of $G$. Then there exists an element $x\in G$ of order $p^{n}$ so that $G/H=\langle xH\rangle$.

Proof.

There are two cases to consider.

Case 1: ${\rm exp}(H)<p^{n}$: Follows from Lemma 2.1.

Case 2: ${\rm exp}(H)=p^{n}$: As $H$ is cocyclic we have that $G/H\cong C_{p^{m}}$ for some $1\leq m\leq n.$ Then there exists $a\in G-H$ so that $G/H=\langle aH\rangle$ and then $a^{p^{m}}\in H$ and $m$ is the smallest integer satisfying this property. If the order of $a^{p^{m}}$ is equal to $p^{n-m}$, then one can take $x=a$. If the order of $a^{p^{m}}$ is less than $p^{n-m}$, as ${\rm exp}(H)=p^{n},$ there exists some $y\in H$ of order $p^{n}$ so one can take $x=ya$. It is clear that $xH$ generates $G/H$.

∎

Lemma 2.3**.**

Let $G$ be an abelian $p$-group of exponent $p^{n}$. Assume that $H$ and $K$ are subgroups of $G$ such that for some $x,y\in G$ of order $p^{n}$, we have that $G/H=\langle xH\rangle\cong C_{p^{m}}\cong G/K=\langle yK\rangle$ where $0\leq m\leq n$. Then, we have that $H/\langle x^{p^{m}}\rangle\cong K/\langle y^{p^{m}}\rangle$.

Proof.

As ${\rm exp}(G)=p^{n}$ and $|x|=|y|=p^{n}$, we can write

[TABLE]

where $A\cong B$, so that $G/\langle x\rangle\cong G/\langle y\rangle$. On the other hand, Second Isomorphism Theorem gives us that

[TABLE]

and similarly $G/\langle y\rangle=\langle y\rangle K/\langle y\rangle\cong K/K\cap\langle y\rangle=K/\langle y^{p^{m}}\rangle$. Therefore, we deduce that

[TABLE]

∎

Lemma 2.4**.**

Let $H$ be a finite abelian $p$-group and let $x$ be a non-generator of $H$. Then there exists a generator $a$ of $H$ such that $\langle x\rangle\leq\langle a\rangle$.

Proof.

Recall that the Frattini subgroup of $H$, denoted by $\Phi(H)$ is the set of all non-generators of $H$. Then, if $H=\langle a_{1}\rangle\times\dots\times\langle a_{r}\rangle$, it is easy to see that $\Phi(H)=\langle{a_{1}}^{p}\rangle\times\dots\times\langle{a_{r}}^{p}\rangle$. As $x$ is a non-generator, $x\in\Phi(H)$ and $x=({a_{1}}^{p})^{i_{1}}\dots({a_{r}}^{p})^{i_{r}}$ where $0\leq i_{j}\leq|a_{j}|$, that is $x=({a_{1}}^{i_{1}}\dots a_{r}^{i_{r}})^{p}$. If at least one of $i_{k}$’s is not a multiple of $p$, set $a={a_{1}}^{i_{1}}\dots a_{r}^{i_{r}}$, then $a\notin\Phi(H)$ which means that $a$ is a generator of $H$ and $\langle a^{p}\rangle=\langle x\rangle\leq\langle a\rangle$. If all $i_{k}^{\prime}$s are multiple of $p$, then for $a={a_{1}}^{\frac{i_{1}}{p}}\dots a_{r}^{\frac{i_{r}}{p}}$ we have that $x=a^{p^{2}}$. So $\langle x\rangle\leq\langle a\rangle$. If one of $\frac{i_{k}}{p}$’s is not a multiple of $p$, then $a\notin\Phi(H)$ and $a$ is a generator. If all $\frac{i_{k}}{p}$’s are a multiple of $p$ then one can continue the process until getting an $\frac{i_{k}}{p}$ which is not a multiple of $p$ for some $1\leq k\leq r$. ∎

Lemma 2.5**.**

Let $G$ be an abelian $p$-group of exponent $p^{n}$. Assume that $H$ and $K$ are isomorphic subgroups of $G$ such that for some $x,y\in G$ of order $p^{n}$, we have that

[TABLE]

where $0\leq m\leq n$. Then $x^{p^{m}}$ is a generator of $H$ if and only if $y^{p^{m}}$ is a generator of $K$.

