The Number of Subgroup Chains of Finite Nilpotent Groups

Abstract

In this paper, we mainly count the number of subgroup chains of a finite nilpotent group. We derive a recursive formula that reduces the counting problem to that of finite p-groups. As applications of our main result, the classification problem of distinct fuzzy subgroups of finite abelian groups is reduced to that of finite abelian p-groups. In particular, an explicit recursive formula for the number of distinct fuzzy subgroups of a finite abelian group whose Sylow subgroups are cyclic groups or elementary abelian groups is given.

1. Introduction

All of the groups considered in this paper are finite. Basic notations and concepts correspond to [1,2,3].

One of the most important problems in combinatorial group theory is to count the number of subgroup chains of a group (see [4]). Many papers have treated various aspects of this problem in the last few years. For example, in [5], the subgroup chains of cyclic group are investigated, and [6] deals with the number of subgroup chains of an elementary abelian p-group. Furthermore, some classes of special subgroup chains are studied in several papers, like, in [4], the authors count the number of maximal subgroup chains of nilpotent groups. In addition, this topic also has close connection with some open questions of other fields of mathematics. For example, in [7,8], the authors study the well-known Delannoy numbers, and prove that they are just the numbers of the subgroup chains of a cyclic group that satisfy a certain property. Another interesting question is the classifying of distinct fuzzy subgroups of abelian groups, which can be translated into a combinatorial problem on the subgroup lattice of a group G: counting the number of some kind of subgroup chains of G. Now, many important results has been given (for example, see [9,10,11,12,13,14,15]).

In the present paper, we are concerned with the number of subgroup chains of nilpotent groups. Let G be a nilpotent group. Subsequently, G can be written as a direct product of its Sylow p-subgroups and all the subgroup chains of G can be constructed from the subgroup chains of its Sylow p-subgroups. We derive a simple recursive formula for counting the number of subgroup chains of a nilpotent group by using that of its Sylow p-subgroups. As applications of our main result, the classification problem of distinct fuzzy subgroups of abelian groups be reduced to the computational problem of the number of subgroup chains of abelian p-groups. In particular, an explicit recursive formula for the number of distinct fuzzy subgroups of abelian groups whose Sylow subgroups are cyclic groups or elementary abelian groups is given. In addition, we make two specific examples of our applications at the end of the paper.

2. Preliminaries

For convenience, we recall some elementary definitions and results in this section.

Let G be a group. A subgroup chain $\Gamma :G_1<G_2<\cdots <G_n$ of G is a set of subgroups of G linearly ordered by set inclusion. In this case the integer $n-1$ is called the length of the subgroup chain $\Gamma$. The subgroup chain $\Gamma$ of G is called rooted (more precisely G-rooted) if $G_n=G$. Otherwise, it is called unrooted. Additionally, note that we say two subgroup chains ${\Gamma }_1$ and ${\Gamma }_2$ of G are same and denoted by ${\Gamma }_1={\Gamma }_2$ if ${\Gamma }_1$ and ${\Gamma }_2$ contain same subgroups of G. Otherwise, we say that ${\Gamma }_1$ and ${\Gamma }_2$ are different and denoted by ${\Gamma }_1\ne {\Gamma }_2$.

Now set $\mathcal{C}\left(G\right)=\left\{\mathrm{the}\mathrm{subgroup}\mathrm{chain}\mathrm{of}\mathrm{G}\right\};$

• $\mathcal{D}\left(G\right)=\left\{\mathrm{the}\mathrm{unrooted}\mathrm{subgroup}\mathrm{chain}\mathrm{of}\mathrm{G}\right\};$
• $\mathcal{F}\left(G\right)=\left\{\mathrm{the}\mathrm{rooted}\mathrm{subgroup}\mathrm{chain}\mathrm{of}\mathrm{G}\right\};$
• $\mathcal{H}\left(G\right)=\{\Gamma \in \mathcal{F}(G)\mid \Gamma \mathrm{doesn}’\mathrm{t}\mathrm{contain}\mathrm{the}\mathrm{identity}\mathrm{subgroup}\left\{\mathrm{e}\right\}\};$
• ${\mathcal{H}}_s\left(G\right)=\{\Gamma \in \mathcal{H}\left(G\right)\mid \mathrm{the}\mathrm{length}\mathrm{of}\Gamma \mathrm{is}s-1,\mathrm{where}s\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer}\}.$

We use $C\left(G\right),D\left(G\right),F\left(G\right),H\left(G\right)$ and $H_s\left(G\right)$ to denote the cardinal numbers of $\mathcal{C}\left(G\right),\mathcal{D}\left(G\right)$, $\mathcal{F}\left(G\right),\mathcal{H}\left(G\right)$ and ${\mathcal{H}}_s\left(G\right)$, respectively.

Remark 1.

$\left(1\right)$ A single subgroup H of G is also a subgroup chain of G, and its length is 0; $\left(2\right)$ Note that $F\left(G\right)=1$ and $H\left(G\right)=0$ when G is the trivial group $\left\{e\right\}$, and $H_1\left(G\right)=1$ when G is any group with $G\ne \left\{e\right\}$.

The following two simple observations are useful for counting subgroup chains of a group.

Proposition 1.

Ref. [8] Let G be a group. Then $F\left(G\right)=D\left(G\right)+1$ and $C\left(G\right)=F\left(G\right)+D\left(G\right)=2F\left(G\right)-1$.

Proposition 2.

Let G be a group with $G\ne \left\{e\right\}$. Then

$$F\left(G\right)=2H\left(G\right)=2\sum _s{H}_s\left(G\right).$$

Proof. 

