Finite non-cyclic $p$-groups whose number of subgroups is minimal

TL;DR

This paper characterizes non-cyclic finite p-groups with the minimal number of subgroups among all p-groups of the same order, complementing previous work on maximal subgroup counts.

Contribution

It provides a complete classification of non-cyclic finite p-groups with the minimal number of subgroups, filling a gap in subgroup enumeration.

Findings

Identifies all non-cyclic p-groups with minimal subgroup counts

Complements existing results on maximal subgroup counts

Provides explicit structural descriptions

Abstract

Recent results of Qu and Tuarnauceanu describe explicitly the finite p-groups which are not elementary abelian and have the property that the number of their subgroups is maximal among p-groups of a given order. We complement these results from the bottom level up by determining completely the non-cyclic finite p-groups whose number of subgroups among p-groups of a given order is minimal.

Equations

$$M_p(1,1,1):=⟨a,b,c\mid a^p=b^p=c^p=1,[a,b]=c,[c,a]=[c,b]=1⟩.$$

$$\prod _{i⩾1}{p}^{{\nu }_{i+1}^{\prime }({\mu }_i^{\prime }-{\nu }_i^{\prime })}{\left[\genfrac{}{}{0ex}{}{{\mu }_i^{\prime }-{\nu }_{i+1}^{\prime }}{{\nu }_i^{\prime }-{\nu }_{i+1}^{\prime }}\right]}_p,$$

$${\left[\genfrac{}{}{0ex}{}{n}{k}\right]}_p=\prod _{i=0}^{k-1}\frac{1-p^{n-i}}{1-p^{i+1}}$$

$$r:={\left[\genfrac{}{}{0ex}{}{{\mu }_1^{\prime }}{1}\right]}_p\cdot \prod _{i⩾2}{p}^{{\nu }_{i+1}^{\prime }({\mu }_i^{\prime }-{\nu }_i^{\prime })}{\left[\genfrac{}{}{0ex}{}{{\mu }_i^{\prime }-{\nu }_{i+1}^{\prime }}{{\nu }_i^{\prime }-{\nu }_{i+1}^{\prime }}\right]}_p,$$

$${\left[\genfrac{}{}{0ex}{}{{\mu }_1^{\prime }}{1}\right]}_p⩾{\left[\genfrac{}{}{0ex}{}{2}{1}\right]}_p=p+1.$$

$${\mu }^{\prime }=(\underset{{\mu }_2}{\underbrace{2,2,\dots ,2}},\underset{{\mu }_1-{\mu }_2}{\underbrace{1,1,\dots ,1}})$$

$${\nu }^{\prime }=(2,\underset{{\mu }_1-1}{\underbrace{1,\dots ,1}}).$$

$$G\cong M_{p^{\lambda }}:=⟨a,b\mid a^{p^{\lambda -1}}=b^p=1,a^b=a^{1+p^{\lambda -2}}⟩.$$

$$|{\mathfrak{s}}_k(G)|\equiv p+1(\mathrm{mod}{p}^2)$$

$$|\mathfrak{s}(G)|=\sum _{k=0}^{\lambda }|{\mathfrak{s}}_k(G)|=2+\sum _{k=1}^{\lambda -1}|{\mathfrak{s}}_k(G)|⩾2+(\lambda -1)(p+1).$$

$$Q_{2^{\lambda }}:=⟨x,y\mid x^{2^{\lambda -1}}=1,y^2=x^{2^{\lambda -2}},x^y=x^{-1}⟩.$$

$$|{\mathfrak{s}}_{\mu }(G)|\equiv 1(\mathrm{mod}2),$$

$$|\mathfrak{s}(Q_{2^{\lambda }})|=|\mathfrak{s}(D_{2^{\lambda -1}})|+1=\tau (2^{\lambda -2})+\sigma (2^{\lambda -2})+1=2^{\lambda -1}+\lambda -1,$$

Full text

1. Introduction

Let

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Write $\mathfrak{s}_{k}(G)$ for the set of subgroups of order $p^{k}$ of the $p$-group $G$ and $\mathfrak{s}(G)$ for the set of all subgroups of $G$. Recently, Qu obtained the following result.

