On the modular isomorphism problem for groups with center of index at most $p^3$

Abstract

Let p be an odd prime number. We show that the modular isomorphism problem has a positive answer for finite p-groups whose center has index $p^3$, which is a strong contrast to the analogous situation for $p = 2$.

1 Introduction

Let p be a prime number and let F be the field with p elements. In this paper, we study the modular isomorphism problem, which can be stated as follows:

Consider finite p-groups G and H such that the group algebras FG and FH are isomorphic. Is it true that G and H are isomorphic?

In the past 50 years, this problem has seen extensive research. There are many classes of finite p-groups for which it is known to have a positive answer, such as the class of abelian p-groups [5], the class of p-groups of nilpotency class 2 with elementary abelian derived subgroup [22], or the class of metacyclic groups [1, 23]. The modular isomorphism problem has a positive answer for groups with a center of index $p^2$. This follows from the earlier-mentioned result by Sandling [22]. Moreover, the generalization of the problem for this class of groups to arbitrary fields of characteristic p has a positive solution, as shown by [6]. The modular isomorphism problem is also known to have a positive answer for groups of order dividing $p^5$ for any prime [17, 20], for groups of order dividing $2^8$, for groups of order dividing $3^7$, and, except for a few groups, for groups of order dividing $5^6$ [7, 14]. Some recent results are for example [3, 15], and a complete overview can be found in the recent survey [12].

Despite all positive results, it was shown in [9] that the general answer to this problem is negative. The smallest counterexample consists of a pair of groups of order $2^9$ with centers of index 8. For $p > 2$, the modular isomorphism problem is still open.

For a p-group G, we write $\textrm{Z}(G)$ for its center, $\Phi (G)$ for its Frattini subgroup, and $(\gamma _i(G))_{i \ge 1}$ for its lower central series. Moreover we write $G^p=\left\langle g^p:g\in G \right\rangle$. Motivated by the counterexample of [9] for $p = 2$, we now study finite p-groups with a center of index $p^3$ for $p > 2$. In contrast to the situation for $p = 2$, it turns out that the modular isomorphism problem has a positive answer for this class of groups. Our main result is the following:

Theorem A

Let $p>2$ be a prime integer, let F be the field with p elements, and let G and H be finite p-groups. Suppose that $|G:\textrm{Z}(G)|=p^3$. If $FG\cong FH$, then $G\cong H$.

Note that the class of p-groups G with $|G:\textrm{Z}(G)| = p^3$ is not contained in the union of the classes of p-groups for which the modular isomorphism problem is known to have a positive answer. Corresponding examples are discussed in Section 4. The key ingredient for the proof of Theorem A is the following result, which might also be of independent interest:

Theorem B

Let $p>2$ be a prime integer, let F be the field with p elements, and consider finite p-groups G and H with $F G\cong F H$. Let $d \in {{\mathbb {Z}}}_{\ge 0}$ with $|G/\Phi (G)\textrm{Z}(G)|=p^d$. If

• (1) $G^p\cap \gamma _2(G)\subseteq \gamma _2(G)^p \gamma _3(G)$ and

• (2) $|\gamma _2(G)/\gamma _2(G)^p\gamma _3(G)|=p^{{d\atopwithdelims ()2}},$

then $G/\gamma _2(G)^p \gamma _4(G)\cong H/\gamma _2(H)^p \gamma _4(H)$.

This statement is a generalization of [14, Theorem 1.2] when $p>2$, as for $d=2$, it is equivalent to the following implication: if $G/\textrm{Z}(G )$ is 2-generated, then the isomorphism class of $G/\gamma _2(G)^p\gamma _4(G)$ is determined by FG (see Corollary 3.5).

This paper is organized in the following way: In Section 2, we introduce the notation used in this paper. In Section 3, we derive results on the so-called small group algebra, which will lead to the proof of Theorem B. Finally, in Section 4, we prove Theorem A.