Let ${\mathcal{H}}^{*}\left(G\right)=\{\Gamma \in \mathcal{F}\left(G\right)\mid \Gamma \mathrm{contains}\mathrm{the}\mathrm{identity}\mathrm{subgroup}\left\{\mathrm{e}\right\}\}.$ Afterwards, it is easy to see that $\mathcal{H}\left(G\right)\cap {\mathcal{H}}^{*}\left(G\right)=\varnothing$ and $\mathcal{F}\left(G\right)=\mathcal{H}\left(G\right)\cup {\mathcal{H}}^{*}\left(G\right)$. Additionally, for any subgroup chain $\Gamma$ in $\mathcal{H}\left(G\right)$, we may construct a subgroup chain ${\Gamma }^{*}$ in ${\mathcal{H}}^{*}\left(G\right)$ by adding identity subgroup in $\Gamma$. Hence, $\left|\mathcal{H}\left(G\right)\right|=|{\mathcal{H}}^{*}\left(G\right)|$. Thus, $F\left(G\right)=2H\left(G\right)$. Subsequently, according to the above definitions, we obtain the following equality immediately:

$$F\left(G\right)=2H\left(G\right)=2\sum _s{H}_s\left(G\right).$$

□

3. The Number of Subgroup Chains of Nilpotent Groups

Let $\pi$ be a set of primes and ${\pi }^{\prime }$ the complement of $\pi$ in the set of all primes. Recall that a $\pi$-number is a positive integer whose prime divisors all belong to $\pi$. A subgroup H of a group G is called a Hall $\pi$-subgroup if $\left|H\right|$ is a $\pi$-number and $|G:H|$ is a ${\pi }^{\prime }$-number.

Now, let G be a group. Let A and B be proper subgroups of G such that $G=A\times B$ with $\left(\right|A|,|B\left|\right)=1$. Let $\pi \left(A\right)$ and $\pi \left(B\right)$ be the sets of prime divisors of $\left|A\right|$ and $\left|B\right|$, respectively. If H is a subgroup of G, then, by [2] (Chapter I. Lemma 3.2), $A\cap H$ is a Hall $\pi \left(A\right)$-subgroup and $B\cap H$ is a Hall $\pi \left(B\right)$-subgroup of H and, therefore, $H=(A\cap H)\times (B\cap H)$. It is clear that this kind of decomposition of H is uniquely determined by H. Thus, if

$$\Gamma :\left\{e\right\}\ne G_1<G_2<\cdots <G_s=G$$

(1)

is a subgroup chain in $\mathcal{H}\left(G\right)$, there exist a unique subgroup $A_t$ of A and a unique subgroup $B_t$ of B, such that $G_t=A_t\times B_t$ for each $t=1,2,\cdots ,s$. Hence, we have subgroup chains $A_1\le A_2\le \cdots \le A_s=A$ and $B_1\le B_2\le \cdots \le B_s=B$. If we remove redundant terms and the identity subgroup $\left\{e\right\}$ in these two subgroup chains, then we have the two subgroup chains in $\mathcal{H}\left(A\right)$ and $\mathcal{H}\left(B\right)$, as follows:

$${\Gamma }_1:\left\{e\right\}\ne A_{k_1}<A_{k_2}<\cdots <A_{k_i}=A,$$

(2)

$${\Gamma }_2:\left\{e\right\}\ne B_{l_1}<B_{l_2}<\cdots <B_{l_j}=B.$$

(3)

Notice that the subgroup chains $\left(2\right)$ and $\left(3\right)$ are uniquely determined by the subgroup chain $\left(1\right)$. We call the subgroup chains $\left(2\right)$ and $\left(3\right)$ the factor chains of the subgroup chain $\left(1\right)$ for the subgroups A and B. By the above discussion, we state the lemma, as follows:

Lemma 1.

Let G be a group with $G\ne \left\{e\right\}$. Let A and B be proper subgroups of G such that $G=A\times B$ with $\left(\right|A|,|B\left|\right)=1$. Then

(1) For any subgroup chain $\Gamma \in {\mathcal{H}}_s\left(G\right)$, there exist two unique subgroup chains ${\Gamma }_1\in {\mathcal{H}}_i\left(A\right)$ and ${\Gamma }_2\in {\mathcal{H}}_j\left(B\right)$, such that ${\Gamma }_1$ and ${\Gamma }_2$ are the factor chains of Γ for the subgroups A and B;

(2) The positive integers $i,j,s$ satisfy that $max\{i,j\}\le s\le i+j$.

Note that, for any given subgroup chains ${\Gamma }_1\in \mathcal{H}\left(A\right)$ and ${\Gamma }_2\in \mathcal{H}\left(B\right)$, there may exist different subgroup chains in $\mathcal{H}\left(G\right)$, such that their factor chains for the subgroups A and B are ${\Gamma }_1$ and ${\Gamma }_2$. For example, the factor chains of the next two subgroup chains are $\left\{e\right\}\ne A_1<A$ and $\left\{e\right\}\ne B_1<B$:

$$A_1\times \left\{e\right\}<A\times \left\{e\right\}<A\times B_1<A\times B=G;$$

$$\left\{e\right\}\times B_1<\left\{e\right\}\times B<A_1\times B<A\times B=G.$$

Now, let ${\Gamma }_1$ and ${\Gamma }_2$ be two subgroup chains in ${\mathcal{H}}_i\left(A\right)$ and ${\mathcal{H}}_j\left(B\right)$, respectively. Assume that positive integers $i,j,s$ satisfy $max\{i,j\}\le s\le i+j$. Let

$${\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)=\{\Gamma \in {\mathcal{H}}_s\left(G\right)\mid \mathrm{the}\mathrm{factor}\mathrm{chains}\mathrm{of}\Gamma \mathrm{are}{\Gamma }_1\mathrm{and}{\Gamma }_2\}.$$

We use $H_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$ to denote the cardinality of ${\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$. Then according to Lemma 1, it is easy to see that

$${\mathcal{H}}_s\left(G\right)=\bigcup _{\begin{array}{c}{\Gamma }_1\in {\mathcal{H}}_i\left(A\right),{\Gamma }_2\in {\mathcal{H}}_j\left(B\right)\\ max\{i,j\}\le s\le i+j\end{array}}{\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right).$$

The next result tells us that the sets ${\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$ are pairwise disjoint sets.

Corollary 1.

Let G be a group with $G\ne \left\{e\right\}$. Let A and B be proper subgroups of G such that $G=A\times B$ with $\left(\right|A|,|B\left|\right)=1$. Let ${\Gamma }_1$, ${\Gamma }_2$ be two subgroup chains in $\mathcal{H}\left(A\right)$ and ${\Gamma }_3$, ${\Gamma }_4$ two subgroup chains in $\mathcal{H}\left(B\right)$. Now, set

$\mathcal{X}=\{\Gamma \in \mathcal{H}\left(G\right)\mid thefactorchainsof\Gamma are{\Gamma }_{1}and{\Gamma }_3\};$

$\mathcal{Y}=\{\Gamma \in \mathcal{H}\left(G\right)\mid thefactorchainsof\Gamma are{\Gamma }_{2}and{\Gamma }_4\}.$

If ${\Gamma }_1$ is different from ${\Gamma }_2$ or ${\Gamma }_3$ is different from ${\Gamma }_4$, then $\mathcal{X}\cap \mathcal{Y}=\varnothing$.