Proof.

Suppose that $x^{p^{m}}$ is a generator of $H$, then $H=\langle x^{p^{m}}\rangle\times H_{1}$ for some $H_{1}\leq H$. Suppose for a contradiction that $y^{p^{m}}$ is not a generator of $K$. Then by Lemma 2.4, there exists a generator $a\in K$ such that $y^{p^{m}}=a^{p^{t}}$ for some $t$ with $t>0$. Note that $K=\langle a\rangle\times K_{1}$ for some $K_{1}\leq K$. On the other hand, since the orders of $x$ and $y$ are equal, we have that $\langle x^{p^{m}}\rangle\not\cong\langle a\rangle$. Hence, since $H\cong K$, there exists $b\in H_{1}$ with $\langle b\rangle\cong\langle a\rangle$ and there exists $z\in K_{1}$ with $\langle z\rangle\cong\langle x^{p^{m}}\rangle$, such that $K=\langle a\rangle\times\langle z\rangle\times K_{2}$ and $H=\langle b\rangle\times\langle x^{p^{m}}\rangle\times H_{2}$ where $K_{2}\cong H_{2}$. It follows that

[TABLE]

But this contradicts with Lemma 2.3. The converse implication is similar. ∎

Proposition 2.6**.**

Let $G$ be an abelian $p$-group of exponent $p^{n}$. Assume that $H$ and $K$ are isomorphic subgroups of $G$ such that for some $x,y\in G$ of order $p^{n}$, $G/H=\langle xH\rangle\cong C_{p^{m}}\cong\langle yK\rangle=G/K$ where $0\leq m\leq n$. Then, there exists an isomorphism $\theta:H\rightarrow K$ such that $\theta(x^{p^{m}})=y^{p^{m}}$.

Proof.

If $m=n$ we have that $x^{p^{m}}=y^{p^{m}}=1$, then for any isomorphism $\theta:H\rightarrow K$ we definitely have $\theta(x^{p^{m}})=y^{p^{m}}$. So we can assume that $m<n$. Then we have that $x^{p^{m}}\in H$ and $y^{p^{m}}\in K$. There are two cases.

Case 1: $x^{p^{m}}$ is a generator of $H$.

By Lemma 2.5, $y^{p^{m}}$ is also a generator of $K$. Then $H=\langle x^{p^{m}}\rangle\times H_{1}$ and $K=\langle y^{p^{m}}\rangle\times K_{1}$ where $H_{1}\cong K_{1}$. So one can choose $\theta_{1}:H_{1}\to K_{1}$ as an isomorphism, and define $\theta:H\rightarrow K$ as

[TABLE]

for $0\leq l\leq p^{n-m}-1$ and $h\in H_{1}$. It is clear that $\theta$ is an isomorphism from $H$ onto $K$ which satisfies $\theta(x^{p^{m}})=y^{p^{m}}.$

Case 2: $x^{p^{m}}$ is a non-generator of $H$.

By Lemma 2.5, $y^{p^{m}}$ is also a non-generator of $K$. In this case, by Lemma 2.4, there exists a generator $a\in H$ such that $\langle x^{p^{m}}\rangle\leq\langle a\rangle$. Similarly, there exists a generator $b\in K$ such that $\langle y^{p^{m}}\rangle\leq\langle b\rangle$. We claim that $\langle a\rangle\cong\langle b\rangle$. Assume that $\langle a\rangle\ncong\langle b\rangle$. Then as $H\cong K$, we have that $H=\langle a\rangle\times\langle b_{1}\rangle\times H_{1}$ and $K=\langle a_{1}\rangle\times\langle b\rangle\times K_{1}$ where $\langle a\rangle\cong\langle a_{1}\rangle$, $\langle b_{1}\rangle\cong\langle b\rangle$, $H_{1}\cong K_{1}$. Note that $\langle a_{1}\rangle\not\cong\langle b_{1}\rangle$ since $\langle a\rangle\not\cong\langle b\rangle$. We have that