2 Preliminaries and notation

Let p be an odd prime number and let G be a finite p-group. We use the standard group-theoretic notation. In particular, as above, Z(G) and $\Phi (G)$ denote the center and the Frattini subgroup of G, respectively. We set $G^p = \langle g^p :g \in G \rangle$. A Burnside basis is a generating system of G of minimal size. For $a,b \in G$, let $[a,b] = a^{-1} b^{-1} a b$. Moreover, we set $\gamma _1(G) = G$ and $\gamma _{i+1}(G) = [\gamma _i(G), G]$ for $i\ge 1$.

Lemma 2.1

Let G be a finite p-group of nilpotency class at most 3. Then for each $x,y\in G$,

$$\begin{aligned}{}[x^p,y] = [x,y]^p [[x,y],x]^{{p\atopwithdelims ()2}} \quad \text {and} \quad [x,y^p]=[x,y]^p [[x,y],y]^{{p\atopwithdelims ()2}}. \end{aligned}$$

Proof

Using several times the identity

$$\begin{aligned}{}[xz,y]=[x,y]^z [z,y]=[x,y][[x,y],z] [z,y] \qquad ( x,y,z\in G), \end{aligned}$$

we derive that

$$\begin{aligned}{}[x^p,y]&= [x,y]^{x^{p-1}}\cdot [x,y]^{x^{p-2}}\cdots [x,y] \\&=[x,y] [[x,y],x]^{p-1}\cdot [x,y]^{x^{p-2}}[[x,y],x]^{p-2}\cdots [x,y] \\&= [x,y]^p [[x,y],x]^{ {p \atopwithdelims ()2}}. \end{aligned}$$

A similar argument shows the second identity. $\square$

As an immediate consequence, we obtain:

Lemma 2.2

Let G be a finite p-group with $p>2$. If $\gamma _2(G)^p\gamma _4(G)=1$, then $G^p\subseteq \textrm{Z}(G)$.

Let F be a field of characteristic p. In this paper, we only consider unitary, finite-dimensional F-algebras. An augmented F-algebra is an F-algebra A endowed with an F-algebra homomorphism $A\rightarrow F$, called the augmentation map. The group algebra FG is an augmented F-algebra with augmentation map $\sum _{g\in G} a_gg\mapsto \sum _{g\in G} a_g$. The kernel I(A) of the augmentation map is called the augmentation ideal of A. For $A=FG$, we abbreviate I(FG) by I(G).

Let A be an augmented local F-algebra, as it is the case for $A=FG$. Then we denote $V(A)= 1+I(A)$, the subgroup of the group of units of A formed by the elements of augmentation 1. Every quotient of A by some ideal J of A is also an augmented algebra with an augmentation map induced by the augmentation of A. We can (and will) make the identifications $I(A/J)=I(A)/J$ and $V(A/J)=1+I(A/J)=V(A)/(1+J)$. We define the set K(A) to be the F-subspace of A generated by the elements of the form $ab-ba$ with $a,b\in A$.

Note that $I(G) I(\gamma _2(G))$ is a 2-sided ideal of FG. A main tool in this paper will be the investigation of the small group algebra $S = FG/I(G) I(\gamma _2(G))$. This object has been extensively studied by R. Sandling and M.A.M. Salim (see [18, 19, 22]). It was used in [20] to positively answer the modular isomorphism problem for groups of order dividing $p^5$. Moreover, the analogous construction for group rings over the integers was used in A. Whitcomb’s thesis [26] to positively answer the isomorphism problem for integral group rings for metabelian groups. Further results about the small group algebra in a wider sense can be found in [10]. By a bar, we denote both the natural projection ${\overline{\cdot }}:FG\rightarrow S$, which is an F-algebra homomorphism, and its restriction ${\overline{\cdot }}:V(FG)\rightarrow V(S)$, which can be viewed as a group homomorphism. A subgroup of V(FG) that is an F-basis for FG is called a group basis of FG. We say a property P is determined by the group algebra if for any pair of p-groups G and H such that $FG\cong FH$, the group G satisfies P if and only if H satisfies P.