Proof. 

This follows immediately from Lemma 1$\left(1\right)$. □

Corollary 2.

Let G be a group with $G\ne \left\{e\right\}$. Let A and B be proper subgroups of group G, such that $G=A\times B$ with $\left(\right|A|,|B\left|\right)=1$. Then

$$H_s\left(G\right)=\sum _{\begin{array}{c}{\Gamma }_1\in {\mathcal{H}}_i\left(A\right),{\Gamma }_2\in {\mathcal{H}}_j\left(B\right)\\ max\{i,j\}\le s\le i+j\end{array}}{H}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right).$$

Next lemma tells us for any two given subgroup chains ${\Gamma }_1\in {\mathcal{H}}_i\left(A\right)$ and ${\Gamma }_2\in {\mathcal{H}}_j\left(B\right)$, how many subgroup chains $\Gamma \in {\mathcal{H}}_s\left(G\right)$ there are the factor chains of $\Gamma$ are ${\Gamma }_1$ and ${\Gamma }_2$.

Lemma 2.

Let G be a group with $G\ne \left\{e\right\}$. Let A and B be proper subgroups of G such that $G=A\times B$ with $\left(\right|A|,|B\left|\right)=1$. Let

$${\Gamma }_1:A_1<A_2<\cdots <A_i=A$$

and

$${\Gamma }_2:B_1<B_2<\cdots <B_j=B$$

be arbitrary subgroup chains in ${\mathcal{H}}_i\left(A\right)$ and ${\mathcal{H}}_j\left(B\right)$, respectively. Additionally, assume that the positive integers $i,j,s$ satisfy that $max\{i,j\}\le s\le i+j$. Then

$$H_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{s}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{i+j-s}}\right).$$

Proof. 

Let

$$\Gamma :G_1<G_2<\cdots <G_s=G$$

be a subgroup chain in ${\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$. Afterwards, by the above discussion, the subgroup chain $\Gamma$ has the decomposition as follows:

$$\Gamma :A_{11}\times B_{21}<A_{12}\times B_{22}<\cdots <A_{1s}\times B_{2s}=A\times B=G.$$

For the next two subgroup chains

$${\Gamma }_3:A_{11}\le A_{12}\le \cdots \le A_{1s}=A$$

and

$${\Gamma }_4:B_{21}\le B_{22}\le \cdots \le B_{2s}=B,$$

we delete the identity subgroup $\left\{e\right\}$ first, and then delete redundant terms. Note that we always delete the right one for repeated two subgroups. For example, if $A_{1t}=A_{1t+1}$, we always delete $A_{1t+1}$. By using this agreement, the subgroup chains ${\Gamma }_3$ and ${\Gamma }_4$ become the following subgroup chains:

$${\Gamma }_5:A_{1x_1}<A_{1x_2}<\cdots <A_{1x_i}=A,$$

$${\Gamma }_6:B_{2y_1}<B_{2y_2}<\cdots <B_{2y_j}=B.$$

It is obvious that ${\Gamma }_5$ and ${\Gamma }_6$ are the factor chains of $\Gamma$. Accordingly, by Lemma 1$\left(1\right)$, we see that ${\Gamma }_1={\Gamma }_5$ and ${\Gamma }_2={\Gamma }_6.$ That is to say $A_h=A_{1x_h}$ for any $h\in \{1,2,\cdots ,i\}$ and $B_f=B_{2y_f}$ for any $f\in \{1,2,\cdots ,j\}$. Now, we consider the $i+j$ dimensional vector ${\alpha }_{\Gamma }=(x_1,x_2,\cdots ,x_i,y_1,y_2,\cdots ,y_j)$. We see that it is uniquely determined by the subgroup chain $\Gamma$. We call ${\alpha }_{\Gamma }$ the location vector of the subgroup chain $\Gamma$ for the subgroups A and B.

Let ${\mathcal{R}}_s^{i+j}$ be the set of all vectors $\alpha$ with $\alpha =(a_1,a_2,\cdots ,a_i,b_1,b_2,\cdots ,b_j)$ satisfying the following conditions:

(i)

$a_1,a_2,\cdots ,a_i,b_1,b_2,\cdots ,b_j\in \{1,2,\cdots ,s\}$;

(ii)

$a_1<a_2<\cdots <a_i$ and $b_1<b_2<\cdots <b_j$;

(iii)

$\{a_1,a_2,\cdots ,a_i\}\cup \{b_1,b_2,\cdots ,b_j\}=\{1,2,\cdots ,s\}$.

We can prove that ${\alpha }_{\Gamma }\in {\mathcal{R}}_s^{i+j}$. In fact, by the above discussion, it is clear that the vector ${\alpha }_{\Gamma }$ satisfies $x_1,x_2,\cdots ,x_i,y_1,y_2,\cdots ,y_j\in \{1,2,\cdots ,s\}$, $x_1<x_2<\cdots <x_i$ and $y_1<y_2<\cdots <y_j.$ Subsequently, we only need to prove that ${\alpha }_{\Gamma }$ satisfies the condition (iii). Assume that $\{x_1,x_2,\cdots ,x_i\}\cup \{y_1,y_2,\cdots ,y_j\}\ne \{1,2,\cdots ,s\}$, there exists a positive integer $r\in \{1,2,\cdots ,s\}$ such that $r\notin \{x_1,x_2,\cdots ,x_i\}$ and $r\notin \{y_1,y_2,\cdots ,y_j\}$. Hence, we must delete the both subgroups $A_{1r}$ in ${\Gamma }_3$ and $B_{2r}$ in ${\Gamma }_4$ when we delate redundant terms in the above. That means $A_{1r-1}=A_{1r}$ and $B_{2r-1}=B_{2r}$, which contradicts that the subgroup chain $\Gamma$ is a proper subgroup chain. Thus, $\{x_1,x_2,\cdots ,x_i\}\cup \{y_1,y_2,\cdots ,y_j\}=\{1,2,\cdots ,s\}$, and, therefore, ${\alpha }_{\Gamma }\in {\mathcal{R}}_s^{i+j}$.

