P. E. Dyubyuk

ON THE NUMBER OF SUBGROUPS OF A FINITE ABELIAN GROUP

(Presented by Academician A. I. Mal’tsev on 24 X 1960)

MATHEMATICS

§ 1. In the present paper we consider the question of the number of subgroups of a given order of the various types of finite abelian \(p\)-groups.

The paper uses the following definitions, previously introduced by the author \((^1)\).

Let \(p\) be a prime number, and let \(\alpha\) and \(\beta\) be integers. The symbol \(P_{\alpha,\beta}\) is defined as follows:
\(P_{\alpha,\beta}=(p^\alpha-1)(p^{\alpha-1}-1)\cdots(p^{\beta+1}-1)\), if \(0\leqslant \beta<\alpha\); \(P_{\alpha,\beta}=1\), if \(\beta=\alpha\) or \(\alpha\leqslant \beta=0\); \(P_{\alpha,\beta}=0\) in all other cases, i.e., if \(0>\beta\ne\alpha\) or \(\alpha<\beta\ne0\).

Put

\[ \varphi_{\alpha,\rho}=\frac{P_{\alpha,\alpha-\rho}}{P_{\rho,0}} =\frac{P_{\alpha,\rho}}{P_{\alpha-\rho,0}}. \]

It follows from this definition that, if the condition \(\alpha\geqslant \rho\geqslant 0\) is not satisfied, then \(\varphi_{\alpha,\rho}=0\) for \(\alpha\ne\rho\) and \(\rho\ne0\); \(\varphi_{\alpha,\rho}=1\) for \(\rho=0\) or \(\alpha=\rho\).

Let \(d_1,d_2,\ldots,d_l\) be nonnegative integers, and suppose that for every \(i=1,2,\ldots,l-1\) we have \(d_i\geqslant d_{i+1}\); let \(a_1,a_2,\ldots,a_l\) be arbitrary integers. Define the symbol

\[ \left\{ \begin{array}{c} d_1,d_2,\ldots,d_l\\ a_1,a_2,\ldots,a_l \end{array} \right\} = \left\{ \begin{array}{c} d_i\\ a_i \end{array} \right\}_{i=1}^{l} \]

as follows:

\[ \left\{ \begin{array}{c} d_i\\ a_i \end{array} \right\}_{i=1}^{l} = p^{\sum\limits_{i=1}^{l}(d_i-a_i)a_{i+1}} \prod_{i=1}^{l} \varphi_{d_i-a_{i+1},\,a_i-a_{i+1}}, \qquad a_{l+1}=0. \]

Let \(P\) be an arbitrary abelian \(p\)-group, \(d\) its rank, and \(p^k\) the greatest of the orders of the elements of the group \(P\). We shall agree to denote by \(d_i\) the number of elements of a minimal basis of the group \(P\) whose orders are not less than \(p^i\). Thus, obviously, \(d_1=d\), \(d_{k+1}=0\), \(d_{k+2}=0\), etc. The numbers \(d_i\) \((i=1,2,\ldots)\) will be called the invariants of the group \(P\), and we shall say, with respect to the group \(P\) itself, that it is of type \(\{d_1,d_2,\ldots,d_k\}\).

In deriving the theorems formulated below, the following propositions, proved by the author in \((^1)\), are used.

Lemma A. If \(a,b,c\) are nonnegative integers, then the identity

\[ \sum_{i=0}^{a} p^{i(b-c+i)}\varphi_{b,c-i}\varphi_{a,i} = \varphi_{a+b,c}. \]

Theorem A. Let \(P\) be an abelian \(p\)-group of type \(\{d_1,d_2,\ldots,d_k\}\). Let \(a_1,a_2,\ldots,a_k\) be integers. The number of subgroups of type \(\{a_1,a_2,\ldots,a_k\}\) of the group \(P\) is equal to

\[ \left\{ \begin{matrix} d_i\\ a_i \end{matrix} \right\}_{i=1}. \]

§ 2.

A direct application of Theorem A makes it possible to prove the following proposition:

Theorem 1. Let \(P\) be an abelian group of order \(p^n\), of rank \(2\), and of type \(\{d_1,d_2,\ldots,d_m\}\). If \(d_l=2,\ d_{l+1}<2\ (l\leq m)\), then the number of subgroups of order \(p^\alpha\)

\[ \left(\alpha \leq \left[\frac{n+1}{2}\right]\right) \]

of the group \(P\) is equal to \(\varphi_{\sigma+1,1}\), where \(\sigma\) is the smaller of the numbers \(\alpha\) and \(l\).

Further, using Lemma A, Theorem A, and applying induction on \(s\), one can justify the following result:

Theorem 2. Let \(P\) be an abelian group of type \(\{d_1,d_2,\ldots,d_s,d_{s+1},\ldots,d_r\}\), with \(d_1=d_2=\cdots=d_s>d_{s+1}\). The number of subgroups of order \(p^\alpha\) \((0\leq \alpha\leq s)\) of the group \(P\) is equal to \(\varphi_{d_1+\alpha-1,\alpha}\).