We have that $(1+I(G)I(\gamma _2(G)))\cap G=\Phi (\gamma _2(G))$ (see the introduction of [19]), so $\overline{G} \cong G/\Phi (\gamma _2(G))$. Note that this holds for any group basis of FG. However, it has been known for a long time that the information encoded by the small group algebra alone is not enough to recover the isomorphism type of $G/\Phi (\gamma _2(G))$ (see [2, Section 4]), even for groups of nilpotency class 3 (see [15, Example 3.11] and the subsequent discussion).

We frequently use the following result due to Sandling:

Lemma 2.3

([22], see also [14, Theorem 3.2]). Let G be a finite p-group. Let $g_1, \ldots , g_m \in \overline{G}$ be elements whose images in $\overline{G}/\gamma _2(\overline{G})$ form an independent generating set as an abelian group. Let A be the subgroup of V(S) generated by the elements $1 + (g_1-1)^{k_1} \cdots (g_m-1)^{k_m}$ with $k_1, \ldots , k_m \in {{\mathbb {Z}}}_{\ge 0}$ and $k_1 + \dots + k_m \ge 2$. Then the following hold:

• (1) $V(S) = \overline{G}A$.

• (2) For $a = 1 + (g_1-1)^{k_1} \cdots (g_m-1)^{k_m} \in A$ and $g \in \overline{G}$, we have

  $$\begin{aligned} \begin{aligned} \left[ g,a\right] = [g, g_1, g_1, \dots , g_1, g_2, \dots , g_2, \dots , g_m, \dots , g_m],\end{aligned} \end{aligned}$$

  where each $g_i$ appears exactly $k_i$ times.

• (3) If additionally $\gamma _4(G) = 1$ holds, then A is abelian and $V(S) = \overline{G} \rtimes A$.

Furthermore, [19, Theorem 1.1] yields that

$$\begin{aligned} \gamma _i(V(S))=\gamma _i(\overline{G}) \end{aligned}$$

(2.1)

for each $i\ge 2$.

3 Results on the small group algebra

Throughout, let F denote the field with p elements for a prime number $p> 2$ and let G be a finite p-group. In this section, we derive results on the small group algebra that will later be used to solve the modular isomorphism problem for groups with a center of index $p^3$.

We use the notation introduced in the preceding section. In particular, let $\overline{\cdot } :FG\rightarrow S$ be the natural projection onto the small group algebra S and let $V(S) = 1 + I(G)/I(G) I(\gamma _2(G))$ denote the group of units of augmentation 1 of S. Furthermore, let A be a subgroup of V(S) as in Lemma 2.3, so $V(S)={\overline{G}} A$.

Remark 3.1

For every normal subgroup N of G contained in $\gamma _2(G)$, it follows easily from the identity $xy-1=x-1+y-1+(x-1)(y-1)$ that

$$\begin{aligned} I(N)FG+ I(\gamma _2(G))I(G)= N-1+I(\gamma _2(G))I(G). \end{aligned}$$

In particular, $\gamma _i({\overline{G}})-1 = \overline{\gamma _i(G)-1}$ is an ideal in S. In the following, we set $\Gamma = \gamma _4(G) -1 \subseteq FG$ and consider the quotient algebra $S_0 = FG/(I(G) I(\gamma _2(G)) + \Gamma ) \cong S/\overline{\Gamma }$ together with the corresponding projection $\pi _0:FG \rightarrow S_0$. Then $\pi _0(G)= G/\gamma _2(G)^p \gamma _4(G)$, and the same holds for every group basis of FG. Moreover, $S_0$ is isomorphic to the small group algebra of $\pi _0(G)$. Let $A_0$ be the image of A in $V(S_0)$. Then by Lemma 2.3, we have that $V(S_0)= \pi _0(G) \rtimes A_0$. More generally, the following holds:

Lemma 3.2

([15, Proposition 3.1]). Let H be any group basis of FG. Then $V(S_0)=\pi _0(H)\rtimes A_0$.