[TABLE]

This contradicts with Lemma 2.3. So $\langle a\rangle\cong\langle b\rangle$. Hence $H=\langle a\rangle\times H_{1}$, $K=\langle b\rangle\times K_{1}$ where $H_{1}\cong K_{1}$. Choose $\theta_{1}:H_{1}\to K_{1}$ as one of those isomorphisms. Then one can define $\theta:H\rightarrow K$ as

[TABLE]

for $0\leq i\leq|a|-1$ and $h\in H_{1}$. Since the orders of $a$ and $b$ are equal, now it is easy to see that $\theta$ is an isomorphism between $H$ and $K$ which takes $x^{p^{m}}$ to $y^{p^{m}}$. ∎

Lemma 2.7**.**

Assume that $G=H\times K$ where $(|H|,|K|)=1$. Then, we have that $G_{1}$ is a cocyclic subgroup of $G$ if and only if $G_{1}=H_{1}\times K_{1}$ where $H_{1}$ is a cocyclic subgroup of $H$ and $K_{1}$ is a cocyclic subgroup of $K$.

Proof.

If $H_{1}$ is a cocyclic subgroup of $H$ and $K_{1}$ is a cocyclic subgroup of $K$, it is easy to see that $H_{1}\times K_{1}$ is a cocyclic subgroup of $H\times K$ since the orders of $H$ and $K$ are coprime. Conversely, if $G_{1}$ is a cocyclic subgroup of $H\times K$. Then, since $H$ and $K$ are groups of coprime order, we have that $G_{1}=H_{1}\times K_{1}$ where $H_{1}\leq H$ and $K_{1}\leq K$. But since $G_{1}$ is a cocyclic subgroup of $H\times K$, we have that $H_{1}$ is a cocyclic subgroup of $H$ and $K_{1}$ is a cocyclic subgroup of $K$. ∎

Now we are ready to prove Proposition 1.1.

Proof of Proposition 1.1.

Since any pair of $G$-isomorphic subgroups of $G$ are isomorphic by definition, it is enough to prove that isomorphic cocyclic subgroups of $G$ are $G$-isomorphic. Let us prove this statement first for $p$-groups. Assume that $G$ is an abelian $p$-group of exponent $p^{n}$ and let $H$ and $K$ be two cocyclic isomorphic subgroups of $G$. If $H=K=G$ there is nothing to do. So assume that $H,K<G$. By Lemma 2.2, there exist $x,y\in G$ of order $p^{n}$ such that $G/H=\langle xH\rangle$ and $G/K=\langle yK\rangle$. Since $H$ and $K$ are isomorphic, their cyclic quotient groups $G/H$ and $G/K$ are isomorphic. Let $G/H=\langle xH\rangle\cong C_{p^{m}}\cong\langle yK\rangle=G/K$ for some $m$ with $0<m\leq n$. By using Proposition 2.6, we can choose an isomorphism $\theta:H\to K$ so that $\theta(x^{p^{m}})=y^{p^{m}}$. For each $g\in G$ there exist a unique $i$ where $0\leq i\leq p^{m}-1$ and a unique $h\in H$ such that $g=x^{i}h$. Now, define $\varphi:G\to G$ as $\varphi(g)=y^{i}\theta(h)$. We claim that $\varphi$ is an automorphism of $G$. It is easy to see that $\varphi$ is a bijection. To show that it is a homomorphism let us choose $g_{1}=x^{i_{1}}h_{1}$ and $g_{2}=x^{i_{2}}h_{2}$ for $0\leq i_{1},i_{2}<p^{m}$ and $h_{1},h_{2}\in H$. Note that $0\leq i_{1}+i_{2}<2p^{m}$, we will show that $\varphi$ is a homomorphism by considering two seperate cases depending on the value of this sum.