Applying the same apparatus, as well as Theorem 2, we prove the following theorem:

Theorem 3. Let \(P\) be an abelian group of order \(p^n\), of type \(\{d_1,d_2,\ldots,\ldots,d_{s-1},d_s\}\), with \(d_1=d_2=\cdots=d_{s-1}\) (so that \(n=(s-1)d_1+d_s\)). The number of subgroups of order \(p^\alpha\) \((0\leq \alpha\leq n)\) of the group \(P\) is congruent to \(\varphi_{\alpha+d_1-1,\alpha}\) modulo \(p^{\,n-\alpha+1}\).

In Theorem 10 of the work (1) an estimate is given for the number of subgroups of order \(p^\alpha\) of an arbitrary abelian group. This estimate is given by means of the function \(\varphi_{n,\alpha}\). It is interesting to note that the estimate of the number of subgroups of order \(p^\alpha\) of the group \(P\) (characterized in the hypothesis of Theorem 3) cannot be improved by introducing any third function of the form \(\varphi_{l,\alpha}\). Indeed, the following proposition holds:

Theorem 4. Let \(P\) be the group defined in the hypothesis of Theorem 3, and let \(N_\alpha\) be the number of subgroups of order \(p^\alpha\) of the group \(P\) \((0\leq \alpha\leq n)\). If \(d_1\ne n-\alpha+1\) and \(\sigma\) is greater than each of the numbers \(d_1\) and \(n-\alpha+1\), then for no \(l\) can the congruence \(N_\alpha\equiv\varphi_{l,\alpha}\pmod {p^\sigma}\) hold.

Relying on Theorem 3, one can prove the following theorems:

Theorem 5. Let \(P\) be an abelian group of order \(p^n\), of type \(\{d_1,d_2,d_3\}\), so that \(n=d_1+d_2+d_3\). The number of subgroups of order \(p^\alpha\) \((0\leq \alpha\leq n)\) of the group \(P\) is congruent to \(\varphi_{d_1+\alpha-1,\alpha}\) modulo \(p^\sigma\), where \(\sigma\) is the smaller of the numbers \(d_1+d_2-1\) and \(n-\alpha+1\).

Theorem 6. Let \(P\) be an abelian group of order \(p^n\), of type \(\{d_1,d_2,d_3,d_4\}\), so that \(n=2d_1+d_3+d_4\). The number of subgroups of order \(p^\alpha\) \((0\leq \alpha\leq n)\) of the group \(P\) is congruent to \(\varphi_{d_1+\alpha-1,\alpha}\) modulo \(p^\sigma\), where \(\sigma\) is the smaller of the numbers \(2d_1+d_3-2\) and \(n-\alpha+1\).

Theorem 7. Let \(P\) be an abelian group of order \(p^n\), of type \(\{d_1,d_2,d_3,d_4\}\), so that \(n=d_1+d_2+d_3+d_4\). The number of subgroups of order \(p^\alpha\) \((0\leq \alpha\leq d_1+2)\) of the group \(P\) is congruent to \(\varphi_{d_1+\alpha-1,\alpha}\) modulo \(p^\sigma\), where \(\sigma\) is the smaller of the numbers \(d_1+d_2-1\) and \(n-\alpha+1\).

It should be noted that in proving the last theorems one has to overcome great and ever-increasing technical difficulties, so that the derivation of each of these propositions takes many pages.

On the other hand, it is clear that one can formulate a proposition of a more general character, pertaining to any abelian group, although in deriving it one would have to overcome very great difficulties. This is the following conjectured result:

Let \(P\) be an abelian group of order \(p^n\) and of type \(\{d_1,d_2,\ldots,d_k,d_{k+1},\ldots,d_s\}\), with

\[ d_2=d_3=\cdots=d_k=d_1 \]

and

\[ d_{k+1}<d_1 \qquad \left(\text{so that } n=kd_1+\sum_{i=k+1}^{s} d_i\right). \]

\((1 \leq k \leq s)\). The number of subgroups of order \(p^\alpha\) \((0 \leq \alpha \leq n)\) of the group \(P\) is congruent to \(\varphi_{d_1+\alpha-1,\alpha}\) modulo \(p^\sigma\), where \(\sigma\) is the smaller of the numbers \(\displaystyle \sum_{i=1}^{k+1} d_i - k\) and \(n-\alpha+1\).

It is not difficult to see that Theorems 5, 6, and 7 are special cases of the proposition just stated. It is easy to show, using Theorem 2, that Theorem 3 is also a special case of this latter, far-reaching result.

Received
22 IX 1960

REFERENCES

1. P. E. Dyubyuk, Izv. AN SSSR, ser. matem., 12, No. 4, 351 (1948).