Finally, we consider the natural projection from $V(S_0)$ onto $V= V(S_0)/ (A_0\cap \textrm{Z}(V(S_0)))$. Composing it with the restriction of $\pi _0$ to V(FG) yields a group homomorphism

$$\begin{aligned} \pi :V(FG)\rightarrow V. \end{aligned}$$

(3.1)

Hence $V=\pi (G)\rtimes E$ follows, where $E=A_0/ (A_0\cap \textrm{Z}(V(S_0)))$ is elementary abelian due to $A_0^p\subseteq \textrm{Z}(V(S_0))$ (see (2.1) and Lemma 2.2). Therefore, by Lemma 3.2, we have

$$\begin{aligned} V = \pi (H) \rtimes E \end{aligned}$$

(3.2)

for any group basis H of FG.

In the next lemma, we derive a presentation for groups satisfying the hypotheses of Theorem B.

Lemma 3.3

Write $|G/\textrm{Z}(G)\Phi (G)|=p^d$, $|\gamma _2(G)^p \gamma _3(G)/\gamma _2(G)^p\gamma _4(G)|=p^m$, and $|G/\Phi (G)|=p^k$. If

• (a) $G^p\cap \gamma _2(G)\subseteq \gamma _2(G)^p \gamma _3(G)$ and

• (b) $|\gamma _2(G)/\gamma _2(G)^p\gamma _3(G)|=p^{{d\atopwithdelims ()2}}$,

then there exist lists of integers

$$\begin{aligned} {\varvec{n}}=(n_i)_{1\le i \le k}, \quad \varvec{\alpha }=(\alpha _{is})_{1\le i \le k; \ 1\le s \le m}, \quad \text {and} \quad \varvec{\beta } = (\beta _{ijls})_{1\le i,j,l\le k; \ 1\le s\le m}\nonumber \\ \end{aligned}$$

(3.3)

with $0<n_i$ and $0\le \alpha _{is}, \beta _{ijls}\le p-1$ such that

$$\begin{aligned} G/\gamma _2(G)^p \gamma _4(G)\cong \left\langle g_1,\dots , g_k,c_1,\dots , c_m \mid {\mathcal {R}}_{ ( {\varvec{n}},\varvec{\alpha },\varvec{\beta } )} ( g_1,\dots , g_k,c_1,\dots , c_m) \right\rangle , \end{aligned}$$

where ${\mathcal {R}}_{( {\varvec{n}},\varvec{\alpha },\varvec{\beta } )} ( g_1,\dots , g_k,c_1,\dots , c_m)$ is the set formed by the following relations:

• 1. $g_i^{p^{n_i}}=\prod _{1\le s \le m}c_s^{\alpha _{is}}$ for each $1\le i \le k$.

• 2. $[[g_i,g_j], g_l]=\prod _{1\le s\le m} c_s^{\beta _{ijls}}$ for $1\le i,j,l\le d$.

• 3. $[g_i,g_j]^p=1$ for $1\le i,j\le d$.

• 4. $g_i$ is central for $d<i\le k$.

• 5. The nilpotency class is at most three (i.e., for $t\ge 4$, all t-fold commutators of the generators vanish).

Moreover, the isomorphism $\phi :G/\gamma _2(G)^p\gamma _4(G)\rightarrow \left\langle g_1,\dots , g_k,c_1,\dots , c_m \right\rangle$ can be chosen such that the elements $\phi ^{-1}(g_{d+1})$,..., $\phi ^{-1}(g_k)\in G/\gamma _2(G)^p \gamma _4(G)$ have central preimages in G.