Case 1: $0\leq i_{1}+i_{2}<p^{m}$

In this case $\varphi(g_{1}g_{2})=\varphi(x^{i_{1}+i_{2}}h_{1}h_{2})=y^{i_{1}+i_{2}}\theta(h_{1}h_{2})=(y^{i_{1}}\theta(h_{1}))(y^{i_{2}}\theta(h_{2}))=\varphi(g_{1})\varphi(g_{2}).$

Case 2: $p^{m}\leq i_{1}+i_{2}<2p^{m}$

In this case, we have that

[TABLE]

from the definition of $\varphi$ and from the fact that $x^{p^{m}}\in H$. But this last expression is equal to $y^{i_{1}+i_{2}}\theta(h_{1}h_{2})$ since $y^{p^{m}}=\theta(x^{p^{m}})$. Moreover $\varphi(g_{1})\varphi(g_{2})=(y^{i_{1}}\theta(h_{1}))(y^{i_{2}}\theta(h_{2}))=y^{i_{1}+i_{2}}\theta(h_{1}h_{2}).$ Hence, $\varphi(g_{1}g_{2})=\varphi(g_{1})\varphi(g_{2})$ in this case, too.

In both of the cases, we have shown that $\varphi$ is an automorphism of $G$ and it is easy to observe that $\varphi$ takes $H$ onto $K$. Hence, $H$ and $K$ are $G$-isomorphic.

Now, let $G$ be a finite abelian group whose order is composite. For $1\leq i\leq r$, let $G_{p_{i}}$ denote a Sylow $p_{i}$-subgroup of $G$. Then $G=G_{p_{1}}\times G_{p_{2}}\times...\times G_{p_{r}}.$ If $H$ and $K$ are two cocyclic subgroups of $G$ which are isomorphic, $H=H_{p_{1}}\times H_{p_{2}}\times...\times H_{p_{r}}$ and $K=K_{p_{1}}\times K_{p_{2}}\times...\times K_{p_{r}}$ where for each $1\leq i\leq r$, one has $H_{p_{i}}\cong K_{p_{i}}$. Moreover, by Lemma 2.7, for each $i$, the groups $H_{p_{i}}$ and $K_{p_{i}}$ are cocyclic subgroups of $G_{p_{i}}$. Hence, by the first part of the proof, $H_{p_{i}}$ and $K_{p_{i}}$ are $G_{p_{i}}$-isomorphic subgroups or equivalently there exists an automorphism $\varphi_{i}$ of $G_{p_{i}}$ which takes $H_{p_{i}}$ onto $K_{p_{i}}$. Now let us set $\varphi:=(\varphi_{1},\varphi_{2},\ldots,\varphi_{r})$, then it is easy to see that $\varphi$ is an automorphism of $G$ which takes $H$ onto $K$. ∎

Proof of Theorem 1.2.

Follows from Proposition III.2, Proposition III.7 and Proposition III.8 in [2], together with Proposition 1.1. ∎

For the proofs of Theorem 1.3, Theorem 1.4 and Theorem 1.7 we need to consider direct products of groups whose orders are relatively prime.

When $G$ is equal to the direct product of subgroups of coprime order, the number of isomorphism classes of cocyclic subgroups, hence the number of non-$G$-equivalent minimal abelian codes is calculated easily as follows.

Theorem 2.8**.**

Let $G=H\times K$ where $(|H|,|K|)=1$. Then we have that $\eta(\mathbb{F}G)=\eta(\mathbb{F}K)\ \eta(\mathbb{F}H).$

Proof.