Now, we claim that there is a one-to-one correspondence between ${\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$ and ${\mathcal{R}}_s^{i+j}$. Consider the map

$$\phi :{\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)⟶{\mathcal{R}}_s^{i+j}$$

$$\Gamma ⟼{\alpha }_{\Gamma },$$

where ${\alpha }_{\Gamma }$ is the location vector of the subgroup chain $\Gamma$. Next we prove that the map $\phi$ is a bijective map.

For any vector $\alpha =(a_1,a_2,\cdots ,a_i,b_1,b_2,\cdots ,b_j)$ of ${\mathcal{R}}_s^{i+j}$, we may construct a subgroup chain ${\Gamma }^{\prime }$ in ${\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$, such that $\phi \left({\Gamma }^{\prime }\right)={\alpha }_{{\Gamma }^{\prime }}=\alpha$ by using the following natural process. If $s=1$, then $i=j=1$ and it is easy to see that $A_1\times B_1=A\times B=G$ is the subgroup chain ${\Gamma }^{\prime }$ we need. If $s\ge 2,$, since $\{a_1,a_2,\cdots ,a_i\}\cup \{b_1,b_2,\cdots ,b_j\}=\{1,2,\cdots ,s\}$, then we have $1\in \{a_1,a_2,\cdots ,a_i\}$ or $1\in \{b_1,b_2,\cdots ,b_j\}$. We set $K_1$, as follows:

$$K_1=\left\{\begin{array}{cc}{A}_1\times B_1,\hfill & \mathrm{if}{a}_1=1\mathrm{and}{b}_1=1;\hfill \\ A_1\times \left\{e\right\},\hfill & \mathrm{if}{a}_1=1\mathrm{and}{b}_1\ne 1;\hfill \\ \left\{e\right\}\times B_1,\hfill & \mathrm{if}{a}_1\ne 1\mathrm{and}{b}_1=1.\hfill \end{array}\right.$$

Now, we assume that $K_{t-1}(2\le t\le s)$ is given and $K_{t-1}=X\times Y$, where $X\in \{\left\{e\right\},A_1,A_2,\cdots ,A_i\}$ and $Y\in \{\left\{e\right\},B_1,B_2,\cdots ,B_j\}$. Then we consider the case t. Additionally, since $t\in \{a_1,a_2,\cdots ,a_i\}$ or $t\in \{b_1,b_2,\cdots ,b_j\}$, there exists $h\in \{1,2,\cdots ,i\}$, such that $t=a_h$ or exists $f\in \{1,2,\cdots ,j\}$ such that $t=b_f$. Accordingly, we may set $K_t$, as follows.

$$K_t=\left\{\begin{array}{cc}{A}_h\times B_f,\hfill & \mathrm{if}t=a_h=b_f;\hfill \\ A_h\times Y,\hfill & \mathrm{if}t=a_h\ne b_f;\hfill \\ X\times B_f,\hfill & \mathrm{if}t=b_f\ne a_h.\hfill \end{array}\right.$$

By using the above methods, we may construct the following subgroup chain of G

$${\Gamma }^{\prime }:K_1<K_2<\cdots <K_s.$$

Additionally, it is easy to see that ${\Gamma }^{\prime }\in {\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$ and $\phi \left({\Gamma }^{\prime }\right)={\alpha }_{{\Gamma }^{\prime }}=\alpha$. Accordingly, the map $\phi$ is a surjective map. Furthermore, let

$${\mathsf{{\rm Y}}}_1:C_1<C_2<\cdots <C_s=G$$

and

$${\mathsf{{\rm Y}}}_2:D_1<D_2<\cdots <D_s=G$$

be two subgroup chains in ${\mathcal{H}}_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$. Let ${\alpha }_{{\mathsf{{\rm Y}}}_1}=(u_1,u_2,\cdots ,u_i,v_1,v_2,\cdots ,v_j)$ and ${\alpha }_{{\mathsf{{\rm Y}}}_2}=(u_1^{\prime },u_2^{\prime },\cdots ,u_i^{\prime },v_1^{\prime },v_2^{\prime },\cdots ,v_j^{\prime })$ be the location vectors of ${\mathsf{{\rm Y}}}_1$ and ${\mathsf{{\rm Y}}}_2$, respectively. If ${\mathsf{{\rm Y}}}_1\ne {\mathsf{{\rm Y}}}_2$, then there exists a positive integer $r\in \{1,2,\cdots ,s\}$ such that $C_r\ne D_r$ and $C_k=D_k$ for any $k=1,2,\cdots ,r-1$. If we assume $C_r=A_{1r}\times B_{1r}$ and $D_r={\tilde{A}}_{1r}\times {\tilde{B}}_{1r}$, then we have $A_{1r}\ne {\tilde{A}}_{1r}$ or $B_{1r}\ne {\tilde{B}}_{1r}$. Without loss of generality, we may assume that $A_{1r}\ne {\tilde{A}}_{1r}$. Subsequently, we see that

$$r\in \{u_1,u_2,\cdots ,u_i\}andr\notin \{u_1^{\prime },u_2^{\prime },\cdots ,u_i^{\prime }\}$$

or

$$r\notin \{u_1,u_2,\cdots ,u_i\}andr\in \{u_1^{\prime },u_2^{\prime },\cdots ,u_i^{\prime }\}.$$

That is to say ${\alpha }_{{\mathsf{{\rm Y}}}_1}\ne {\alpha }_{{\mathsf{{\rm Y}}}_2}$. Hence, the map $\phi$ is also an injective map. Hence, the claim is proved.

By the above claim, we have

$$H_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)=\mid {\mathcal{R}}_s^{i+j}\mid .$$

Now, we calculate the $|{\mathcal{R}}_s^{i+j}|$. For any vector $\alpha =(a_1,a_2,\cdots ,a_i,b_1,b_2,\cdots ,b_j)\in {\mathcal{R}}_s^{i+j}$, it is easy to see that $\alpha$ can be determined once $\{a_1,a_2,\cdots ,a_i\}$ and $\{a_1,a_2,\cdots ,a_i\}\cap \{b_1,b_2,$$\cdots ,b_j\}$ are given. Notice that

$$|\{a_1,a_2,\cdots ,a_i\}\cap \{b_1,b_2,\cdots ,b_j\}|=i+j-s.$$

Hence,

$$|{\mathcal{R}}_s^{i+j}|=\left({\displaystyle \genfrac{}{}{0pt}{}{s}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{i+j-s}}\right).$$

Therefore,

$$H_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{s}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{i+j-s}}\right).$$

The lemma is proved. □

From Lemma 2, we can see that for another different two given subgroup chains ${\Gamma }_1^{\prime }\in {\mathcal{H}}_i\left(A\right)$ and ${\Gamma }_2^{\prime }\in {\mathcal{H}}_j\left(B\right)$, the number $H_s^{{\Gamma }_1,{\Gamma }_2}\left(G\right)$ is equal to the number $H_s^{{\Gamma }_1^{\prime },{\Gamma }_2^{\prime }}\left(G\right)$.