Theorem 1.1** ([9, Thm. 1.4]).**

Let $G$ be a group of order $p^{n}$, where $p$ is an odd prime, and $\widetilde{G}=M_{p}(1,1,1)\times C_{p}^{n-3}$. If G is not elementary abelian, then for all $k$ such that $1\leqslant k\leqslant n$, we have $\left\lvert\mathfrak{s}_{k}(G)\right\rvert\leqslant\lvert\mathfrak{s}_{k}(\widetilde{G})\rvert$. In particular, if $2\leqslant k\leqslant n-2$, then $\lvert\mathfrak{s}_{k}(G)\rvert<\lvert\mathfrak{s}_{k}(\widetilde{G})\rvert$.

Thus Qu finds the $p$-groups whose number of subgroups of possible order is maximal except for elementary abelian $p$-groups when $p>2$. The analogue of Qu’s result for $p=2$ was recently obtained by Tărnăuceanu (cf. [10]). We wish to complement Qu’s and Tărnăuceanu’s results from the bottom level up. In particular, in Theorems A and B we determine completely the structure of those finite non-cyclic $p$-groups whose number of subgroups is minimal.

2. Finite $p$-groups realising the second minimal level of subgroup numbers

We begin with the following preparatory lemma.

Lemma 2.1**.**

Suppose that $G$ is an abelian $p$-group of order $p^{\lambda}$, and that $G$ has $p+1$ subgroups of order $p^{r}$ for all $r$ such that $1\leqslant r\leqslant\lambda-1$. Then $G\cong C_{p^{\lambda-1}}\times C_{p}$.

**Proof. ** By a well-known result, the number of subgroups of type $\nu=(\nu_{1},\ldots,\nu_{\ell})$ in an abelian $p$-group of type $\mu=(\mu_{1},\ldots,\mu_{m})$ is given by the formula

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where $\mu^{\prime},\nu^{\prime}$ are the conjugates of the partitions $\mu$ and $\nu$, respectively, and

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is the number of $k$-dimensional subspaces of an $n$-dimensional vector space over the field $\mathbb{Z}/p\mathbb{Z}$; see, for instance, [4, Eqn. (1)]. Since $G$ is non-cyclic of type $\mu=(\mu_{1},\ldots,\mu_{m})$ say, it follows that $m\geqslant 2$ (and $\mu_{2}>0$), thus $\mu_{1}^{\prime}\geqslant 2$. Now, let us the count the number of subgroups of type $\nu=(1,0,\ldots,0)$. Observe that $\nu^{\prime}=(1,0,\ldots,0)$, so according to formula (2.1) there are

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such subgroups. However, $r\leqslant p+1$ since by assumption the number of subgroups of order $p$ in $G$ is $p+1$, and $r\geqslant p+1$, owing to

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Thus $\mu_{1}^{\prime}=2$, which implies that $m=2$ and that $G$ is of type $\mu=(\mu_{1},\mu_{2})$. It follows that

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Suppose that $\mu_{2}\geqslant 2$. Next, we count subgroups of type $\nu=(\mu_{1}-1,1)$ in $G$. Note that

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Moreover, the second term in the product formula for $\nu=(\mu_{1}-1,1)$ evaluates to $p$, thus the number of subgroups of type $\nu=(\mu_{1}-1,1)$ in $G$ is at least $p$. Further, the $(\mu_{2}-1)$-term in the product formula for $\nu=(\mu_{1},0)$ evaluates to $p$ as well, and thus we deduce that the number of subgroups of type $\nu=(\mu_{1},0)$ in $G$ is, again, at least $p$. Therefore, we get at least $2p>p+1$ subgroups of order $p^{\mu_{1}}$ in $G$, contrary to our assumption that the number of subgroups of order $p^{\mu_{1}}$ is $p+1$. This contradiction shows that $\mu_{2}=1$ and $\mu_{1}=\lambda-1$, so $G=C_{p^{\lambda-1}}\times C_{p}$ is of the type we asserted. $\blacksquare$

We shall present our main result as two separate theorems, dealing with the cases $p>2$ and $p=2$ respectively, since the $p=2$ case, although ultimately similar to the $p>2$ case, presents a somewhat erratic behaviour at small values.