Proof

Set ${\widetilde{G}}=G/\gamma _2(G)^p \gamma _4(G)$ and, for $g \in G$, write ${\widetilde{g}} = g \gamma _2(G)^p \gamma _4(G) \in {\widetilde{G}}$ (similarly for subsets of G). Observe that $|\gamma _3({\widetilde{G}})|=p^m$ and $|{\widetilde{G}}/\Phi ({\widetilde{G}})|=p^k$. Write $|{\widetilde{G}}/\textrm{Z}({\widetilde{G}}) \Phi ({\widetilde{G}})|=p^t$. As the image of $Z(G) \Phi (G)$ in ${\widetilde{G}}$ has index at most $p^d$, we obtain $p^t\le p^d$. Moreover, $\gamma _2({{\widetilde{G}}})$ is elementary abelian and the hypotheses imply that ${\widetilde{G}}^p \cap \gamma _2({{\widetilde{G}}})\subseteq \gamma _3({{\widetilde{G}}})$ and $|\gamma _2({{\widetilde{G}}})/\gamma _3({{\widetilde{G}}})|=p^{d\atopwithdelims ()2}$. Since ${{{\widetilde{G}}}}^p\subseteq \textrm{Z}({\widetilde{G}})$ by Lemma 2.2, we have a well-defined map ${\mathfrak {c}} :{\widetilde{G}}/\textrm{Z}({\widetilde{G}}) \Phi ({\widetilde{G}}) \times {\widetilde{G}}/\textrm{Z}({\widetilde{G}}) \Phi ({\widetilde{G}}) \rightarrow \gamma _2({{\widetilde{G}}})/\gamma _3({{\widetilde{G}}})$ induced by the group commutator map, which is bilinear and anti-symmetric. If $\{x_1,\dots , x_t\}$ is a basis of ${\widetilde{G}}/\textrm{Z}({\widetilde{G}}) \Phi ({\widetilde{G}})$, then ${\mathfrak {c}}(x_i,x_j)$ with $1\le i <j \le t$ generates $\gamma _2({{\widetilde{G}}})/\gamma _3({{\widetilde{G}}})$ as an F-vector space. Thus $p^{{d\atopwithdelims ()2}}=|\gamma _2(\widetilde{G})/\gamma _3({{\widetilde{G}}})| \le p^{{t\atopwithdelims ()2}}$. This yields $t=d$.

Let $\{x_1,\dots , x_d\}$ be a set of preimages of a basis of $G/\textrm{Z}( G)\Phi ( G)$ in G, and let $\{x_{d+1},\dots x_k$} be a set of central preimages in G of a basis of $\textrm{Z}(G)\Phi (G)/\Phi (G)$. Thus $\{x_1,\dots , x_k\}$ is a Burnside basis of G, and its image is a Burnside basis of ${{\widetilde{G}}}$. By abuse of notation, we regard $\{x_1,\dots , x_k\}$ as a subset of ${{\widetilde{G}}}$. Let $\{z_1,\dots , z_m\}$ be a basis of $\gamma _3({{\widetilde{G}}})$. Write $p^{n_i}$ for the order of $x_i \gamma _2({\widetilde{G}})$ in ${\widetilde{G}}/\gamma _2({{\widetilde{G}}})$. As ${\widetilde{G}}^p\cap \gamma _2({{\widetilde{G}}}) \subseteq \gamma _3({{\widetilde{G}}})$, we have that $x_i^{p^{n_i}}=\prod _{1\le s \le m} z_s^{\alpha _{i s}}$ for uniquely determined $\alpha _{is} \in \{0, \dots , p-1\}$. Moreover, for $1\le i,j,l\le d$, we can write $[[x_i,x_j], x_l]=\prod _{1\le s\le m} z_s^{\beta _{ijls}}$ for uniquely determined $\beta _{ijls} \in \{0, \dots , p-1\}$. As $\{z_1, \ldots , z_m\}$ is a basis of $\gamma _3({{\widetilde{G}}})$, the matrix $(\beta _{ijls})_{ (i,j,l), s}$, with rows indexed by triples (i, j, l) and columns indexed by s, has rank m.