Now, we can give the main result of this paper.

Theorem 1.

Let $s,m,i,j$ be positive integers with $m\ge 2$, and let $p_1,p_2,\cdots ,p_m$ be different primes. If $G=P_1\times P_2\times \cdots \times P_m$ is a nilpotent group with $P_t$, the Sylow $p_t$-subgroup of G and $P_t\ne 1$ for $t=1,2,\cdots m$, then

$$H_s(P_1\times P_2\times \cdots \times P_k)=\sum _{\begin{array}{c}max\{i,j\}\le s\le i+j\end{array}}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{i+j-s}}\right)H_i(P_1\times P_2\times \cdots \times P_{k-1})H_j\left(P_k\right)$$

for any $k=2,\cdots ,m$.

Proof. 

For $k=2,\cdots ,m$, since $\left(\right|P_1\times P_2\times \cdots \times P_{k-1}|,|P_k\left|\right)=1$, according to Corollarys 1 and 2, and Lemma 2, we can see that

$$H_s(P_1\times P_2\times \cdots \times P_k)=\sum _{\begin{array}{c}max\{i,j\}\le s\le i+j\end{array}}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{i+j-s}}\right)H_i(P_1\times P_2\times \cdots \times P_{k-1})H_j\left(P_k\right).$$

The theorem is proved. □

4. The Number of Distinct Fuzzy Subgroups of Abelian Groups

A recent problem that involves some combinatorial aspects on subgroup chains is the classifying of distinct fuzzy subgroups of abelian groups. This topic has enjoyed a rapid evolution in the last years and many results have been given. For example, the authors in [10] determine the number of distinct fuzzy subgroups of a cyclic group of square-free order, and the authors in [11,15] deal with the number for cyclic groups of order $p^n{q}^m$($p,q$ are primes). In addition, Tărnăuceanu and Bentea in [14] give an explicit formula for the number of fuzzy subgroups of a cyclic group and establish a recurrence relation verified by the number of fuzzy subgroups of an elementary abelian p-group. Ngcibi, Murali, and Makamba in [13] obtain a formula for the number of fuzzy subgroups of the group ${\mathbb{Z}}_{p^m}\times {\mathbb{Z}}_{p^n}$ (${\mathbb{Z}}_{p^n}$ is a cyclic p-group of order $p^n$) for $n=1,2,3$, which has been extended by Oh in [12] for all values of n. Additionally, in [9], the authors count the distinct fuzzy subgroups of some rank-3 abelian p-groups. Although the classification problem of distinct fuzzy subgroups of cyclic groups, elementary abelian p-groups, and some special abelian p-groups has been solved, it is still an open question for an arbitrary abelian group.

Now, we recall some basic notions and the results of fuzzy groups.

Definition 1.

Ref. [14] Let S be a set. A mapping $\mu :S\to [0,1]$ is called a fuzzy subset of S.

Definition 2.

Ref. [14] Let G be a group and $\mu :G\to [0,1]$ be a fuzzy subset of G. We say that μ is a fuzzy subgroup of G if it satisfies the next two conditions:

(1) 

$\mu \left(xy\right)\ge min\left\{\mu \right(x),\mu (y\left)\right\}$, for all $x,y\in G;$

(2) 

$\mu \left(x^{-1}\right)\ge \mu \left(x\right)$, for any $x\in G.$

The fuzzy subgroups of a group can be classified up to some natural equivalence relations. A widely used equivalence relation is shown below: let $\mu$ and $\eta$ be two fuzzy subgroups of a group G. We say $\mu$ and $\eta$ are equivalent if $\mu \left(x\right)>\mu \left(y\right)⟺\eta \left(x\right)>\eta \left(y\right)\mathrm{for}\mathrm{all}x,y\in G$. Additionally, we say $\mu$ and $\eta$ are distinct if $\mu$ and $\eta$ are not equivalent. In [15], a necessary and sufficient condition for $\mu$ and $\eta$ to be equivalent was given, where it is proved that

“μ and η are equivalent if and only if μ and η determine the same rooted subgroup chains of G”.

That is to say there exists a one-to-one correspondence between the set of distinct fuzzy subgroups of G and the set of rooted subgroup chains of G. Therefore, the number of distinct fuzzy subgroups of G is equal to the number $F\left(G\right)$ of the rooted subgroup chains of G. According to this result, we can apply our main result in the classifying of distinct fuzzy subgroups of abelian groups.

It is known that abelian groups are nilpotent groups, so, by Proposition 2 and Theorem 1, we obtain the next theorem.

Theorem 2.

Let $s,m,i,j,{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_m$ be positive integers with $m\ge 2$, and let $p_1,p_2,\cdots ,p_m$ be different primes. If $G=P_1\times P_2\times \cdots \times P_m$ is an abelian group with $P_t$ the Sylow $p_t$-subgroup of G and $\mid P_t\mid =p_t^{{\alpha }_t}$ for $t=1,2,\cdots m$, then the number of distinct fuzzy subgroups of G is

$$F\left(G\right)=2\sum _{s=1}^{{\alpha }_1+{\alpha }_2+\cdots +{\alpha }_m}{H}_s\left(G\right),$$

where the $H_s\left(G\right)$ satisfy the following recursive formula

$$H_s(P_1\times P_2\times \cdots \times P_k)=\sum _{\begin{array}{c}max\{i,j\}\le s\le i+j\end{array}}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{i+j-s}}\right)H_i(P_1\times P_2\times \cdots \times P_{k-1})H_j\left(P_k\right)$$

for any $k=2,\cdots ,m$.

Now, assume that the Sylow subgroups of an abelian group G are cyclic groups or elementary abelian groups. Subsequently, according to Theorem 2, if we can clarify the number of subgroup chains of cyclic p-groups and elementary abelian p-groups for any prime p, then we may count the number of distinct fuzzy subgroups of the abelian group G.