Theorem A**.**

Let $p$ be an odd prime, $G$ a $p$-group of order $p^{\lambda}$. If $G$ is not the cyclic group $C_{p^{\lambda}}$, then $\lvert\mathfrak{s}(G)\rvert\geqslant(p+1)(\lambda-1)+2$, with equality if and only if $G\cong C_{p^{\lambda-1}}\times C_{p}$ or

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**Proof. ** Given a non-cyclic $p$-group $G$, $p$ odd, a well-known theorem due to Kulakoff [7, Satz 1] asserts that

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for all $k$ such that $1\leqslant k\leqslant\lambda-1$. Thus, in particular, $\lvert\mathfrak{s}_{k}(G)\rvert\geqslant p+1$, and therefore

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This proves the first part of our assertion.

Now, we assume (as we may) that $\lambda\geqslant 3$. Applying a result of Lindenberg [8, Folgerung 3.4], we have that if $G=C_{p^{\lambda-1}}\rtimes C_{p}$ (the implied action of $C_{p}$ on $C_{p^{\lambda-1}}$ may well be trivial; we only require that $G$ be a split extension) then $G$ has, apart from the trivial subgroup and the whole group $G$, $p+1$ subgroups of order $p^{j}$, $1\leqslant j\leqslant\lambda-1$, and thus $\lvert\mathfrak{s}(C_{p^{\lambda-1}}\rtimes C_{p})\rvert=(p+1)(\lambda-1)+2$ for any such split extension.

Our goal now is to establish that if a (necessarily non-cyclic) $p$-group $G$ of order $|G|=p^{\lambda}$ has $p+1$ subgroups of order $p^{j}$, for all $j$ such that $1\leqslant j\leqslant\lambda-1$, then $G$ is a split extension $C_{p^{\lambda-1}}\rtimes C_{p}$.

If $G$ is abelian, then the claim follows from Lemma 2.1. Now, assume that $G$ is non-abelian. Since $G$ is a $p$-group, $p>2$, this implies that $G$ is not a Dedekind group. Of the $p+1$ subgroups of order $p^{j}$, for each $j$ such that $1\leqslant j\leqslant\lambda-1$, one is certainly normal, since $p$-groups have normal subgroups of each possible order. The other $p$ are either all normal, or lie in the same conjugacy class. It follows that the non-normal subgroups of $G$ for each possible order are all conjugate, and thus $G$ is a CO-group; cf. top of [2, Sec. 58]. Janko’s theorem (see [2, Thm. 58.3]) now yields $G\cong M_{p^{\lambda}}$, where $M_{p^{\lambda}}$ was defined in (2.3). Our proof is complete. $\blacksquare$

Remark 2.2**.**

We note here that there may, in principle, exist many non-isomorphic semidirect products $C_{p^{\lambda-1}}\rtimes C_{p}$ which are not direct products. However, an early theorem due to Burnside asserts that the only non-abelian $p$-group, $p$ odd, which has a cyclic maximal subgroup is $M_{p^{\lambda}}$; see [3, Chap. VIII, Sec. 109].

Recall that the generalised quaternion group of order $2^{\lambda}$ is the group defined by the presentation

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Next, we address the $p=2$ case.