Define ${\varvec{n}}, \varvec{\alpha }$, and $\varvec{\beta }$ as in (3.3) and let

$$\begin{aligned} H=\left\langle g_1,\dots , g_k,c_1,\dots ,c_m \mid {\mathcal {R}}_{( {\varvec{n}},\varvec{\alpha },\varvec{\beta } )} ( g_1,\dots , g_k,c_1,\dots , c_m) \right\rangle . \end{aligned}$$

The assignment $g_i \mapsto x_i$ ($i = 1, \ldots , k$) and $c_j \mapsto z_j$ ($j = 1, \ldots , m$) defines a surjective group homomorphism $H\rightarrow {{\widetilde{G}}}$. To complete the proof, it suffices to show that it is an isomorphism, or equivalently that $|H|\le |{{\widetilde{G}}}|$.

Observe that since the matrix $(\beta _{ijls})_{ (i,j,l),s}$ has rank m, relation (2) implies that $c_1, \ldots , c_m \in \gamma _3(H)$. In particular, we have $c_1, \ldots , c_m \in \Phi (H)$ and hence $|H/\Phi (H)|\le p^k$.

As H has nilpotency class at most 3, $\gamma _2(H)$ is abelian. The group $\gamma _2(H)$ is generated by $\gamma _3(H)$ and elements of order p by relation (3), and hence $\gamma _2(H)/\gamma _3(H)$ is elementary abelian. Thus so is $\gamma _2(H)/\gamma _2(H)\cap \textrm{Z}(H)$. On the other hand, the image of the map $H/\textrm{Z}(H)\times \gamma _2(H)/\gamma _2(H) \cap \textrm{Z}(H)\rightarrow \gamma _3(H)$ induced by the commutator map generates $\gamma _3(H)$. For $h \in H$ and $h' \in \gamma _2(H)$, we have $[h \textrm{Z}(H), h' \gamma _2(H)\cap \textrm{Z}(H)]^p= [h\textrm{Z}(H), (h')^p \gamma _2(H)\cap \textrm{Z}(H)]=1$ (see Lemma 2.1). Thus $\gamma _3(H)$ is elementary abelian. This implies that $\gamma _2(H)$ is elementary abelian (as an abelian group generated by elements of order p).

By relations (2) and (5), we have $\gamma _3(H) = \langle c_1, \dots , c_m\rangle$, so

$$\begin{aligned} |\gamma _3(H)|\le p^m=|\gamma _3({{\widetilde{G}}})|. \end{aligned}$$

(3.4)

Moreover, using (4), we derive that $|H/\textrm{Z}(H)\Phi (H)|\le p^d$. Observe that $[H,\Phi (H)] \subseteq \gamma _3(H)$ due to $[x^p,y] \gamma _3(H) = [x,y]^p \gamma _3(H)$ for all $x,y \in H$ (see Lemma 2.1) and $\gamma _2(H)$ being elementary abelian. Then the commutator in the group defines a bilinear map $H/Z(H) \Phi (H) \times H/Z(H) \Phi (H)\rightarrow \gamma _2(H)/\gamma _3(H)$, $(x\textrm{Z}(H)\Phi (H), y\textrm{Z}(H)\Phi (H))\mapsto [x,y] \gamma _3(H)$. As for ${\widetilde{G}}$, we derive that

$$\begin{aligned} |\gamma _2(H)/\gamma _3(H)|\le p^{{d\atopwithdelims ()2}}=|\gamma _2(\widetilde{G})/\gamma _3({{\widetilde{G}}})|. \end{aligned}$$

(3.5)

Finally, $H/\gamma _2(H)=\left\langle g_1 \gamma _2(H),\dots , g_k \gamma _2(H) \right\rangle$ has order at most $\prod _{i=1 }^k p^{n_i}=|\widetilde{G}/\gamma _2({{\widetilde{G}}})|$ by relation (1). Since $|{{\widetilde{G}}}|= |{{\widetilde{G}}}/\gamma _2({{\widetilde{G}}})|\cdot |\gamma _2({{\widetilde{G}}})/\gamma _3({{\widetilde{G}}}) |\cdot |\gamma _3({{\widetilde{G}}})|$, combining this with (3.4) and (3.5) yields that $|H|\le |\widetilde{G}|$. This proves the statement. $\square$

Theorem 3.4

Let p be an odd prime number and let F be the field with p elements. Consider finite p-groups G and H with $FG \cong FH$ and let $|G/\Phi (G)\textrm{Z}(G)|=p^d$. If

• (a) $G^p\cap \gamma _2(G)\subseteq \gamma _2(G)^p \gamma _3(G)$ and

• (b) $|\gamma _2(G)/\gamma _2(G)^p\gamma _3(G)|=p^{{d\atopwithdelims ()2}},$

then $G/\gamma _2(G)^p \gamma _4(G)\cong H/\gamma _2(H)^p \gamma _4(H)$.