By Lemma 2.7, any cocylic subgroup of $G$ is of the form $H_{1}\times K_{1}$ where $H_{1}$ is a cocyclic of $H$ and $K_{1}$ is a cocyclic of $K$. So the number of isomorphism classes of cocyclic subgroups of $G$ is the product of number of isomorphism classes of cocyclic subgroups of $H$ and the number of isomorphism classes of cocyclic subgroups of $K$. Now the result follows from Theorem 1.2. ∎

Now, thanks to Theorem 1.2, to count the number of non-$G$-equivalent minimal abelian codes over $\mathbb{F}$, we just need to count the number of isomorphism classes of cocyclic subgroups of $G$.

Proof of Theorem 1.3.

By using classification of finitely generated abelian groups and Theorem 2.8 it is enough to prove the result when $H$ is a finite $p$-group. Let $H=C_{p^{a_{1}}}\times\ldots\times C_{p^{a_{n}}}$ where $a_{i}\geq 1$ are integers and $G=H^{k}$ for $k\geq 1$, then $G=G_{1}\times\ldots\times G_{n}$ where $G_{i}=(C_{p^{a_{i}}})^{k}$ for $i=1,\ldots,n$. Let $L$ be a cocyclic subgroup of $G$. For each $i$, we have that

[TABLE]

which implies that $G_{i}/G_{i}\cap L$ is cyclic. So $G_{i}\cap L$ should contain a subgroup $L_{i}$ which is isomorphic to $(C_{p^{a_{i}}})^{k-1}$ (for example by [2, Theorem V.2] ). Moreover, it is easy to see that for each $i$, there exists an element $x_{i}\in G_{i}$ of order $p^{a_{i}}$ such that $G_{i}=L_{i}\times\langle x_{i}\rangle$. Hence

[TABLE]

where the first term of the product is isomorphic to $(H)^{k-1}$ and the second is isomorphic to $H$. Now, by the use of Correspondence Theorem, there is a bijection between the subgroups of $G$ containing $\prod_{i=1}^{n}L_{i}$ and the subgroups of $\prod_{i=1}^{n}\langle x_{i}\rangle$. Under this bijection, $L$ corresponds to a cocyclic subgroup $C_{L}$ of $\prod_{i=1}^{n}\langle x_{i}\rangle$ where $L=(\prod_{i=1}^{n}L_{i})\times C_{L}$. By Theorem 1.2, the result follows since $\prod_{i=1}^{n}\langle x_{i}\rangle$ is isomorphic to $H$. ∎

Proof of Theorem 1.4.

It is enough to prove the result when $H$ and $K$ are finite $p$-groups by the classification of finitely generated abelian groups and Lemma 2.8. Let $p^{n}$ be the exponent of $H$ and $K$. Then there exists $x\in H$ of order $p^{n}$ such that $H=\langle x\rangle\times\hat{H}$ where $\hat{H}$ is a finite $p$-group of exponent less or equal than $p^{n}$ and $K\cong(C_{p^{n}})^{r}$ for some positive integer $r$. Let $G_{1}\leq G=H\times K$ with $G_{1}\cong(C_{p^{n}})^{r+1}$ so that $G=G_{1}\times\hat{H}$ and let $L$ be a cocyclic subgroup of $G$. Then by a similar reasoning as in the proof of Theorem 1.3 we deduce that $G_{1}/G_{1}\cap L$ is cyclic and so $G_{1}\cap L$ contains a subgroup isomorphic to $(C_{p^{n}})^{r}$, call this subgroup as $K_{L}$. Then there exists an element $x_{1}\in G_{1}$ of order $p^{n}$ such that $G_{1}=K_{L}\times\langle x_{1}\rangle$. So $G=K_{L}\times\langle x_{1}\rangle\times\hat{H}$ and letting $H_{L}=\langle x_{1}\rangle\times\hat{H},G$ is equal to $K_{L}\times H_{L}$ where $K_{L}$ and $H_{L}$ are isomorphic to $K$ and $H$, respectively. Since $L$ is a cocyclic subgroup of $K_{L}\times H_{L}$ containing $K_{L}$, by the Correspondence Theorem $L$ corresponds to a cocyclic subgroup $C_{L}$ of $H_{L}$. Therefore $L=K_{L}\times C_{L}$ where $K_{L}$ is isomorphic to $K$ and $C_{L}$ is cocyclic subgroup of $H_{L}$. Now, the result follows from Theorem 1.2. ∎

In Theorem 1.4, the assumption on the exponents of the groups is important. We end this section by presenting the significance of this assumption with the following examples.