Let p be a prime and n a positive integer. We use ${\mathbb{Z}}_{p^n}$ to denote a cyclic p-group of order $p^n$ and use ${\mathbb{Z}}_p^n$ to denote an elementary abelian p-group of order $p^n$. We need to recall two well-known results in group theory first in order to count the number of subgroup chains of ${\mathbb{Z}}_{p^n}$ and ${\mathbb{Z}}_p^n$.

Lemma 3.

[3] (Chapter I. Theorem 2.20) Let G be a cyclic group of order n. Then there is a unique subgroup of G of order d for any positive divisor d of n.

Lemma 4.

[3] (Chapter III. Theorem 8.5) Let p be a prime and let G be an elementary abelian p-group of order $p^n$ with $n\ge 1$. We denote the number of subgroups of G of order $p^m$$(1\le m\le n)$ by ${\left[\begin{array}{c}n\\ m\end{array}\right]}_p$.Then

$${\left[\begin{array}{c}n\\ m\end{array}\right]}_p=\frac{(p^n-1)(p^{n-1}-1)\cdots (p^{n-m+1}-1)}{(p^m-1)(p^{m-1}-1)\cdots (p-1)}.$$

The above two lemmas allow for us to obtain the formulas for calculating the number of subgroup chains of cyclic p-groups and elementary abelian p-groups.

Lemma 5.

Let $G={\mathbb{Z}}_{p^n}$ with $n\ge 1$. Then

$$H_s\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{s-1}}\right),1\le s\le n.$$

Proof. 

By Lemma 3, we can see that all the subgroups of G are $\left\{e\right\},G_1,G_2,\cdots ,G_n$, where $|G_s|=p^s$ for $s=1,2,\cdots ,n$. Additionally, these subgroups satisfy

$$\left\{e\right\}<G_1<G_2<\cdots <G_n=G.$$

Now, it easily is verified that

• ${\mathcal{H}}_1\left(G\right)=\{G_n\mid G_n\mathrm{is}\mathrm{a}\mathrm{subgroup}\mathrm{chain}\};$
• ${\mathcal{H}}_s\left(G\right)=\{G_{i_1}<G_{i_2}<\cdots <G_{i_{s-1}}<G_n\mid 1\le i_1<i_2<\cdots <i_{s-1}\le n-1\}\mathrm{for}2\le s\le n.$

Subsequently, we have

$H_s\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{s-1}}\right),1\le s\le n.$

The lemma is proved. □

Lemma 6.

Let $G={\mathbb{Z}}_p^n$ with $n\ge 1$. Then

$$H_s\left(G\right)=\left\{\begin{array}{cc}1,\hfill & s=1;\hfill \\ {\displaystyle \sum _{1\le i_1<i_2<\cdots <i_{s-1}\le n-1}}{\left[\begin{array}{c}n\\ i_{s-1}\end{array}\right]}_p{\left[\begin{array}{c}{i}_{s-1}\\ i_{s-2}\end{array}\right]}_p\cdots {\left[\begin{array}{c}{i}_2\\ i_1\end{array}\right]}_p,\hfill & 2\le s\le n.\hfill \end{array}\right.$$

Proof. 

It is clear that $H_1\left(G\right)=1.$ Let $\Gamma$ be a subgroup chain in ${\mathcal{H}}_s\left(G\right)$ with $2\le s\le n$ as follows:

$$\Gamma :G_1<G_2<\cdots <G_s=G.$$

Afterwards, we can naturally have the following s-dimensional vector

$$\alpha =\left(\right|G_1|,|G_2|,\cdots ,|G_s\left|\right).$$

For convenience, we call $\alpha$ the order vector of $\Gamma$. It is clear that

$$|G_i\left|\right||G_{i+1}|\mathrm{for}i=1,2,\cdots ,s-1,\mathrm{and}|G_s|=p^n.$$

Now, set

$$\Lambda =\{\alpha \mid \alpha \mathrm{is}\mathrm{the}\mathrm{order}\mathrm{vector}\mathrm{of}\Gamma ,\Gamma \in {\mathcal{H}}_s\left(G\right)\}$$

and for any $\alpha \in \Lambda$,

$${\mathcal{H}}_s^{\alpha }\left(G\right)=\{\Gamma \in {\mathcal{H}}_s\left(G\right)\mid \mathrm{the}\mathrm{order}\mathrm{vector}\mathrm{of}\Gamma \mathrm{is}\alpha \}.$$

Subsequently, it is clear that

$${\mathcal{H}}_s\left(G\right)=\bigcup _{\alpha \in \Lambda }{\mathcal{H}}_s^{\alpha }\left(G\right),$$

and

$${\mathcal{H}}_s^{\alpha }\left(G\right)\cap {\mathcal{H}}_s^{\beta }\left(G\right)=\varnothing \mathrm{if}\alpha \ne \beta .$$

Thus

$$H_s\left(G\right)=\sum _{\alpha \in \Lambda }{H}_s^{\alpha }\left(G\right).$$

It is easy to see that

$$\Lambda =\{\alpha =(p^{i_1},p^{i_2},\cdots ,p^{i_{s-1}},p^n)\mid 1\le i_1<i_2<\cdots <i_{s-1}\le n-1\}.$$

Notice that the subgroups of an elementary abelian p-group are still elementary abelian p-groups, we see, by Lemma 4, that

$$H_s\left(G\right)=\sum _{1\le i_1<i_2<\cdots <i_{s-1}\le n-1}{\left[\begin{array}{c}n\\ i_{s-1}\end{array}\right]}_p{\left[\begin{array}{c}{i}_{s-1}\\ i_{s-2}\end{array}\right]}_p\cdots {\left[\begin{array}{c}{i}_2\\ i_1\end{array}\right]}_p,2\le s\le n.$$

The lemma is proved. □

Now, by Theorem 2, Lemmas 5 and 6, we can obtain a formula for counting the number of distinct fuzzy subgroups of the abelian groups, whose Sylow subgroups are cyclic groups or elementary abelian groups.

Theorem 3.