Theorem B**.**

Let $G$ be a $2$-group of order $2^{\lambda}$. If $G$ is not the cyclic group $C_{2^{\lambda}}$ and $\lambda\geqslant 5$, then $\lvert\mathfrak{s}(G)\rvert\geqslant 3\lambda-1$, with equality if and only if $G\cong C_{2^{\lambda-1}}\times C_{2}$ or $G\cong M_{2^{\lambda}}$, where $M_{2^{\lambda}}$ is as in (2.3) with $p=2$. If $\lambda=3$ and $G$ is not $C_{8}$, then $\lvert\mathfrak{s}(G)\rvert\geqslant 6$ with equality if and only if $G\cong Q_{8}$, while if $\lambda=4$ and $G$ is not $C_{16}$, then $\lvert\mathfrak{s}(G)\rvert\geqslant 11$, with equality if and only if $G\cong Q_{16}$, or $G\cong M_{16}$, or $G\cong C_{8}\times C_{2}$.

**Proof. ** There are 5 groups of order $8=2^{3}$, and 14 groups of order $16=2^{4}$. We use GAP [6] to obtain a full list of the isomorphism classes of groups in each case, and ask GAP for the total number of subgroups of each group in the list. Our claim for $\lambda=3$ and $\lambda=4$ is now a simple matter of inspection.

We may thus assume that $\lambda\geqslant 5$. By Frobenius’ generalisation of Sylow’s theorem, we have

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for all $\mu$ such that $0\leqslant\mu\leqslant\lambda$.

Moreover, since $G\ncong C_{2^{\lambda}}$, we have $\lvert\mathfrak{s}_{\mu}(G)\rvert\neq 1$, $1\leqslant\mu\leqslant\lambda-1$ by [1, Prop. 1.3] unless $G\cong Q_{2^{\lambda}}$ and $\mu=1$. It is well-known that $Q_{2^{\lambda}}$ has a unique involution $t$, which generates its centre, and affords the quotient Q_{2^{\lambda}}\big{/}\langle t\rangle\cong D_{2^{\lambda-1}}. Since $\langle t\rangle$ is the unique subgroup of order 2 in $Q_{2^{\lambda}}$, it follows easily that

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where $\lvert\mathfrak{s}(D_{2m})\rvert=\tau(m)+\sigma(m)$ follows from [5, Ex. 1]. Moreover, notice that $\lvert\mathfrak{s}(Q_{2^{\lambda}})\rvert\geqslant 3\lambda-1$ for all $\lambda\geqslant 4$, with equality precisely when $\lambda=4$. Since $\lambda\geqslant 5$, it follows that $G\ncong Q_{2^{\lambda}}$. This shows that $\lvert\mathfrak{s}(G)\rvert\geqslant 3\lambda-1$ for all $\lambda\geqslant 5$, and by Lemma 2.1 the only abelian group of order $2^{\lambda}$ realising this bound is $C_{2^{\lambda-1}}\times C_{2}$.

It is well known that a 2-group (of order $2^{\lambda}$, $\lambda\geqslant 5$) is Dedekind non-abelian if and only if $G\cong Q_{8}\times C_{2}^{\lambda-3}$. Note here that every group of this type has $G_{0}\coloneqq Q_{8}\times(C_{2}\times C_{2})$ as a direct factor, and that $\lvert\mathfrak{s}_{2}(Q_{8}\times C_{2}^{2})\rvert=7$. Hence, a non-abelian group $G$ realising the bound $\lvert\mathfrak{s}(G)\rvert=3\lambda-1$ cannot be Dedekind. By Janko’s theorem mentioned previously, the only possible non-abelian group realising the bound $\lvert\mathfrak{s}(G)\rvert=3\lambda-1$ is $M_{2^{\lambda}}$. But Lindenberg’s result applies in the case of $M_{2^{\lambda}}$, since $1+2^{\lambda-2}\not\equiv-1\;(\bmod\;4)$. We conclude that the only groups of order $2^{\lambda}$, $\lambda\geqslant 5$, which realise the bound $\lvert\mathfrak{s}(G)\rvert=3\lambda-1$ are $C_{2^{\lambda-1}}\times C_{2}$ and $M_{2^{\lambda}}$. This completes our proof. $\blacksquare$