Example 2.9**.**

For an odd prime $p$, if we take $H=C_{p}\times C_{p}$ and $K=C_{p^{2}}\times C_{p^{2}}$, then $\eta(\mathbb{F}H)=2$, $\eta(\mathbb{F}K)=3$ and $\eta(\mathbb{F}(H\times K))=4$ where the characteristic of $\mathbb{F}$ is coprime to $p$.

Example 2.10**.**

Let $K$ be a finite homocyclic group and $H$ be a finite abelian group which is not homocyclic and $exp(K)>exp(H)$. Let $\mathbb{F}$ be a finite field of characteristic coprime to $3$. If $G=K\times H$, $\eta(\mathbb{F}G)=\eta(\mathbb{F}H)$ is no longer true. Indeed, if $K=C_{27}\times C_{27}$ and $H=C_{9}\times C_{3}$, then $\eta(\mathbb{F}H)=4$, $\eta(\mathbb{F}K)=4$ and $\eta(\mathbb{F}G)=8$. However, for $G=C_{27}\times C_{27}\times C_{9}\times C_{3}$ if we write $G=H\times K$ where $K=C_{27}$ and $H=C_{27}\times C_{9}\times C_{3}$ then $\eta(\mathbb{F}G)=\eta(\mathbb{F}H)$.

Example 2.11**.**

Consider $G=C_{27}\times C_{9}\times C_{3}\times C_{3}$ and take $H=C_{27}\times C_{9}$ and $K=C_{3}\times C_{3}$. $\eta(\mathbb{F}H)=6$, $\eta(\mathbb{F}K)=2$ and $\eta(\mathbb{F}G)=8$ where the characteristic of $\mathbb{F}$ is coprime to $3$.

3. Calculation for $C_{p^{n}}\times C_{p^{m}}$

For the proof of Theorem 1.5, we need the following lemma.

Lemma 3.1**.**

Let $G=\langle a\rangle\times\langle b\rangle$ where $\langle a\rangle\cong C_{p^{n}}$ and $\langle b\rangle\cong C_{p^{m}}$ with $n>m\geq 1$. Assume that $L$ is a cocyclic subgroup of $G$ which is not cyclic. Then $L\cong C_{p^{i}}\times C_{p^{j}}$ where $m\leq i\leq n$ and $1\leq j\leq m$.

Proof.

Let $L$ be such a cocyclic subgroup of $G$. If the exponent of $L$ is $p^{n}$, then $L\cong C_{p^{n}}\times C_{p^{j}}$ where $1\leq j\leq m$. If the exponent of $L$ is strictly less than $p^{n}$, by Lemma 2.1, $G/L=\langle aL\rangle$, so that $G=L\langle a\rangle$. Thus, we have that

[TABLE]

that is $L$ has a quotient isomorphic to $C_{p^{m}}$, hence $L$ has a subgroup isomorphic to $C_{p^{m}}$. So the exponent of $L$ is at least $p^{m}$, in this case. Hence, $L\cong C_{p^{i}}\times C_{p^{j}}$ for $i\geq m$ and $j\geq 1$. As $|L|\geq p^{m+1}$, $|G/L|\leq p^{(m+n)-(m+1)}=p^{n-1}$. Therefore, the index of $L$ in $G$ is at most $p^{n-1}$.