Proof

Consider the isomorphism of elementary abelian p-groups

$$\begin{aligned} \varphi :G/\Phi (G)\rightarrow I(G)/I(G)^2, \quad x\Phi (G)\mapsto (x-1) + I(G)^2 \end{aligned}$$

(see [24, Propositions III.1.14 and III.1.15]). We have an F-vector space decomposition $\textrm{Z}(FG) = F\textrm{Z}(G) + \textrm{Z}(FG) \cap K(FG)$ (see [21, Lemma 6.10]). It is well-known that $FG \cdot K(FG) = K(FG) \cdot FG = I(\gamma _2(G)) FG$. As $\gamma _2(G) \subseteq \Phi (G) = (1 + I(G)^2) \cap G$ holds (see [11, Theorem 5.5]), we have

$$\begin{aligned} K(FG) \subseteq I(\gamma _2(G)) FG \subseteq I(G)^2. \end{aligned}$$

(3.6)

In particular, $\varphi$ restricts to an isomorphism

$$\begin{aligned} \textrm{Z}(G)\Phi (G)/\Phi (G) \rightarrow \left( \textrm{Z}(FG)\cap I(G)+I(G)^2 \right) /I(G)^2. \end{aligned}$$

(3.7)

Of course, the analogous reasoning applies to H. In particular, we obtain

$$\begin{aligned} p^d=|G/\Phi (G)\textrm{Z}(G)|=|I(G)/(\textrm{Z}(FG) \cap I(G) + I(G)^2)|= |H/\Phi (H)\textrm{Z}(H)|. \end{aligned}$$

Moreover, note that the conditions (a) and (b) depend only on the quotient $G/\gamma _2(G)^p \gamma _3(G)$, which is determined by FG by [22]. Thus, if G satisfies (a) and (b), then the analogous statements hold for H.

Consider the group V and the projection $\pi :V(FG) \rightarrow V$ introduced in (3.1). Let $\{g_1,g_2,\dots , g_k, c_1,\dots ,c_m\}$ be a set of generators of $\pi (G)$ satisfying relations

$$\begin{aligned} {\mathcal {R}}_{( {\varvec{n}},\varvec{\alpha },\varvec{\beta } )} ( g_1,\dots , g_k,c_1,\dots , c_m) \end{aligned}$$

as in Lemma 3.3. For $1\le i \le k$, the decomposition $V=\pi (H)\rtimes E$ (see (3.2)) guarantees that $g_i=h_i e_i$ for some unique $h_i\in \pi (H)$ and $e_i\in E$. Moreover $c_j \in \gamma _2(\pi (G))=\gamma _2(S)=\gamma _2(\pi (H))$ for $j=1,\dots , m$ (see (2.1)). Now let $d<i \le k$, so $g_i \in Z(\pi (G))$. We have

$$\begin{aligned} (g_i -1) + I(G)^2= & {} (h_i e_i-1)+ I(G)^2 = h_i -1 + e_i - 1 + I(G)^2 \\= & {} (h_i-1) + I(G)^2, \end{aligned}$$