Let $s,m,i,j,{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_m$ be positive integers with $m\ge 2$, and let $p_1,p_2,\cdots ,p_m$ be different primes. If $G=P_1\times P_2\times \cdots \times P_m$ is an abelian group with $P_t$ the Sylow $p_t$-subgroup of G and $\mid P_t\mid =p_t^{{\alpha }_t}$ for $t=1,2,\cdots m$. Assume that $P_1,P_2,\cdots ,P_m$ are cyclic groups or elementary abelian groups. Then the number of distinct fuzzy subgroups of G is

$$F\left(G\right)=2\sum _{s=1}^{{\alpha }_1+{\alpha }_2+\cdots +{\alpha }_m}{H}_s\left(G\right),$$

where $H_s\left(G\right)$ satisfy the following recursive formula

$$H_s(P_1\times P_2\times \cdots \times P_k)=\sum _{\begin{array}{c}max\{i,j\}\le s\le i+j\end{array}}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{i+j-s}}\right)H_i(P_1\times P_2\times \cdots \times P_{k-1})H_j\left(P_k\right)$$

for any $k=2,\cdots ,m$, and for any $k=1,2,\cdots ,m$, if $H_j\left(P_k\right)$ is a cyclic group of order $p_k^{{\alpha }_k}$, then

$$H_j\left(P_k\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{{\alpha }_k-1}{j-1}}\right).$$

If $H_j\left(P_k\right)$ is an elementary abelian group of order $p_k^{{\alpha }_k}$, then

$$H_j\left(P_k\right)=\left\{\begin{array}{cc}1,\hfill & j=1;\hfill \\ {\displaystyle \sum _{1\le i_1<i_2<\cdots <i_{j-1}\le {\alpha }_k-1}}{\left[\begin{array}{c}{\alpha }_k\\ i_{j-1}\end{array}\right]}_{p_k}{\left[\begin{array}{c}{i}_{j-1}\\ i_{j-2}\end{array}\right]}_{p_k}\cdots {\left[\begin{array}{c}{i}_2\\ i_1\end{array}\right]}_{p_k},\hfill & j\ge 2.\hfill \end{array}\right.$$

5. Examples

In this section, we provide two examples to illustrate our results.

Example 1.

Let $p,q,r$ be different primes. Let $G={\mathbb{Z}}_{p^3}\times {\mathbb{Z}}_q^2\times {\mathbb{Z}}_{r^2}$. Then the number $F\left(G\right)$ of distinct fuzzy subgroups of G is

$$\begin{array}{cc}\hfill F\left(G\right)& =5136(q+1)+1208.\hfill \end{array}$$

Proof. 

According to Theorem 3, it is easy to obtain

$H_1\left({\mathbb{Z}}_{p^3}\right)=1,H_2\left({\mathbb{Z}}_{p^3}\right)=2,H_3\left({\mathbb{Z}}_{p^3}\right)=1;$

$H_1\left({\mathbb{Z}}_q^2\right)=1,H_2\left({\mathbb{Z}}_q^2\right)=q+1;$

$H_1\left({\mathbb{Z}}_{r^2}\right)=1,H_2\left({\mathbb{Z}}_{r^2}\right)=1.$

For convenience we define $A={\mathbb{Z}}_{p^3}\times {\mathbb{Z}}_q^2$. Then

$H_1\left(A\right)=1$;

$H_2\left(A\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left({\mathbb{Z}}_{p^3}\right)·H_1\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{1}}\right)·H_1\left({\mathbb{Z}}_{p^3}\right)·H_2\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left({\mathbb{Z}}_{p^3}\right)·H_1\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·H_2\left({\mathbb{Z}}_{p^3}\right)·H_2\left({\mathbb{Z}}_q^2\right)=4(q+1)+6$;

$H_3\left(A\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left({\mathbb{Z}}_{p^3}\right)·H_2\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left({\mathbb{Z}}_{p^3}\right)·H_1\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left({\mathbb{Z}}_{p^3}\right)·H_2\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)·H_3\left({\mathbb{Z}}_{p^3}\right)·H_1\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)·H_3\left({\mathbb{Z}}_{p^3}\right)·H_2\left({\mathbb{Z}}_q^2\right)=18(q+1)+9$;

$H_4\left(A\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left({\mathbb{Z}}_{p^3}\right)·H_2\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{0}}\right)·H_3\left({\mathbb{Z}}_{p^3}\right)·H_1\left({\mathbb{Z}}_q^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)·H_3\left({\mathbb{Z}}_{p^3}\right)·H_2\left({\mathbb{Z}}_q^2\right)=24(q+1)+4$;

$H_5\left(A\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{5}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{0}}\right)·H_3\left({\mathbb{Z}}_{p^3}\right)·H_2\left({\mathbb{Z}}_q^2\right)=10(q+1)$.

Afterwards,

$H_1\left(G\right)=1$;

$H_2\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{1}}\right)·H_1\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·H_2\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)=12(q+1)+22$;

$H_3\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)·H_3\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)·H_3\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)=144(q+1)+111$;

$H_4\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{0}}\right)·H_3\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)·H_3\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)·H_4\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)·H_4\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)=552(q+1)+220$;

$H_5\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{5}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{0}}\right)·H_3\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{5}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{0}}\right)·H_4\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{5}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)·H_4\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{5}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{1}}\right)·H_5\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{5}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{2}}\right)·H_5\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)=930(q+1)+190$;

$H_6\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{6}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{0}}\right)·H_4\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{6}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{0}}\right)·H_5\left(A\right)·H_1\left({\mathbb{Z}}_{r^2}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{6}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{1}}\right)·H_5\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)=720(q+1)+60$;

$H_7\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{7}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{0}}\right)·H_5\left(A\right)·H_2\left({\mathbb{Z}}_{r^2}\right)=210(q+1)$.

Then

$$\begin{array}{cc}\hfill F\left(G\right)& =2\left(H_1\left(G\right)+H_2\left(G\right)+H_3\left(G\right)+H_4\left(G\right)+H_5\left(G\right)+H_6\left(G\right)+H_7\left(G\right)\right)\hfill \\ \hfill & =5136(q+1)+1208.\hfill \end{array}$$

□

Example 2.

Let $p,q,r$ be different primes. Let $G={\mathbb{Z}}_p^2\times {\mathbb{Z}}_q^3\times {\mathbb{Z}}_r^2$. Then the number $F\left(G\right)$ of distinct fuzzy subgroups of G is

$$\begin{array}{cc}\hfill F\left(G\right)& =2354(p+1)(q^2+q+1)(q+1)(r+1)+1636(p+1)(q^2+q+1)(r+1)\hfill \\ \hfill & +478(p+1)(q^2+q+1)(q+1)+478(q^2+q+1)(q+1)(r+1)+404(p+1)(q^2+q+1)\hfill \\ \hfill & +404(q^2+q+1)(r+1)+114(q^2+q+1)(q+1)+202(p+1)(r+1)+124(q^2+q+1)\hfill \\ \hfill & +62(p+1)+62(r+1)+26.\hfill \end{array}$$

Proof. 