We prove the required result by induction on the index of the cocyclic subgroup $L$ in $G$. Clearly, the statement holds when $|G/L|=1$. If $|G/L|=p$, then either $L\cong C_{p^{n-1}}\times C_{p^{m}}$ or $L\cong C_{p^{n}}\times C_{p^{m-1}}$. Assume the statement holds for any non-cyclic cocyclic subgroup of $G$ with index strictly less than $p^{s}$ where $1\leq s\leq n-1$. Now let $L$ be a cocyclic subgroup of $G$ such that $|G/L|=p^{s}$. Then there exists a cocyclic subgroup $L_{1}$ of $G$ such that $L<L_{1}\leq G$ where $|G/L_{1}|=p^{s-1}$. By our induction hypothesis, $L_{1}\cong C_{p^{i}}\times C_{p^{j}}$ where $m\leq i\leq n$ and $1\leq j\leq m$. Moreover, $i$ and $j$ satisfy $m+n-(i+j)=s-1\leq n-2$. We also have that $L_{1}/L\cong C_{p}$. Thus, if $i\neq m$ and $j\neq 1$, then we deduce that $L\cong C_{p^{i-1}}\times C_{p^{j}}$ or $L\cong C_{p^{i}}\times C_{p^{j-1}}$. If $i=m$, then $j\geq 2$ by the inequality $m+n-(i+j)=s-1\leq n-2$. Since the exponent of $L$ is at least $p^{m}$, we shall have that $L\cong C_{p^{m}}\times C_{p^{j-1}}$. If $j=1$, then by the same inequality we have that $i\geq m+1$. As $L$ is not cyclic, we shall have that $L\cong C_{p^{i-1}}\times C_{p}$. Therefore, we deduce that $L\cong C_{p^{i}}\times C_{p^{j}}$ where $m\leq i\leq n$ and $1\leq j\leq m$.

∎

Proposition 3.2**.**

Let $G=\langle a\rangle\times\langle b\rangle$ where $\langle a\rangle\cong C_{p^{n}}$ and $\langle b\rangle\cong C_{p^{m}}$ with $n>m\geq 1$. Then any cocyclic subgroup is isomorphic to one of the following subgroups in the following set

[TABLE]

Proof.

There are two cases to consider.

Case 1 (Cocyclic subgroups of $G$ which are cyclic): For each $k\in\{0,\dots n-m\}$, $H_{k}$ is a cocyclic subgroup of $G$, because $G/H_{k}=\langle aH_{k}\rangle\cong C_{p^{m+k}}$. Notice that there are exactly $n-m+1$ such subgroups of $G$. Up to isomorphism, there is no other cyclic cocyclic subgroup of $G$. Indeed, if there is one such subgroup $H$ which is not isomorphic to any $H_{k}$ for $0\leq k\leq n-m$, then $|H|=p^{s}$ where $s\in\{0\dots m-1\}$. In this case, $G/H\cong C_{{p}^{n+m-s}}$ where $n+m-s\geq n+1$, but this is impossible since the exponent of $G$ is equal to $p^{n}$. Note that $H_{k}\cong H_{k}\times K_{m}$ because $K_{m}=1$.

Case 2 (Cocyclic subgroups of $G$ which are not cyclic): From Case 1, we know that $H_{k}$ is a cocyclic subgroup of $G$, for each $k$, so that $G/H_{k}$ is cyclic. Since for each $j$, the quotient $G/(H_{k}\times K_{j})$ is isomorphic to a subgroup of $G/H_{k}$, we have that $H_{k}\times K_{j}$ is a cocyclic subgroup of $G$ for any $j\in\{0,\dots m\}$. Conversely, using Lemma 3.1, we deduce that any cocyclic subgroup is isomorphic to one of $H_{k}\times K_{j}$ since $H_{k}\cong C_{p^{n-k}}$ where $k\in\{0,\dots n-m,\}$ and $K_{j}\cong C_{p^{m-j}}$ for $j\in\{0,\dots m-1\}$. ∎

Proof of Theorem 1.5.