According to Theorem 3, it is easy to obtain

• $H_1\left({\mathbb{Z}}_p^2\right)=1,H_2\left({\mathbb{Z}}_p^2\right)=p+1;$
• $H_1\left({\mathbb{Z}}_q^3\right)=1,H_2\left({\mathbb{Z}}_q^3\right)=2(q^2+q+1),H_3\left({\mathbb{Z}}_q^3\right)=(q^2+q+1)(q+1);$
• $H_1\left({\mathbb{Z}}_r^2\right)=1,H_2\left({\mathbb{Z}}_r^2\right)=r+1.$

For convenience we define $A={\mathbb{Z}}_p^2\times {\mathbb{Z}}_q^3$. Then

$H_1\left(A\right)=1$;

$H_2\left(A\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left({\mathbb{Z}}_p^2\right)·H_1\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{1}}\right)·H_1\left({\mathbb{Z}}_p^2\right)·H_2\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left({\mathbb{Z}}_p^2\right)·H_1\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·H_2\left({\mathbb{Z}}_p^2\right)·H_2\left({\mathbb{Z}}_q^3\right)=2(p+1)(q^2+q+1)+4(q^2+q+1)+2(p+1)+2$;

$H_3\left(A\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left({\mathbb{Z}}_p^2\right)·H_2\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{1}}\right)·H_1\left({\mathbb{Z}}_p^2\right)·H_3\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left({\mathbb{Z}}_p^2\right)·H_1\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left({\mathbb{Z}}_p^2\right)·H_2\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·H_2\left({\mathbb{Z}}_p^2\right)·H_3\left({\mathbb{Z}}_q^3\right)=3(p+1)(q^2+q+1)(q+1)+12(p+1)(q^2+q+1)+3(q^2+q+1)(q+1)+6(q^2+q+1)+3(p+1)$;

$H_4\left(A\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left({\mathbb{Z}}_p^2\right)·H_3\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left({\mathbb{Z}}_p^2\right)·H_2\left({\mathbb{Z}}_q^3\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left({\mathbb{Z}}_p^2\right)·H_3\left({\mathbb{Z}}_q^3\right)=12(p+1)(q^2+q+1)(q+1)+12(p+1)(q^2+q+1)+4(q^2+q+1)(q+1)$;

$H_5\left(A\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{5}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left({\mathbb{Z}}_p^2\right)·H_3\left({\mathbb{Z}}_q^3\right)=10(p+1)(q^2+q+1)(q+1)$.

Subsequently,

$H_1\left(G\right)=1$;

$H_2\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{1}}\right)·H_1\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·H_2\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)=2(p+1)(q^2+q+1)(r+1)+4(p+1)(q^2+q+1)+4(q^2+q+1)(r+1)+2(p+1)(r+1)+8(q^2+q+1)+4(p+1)+4(r+1)+6$;

$H_3\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)·H_1\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·H_2\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)·H_3\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)·H_3\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)=9(p+1)(q^2+q+1)(q+1)(r+1)+48(p+1)(q^2+q+1)(r+1)+9(p+1)(q^2+q+1)(q+1)+9(q^2+q+1)(q+1)(r+1)+42(p+1)(q^2+q+1)+42(q^2+q+1)(r+1)+9(q^2+q+1)(q+1)+21(p+1)(r+1)+30(q^2+q+1)+15(p+1)+15(r+1)+6$;

$H_4\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{0}}\right)·H_2\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{0}}\right)·H_3\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right)·H_3\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)·H_4\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{4}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)·H_4\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)=108(p+1)(q^2+q+1)(q+1)(r+1)+228(p+1)(q^2+q+1)(r+1)+60(p+1)(q^2+q+1)(q+1)+60(q^2+q+1)(q+1)(r+1)+96(p+1)(q^2+q+1)+96(q^2+q+1)(r+1)+28(q^2+q+1)(q+1)+48(p+1)(r+1)+24(q^2+q+1)+12(p+1)+12(r+1)$;

$H_5\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{5}{3}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3}{0}}\right)·H_3\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{5}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{0}}\right)·H_4\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{5}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)·H_4\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{5}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{1}}\right)·H_5\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{5}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{2}}\right)·H_5\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)=370(p+1)(q^2+q+1)(q+1)(r+1)+360(p+1)(q^2+q+1)(r+1)+110(p+1)(q^2+q+1)(q+1)+110(q^2+q+1)(q+1)(r+1)+60(p+1)(q^2+q+1)+60(q^2+q+1)(r+1)+20(q^2+q+1)(q+1)+30(p+1)(r+1)$;

$H_6\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{6}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{0}}\right)·H_4\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{6}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{0}}\right)·H_5\left(A\right)·H_1\left({\mathbb{Z}}_r^2\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{6}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{1}}\right)·H_5\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)=480(p+1)(q^2+q+1)(q+1)(r+1)+180(p+1)(q^2+q+1)(r+1)+60(p+1)(q^2+q+1)(q+1)+60(q^2+q+1)(q+1)(r+1)$;

$H_7\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{7}{5}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{5}{0}}\right)·H_5\left(A\right)·H_2\left({\mathbb{Z}}_r^2\right)=210(p+1)(q^2+q+1)(q+1)(r+1)$.

Afterwards,

$$\begin{array}{cc}\hfill F\left(G\right)& =2\left(H_1\left(G\right)+H_2\left(G\right)+H_3\left(G\right)+H_4\left(G\right)+H_5\left(G\right)+H_6\left(G\right)+H_7\left(G\right)\right)\hfill \\ \hfill & =2354(p+1)(q^2+q+1)(q+1)(r+1)+1636(p+1)(q^2+q+1)(r+1)\hfill \\ \hfill & +478(p+1)(q^2+q+1)(q+1)+478(q^2+q+1)(q+1)(r+1)+404(p+1)(q^2+q+1)\hfill \\ \hfill & +404(q^2+q+1)(r+1)+114(q^2+q+1)(q+1)+202(p+1)(r+1)+124(q^2+q+1)\hfill \\ \hfill & +62(p+1)+62(r+1)+26.\hfill \end{array}$$

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