By Proposition 3.2, the number of isomorphism classes of cocyclic subgroups of $G$ is $(n-m+1)(m+1)$. By Theorem 1.2, $\eta(\mathbb{F}G)=(n-m+1)(m+1).$ ∎

An immediate consequence of Theorem 1.5 and Theorem 1.4 is the following result.

Corollary 3.3**.**

Let $n,m$ and $s$ be positive integers such that $n>m$. If $G=(C_{p^{n}}\times C_{p^{m}})^{s}$ for $s\in\mathbb{N}$, then $\eta(\mathbb{F}G)=(n-m+1)(m+1)$. Moreover if $G=C_{p^{n}}\times C_{p^{m}}\times(C_{p^{n}})^{s}$, then $\eta(\mathbb{F}G)=(n-m+1)(m+1)$.

4. Proof of Theorem 1.7

Let $\tau(G)$ denote the number of divisors of the exponent of $G$. It is not difficult to see that the number of non-$G$-equivalent minimal abelian codes is greater than or equal to $\tau(G)$ when $G$ is a finite abelian group. Therefore, if the exponent of $G$ is given, Theorem 1.7 gives a complete characterization of the groups having $\tau(G)$ non-$G$-equivalent minimal abelian codes, that is having the least possible number of non-$G$-equivalent abelian codes. For the proof of Theorem 1.7, first of all we find the number of non-$G$-equivalent minimal abelian group codes for homocyclic $p$-groups and prove the following.

Theorem 4.1**.**

Let $G$ be a finite abelian $p$-group. The number of non-$G$-equivalent minimal abelian codes is equal to $\tau(G)$ if and only if $G$ is homocyclic.

Proof.

Assume that $G$ is homocyclic, that is $G\cong(C_{p^{n}})^{s}$. Then by Theorem 1.3, $\eta(\mathbb{F}G)=\eta(\mathbb{F}C_{p^{n}})$. Now it is clear that the number of isomorphism classes of subgroups of $C_{p^{n}}$ is equal to the number of divisors of $p^{n}$. For the converse implication, assume that $G$ is not homocyclic. If the exponent of $G$ is $p^{r}$ for some $r\geq 1$, then $G\cong C_{p^{r}}\times H$ where $H\cong K\times C_{p^{i}}$ for some $1\leq i\leq r-1$ for some subgroup $K$ of $H$. Then $H,H\times C_{p},H\times C_{p^{2}},...,H\times C_{p^{r-1}}$ is a family of non-isomorphic cocyclic subgroups of $G$. Obviously $K\times C_{p^{r}}$ is another cocyclic subgroup which is not isomorphic to none of the elements of this family. So we have at least $r+2$ non-isomorphic cocyclic subgroups. Hence, $\eta(\mathbb{F}G)$ is at least $r+2$. This leads to a contradiction because $\tau(G)=r+1$. ∎

Proof of Theorem 1.7.

Let $G=G_{p_{1}}\times G_{p_{2}}\times...\times G_{p_{k}}$ where each $G_{p_{i}}$ is a homocyclic Sylow $p_{i}$-subgroup of $G$. If the exponent of each $G_{p_{i}}$ is $p^{e_{i}}$, then by Theorem 4.1, $\eta(\mathbb{F}G_{p_{i}})$ is equal to $\tau(G_{p_{i}})=e_{i}+1$. By Lemma 2.8, $\eta(\mathbb{F}G)$ is equal to $\prod_{i=1}^{k}(e_{i}+1)$ which is equal to $\tau(G).$

For the converse implication, assume for some $i$, a Sylow $p_{i}$-subgroup $G_{p_{i}}$ is not homocyclic. Then by Theorem 4.1, $\eta(\mathbb{F}G_{p_{i}})>\tau(G_{p_{i}})=e_{i}+1$ which gives a contradiction. ∎

Acknowledgements. The authors were partially supported by Mimar Sinan Fine Arts University Scientific Research Unit with project number 2019-27.