using $e_i -1 \in I(G)^2$ (see Lemma 2.3). In particular, this yields $h_i -1 + I(G)^2 \in (Z(FG) \cap I(G) + I(G)^2)/I(G)^2$. Using the isomorphism in (3.7), and the analogue for H, we derive that $h_i ={\hat{h}}_i {{\widetilde{h}}}_i$, with ${\hat{h}}_i\in \textrm{Z}(\pi (H))$ and ${{\widetilde{h}}}_i\in \Phi (\pi ( H)) = \gamma _2(\pi (H)) \pi (H)^p$. Observe that $\pi (H)^p$ is central in $\pi (H)$ by Lemma 2.2, so we can furthermore assume that ${{\widetilde{h}}}_i \in \gamma _2(\pi (H))$. Then $\{h_1,\dots , h_d, {\hat{h}}_{d+1}, \dots , {\hat{h}}_k,c_1,\dots , c_m\}$ forms a generating set of $\pi (H)$. We now show that it satisfies the relations

$$\begin{aligned} {\mathcal {R}}_{ ( {\varvec{n}},\varvec{\alpha },\varvec{\beta } )} ( h_1,\dots , h_d, {\hat{h}}_{d+1}, \dots , {\hat{h}}_k,c_1,\dots , c_m). \end{aligned}$$

Then H is an epimorphic image of G of the same size and we obtain $G\cong H$ as desired.

We check the relations one by one. Relations (3), (4), and (5) are immediate. For $1\le i \le k$, one has

$$\begin{aligned} g_i^p=(h_ie_i)^p=h_i^pe_i^p [h_i,e_i]^p=h_i^p, \end{aligned}$$

and, in particular, for $d<i\le k$, one has $g_i^p=h_i^p= {\hat{h}}_i^p {{\widetilde{h}}}_i^p = {\hat{h}}_i ^p$. Then condition (1) follows. Condition (2) also follows readily because, for $1\le i \le d$ and arbitrary $x_i,x_j,x_l\in A$, we have that

$$\begin{aligned}{}[[g_ix_i, g_jx_j], g_lx_l] =[[g_i,g_j] [x_j, g_j][g_i,x_j], g_l x_l ] =[[g_i,g_j], g_l x_l]=[[g_i,g_j],g_l].\nonumber \\ \end{aligned}$$

(3.8)

This completes the proof. $\square$

The next result shows that Theorem 3.4 is a generalization of [14, Theorem 1.2].

Corollary 3.5

Let p be an odd prime number, let F be the field with p elements, and let G and H be finite p-groups such that $F G\cong F H$. If $|G/\Phi (G)\textrm{Z}(G)|=p^2$, then

$$\begin{aligned} G/\gamma _2(G)^p \gamma _4(G)\cong H/\gamma _2(H)^p\gamma _4(H). \end{aligned}$$

Proof

If $G/ \gamma _2(G)^p \gamma _4(G)$ has nilpotency class at most 2, then so does $H/\gamma _2(H)^p \gamma _4(H)$ (see (2.1)) and hence the result follows by [22]. Thus we assume that G has class 3. Observe that $\gamma _2(G)/\gamma _2(G)^p\gamma _3(G)$ has order p, as it is generated by $[g_1,g_2]\gamma _2(G)^p\gamma _3(G)$ if we consider a set of generators $\{g_1,\dots ,g_k,c_1,\dots , c_m\}$ as in Lemma 3.3. Observe that every p-power is central in $G/ \gamma _2(G)^p \gamma _4(G)$ by Lemma 2.2. On the other hand, the image of $[g_1,g_2]$ in $G/\gamma _2(G)^p \gamma _4(G)$ is not central as G does not have nilpotency class 2. This yields that $G^p \cap \gamma _2(G)\subseteq \gamma _2(G)^p \gamma _3(G)$. Thus the hypotheses of Theorem 3.4 with $d=2$ hold, and the statement follows. $\square$

4 Groups with a center of index $p^3$

Let p be a prime number. In this section, we use the results about the small group algebra from Section 3 to give a positive answer to the modular isomorphism problem for groups with center of odd index $p^3$.

Lemma 4.1

Let p be an odd prime number and let G be a finite p-group with $|G:\textrm{Z}(G)|=p^3$. Then:

• (1) The nilpotency class of G is at most 3.

• (2) $\gamma _2(G)$ is elementary abelian.

• (3) Either G has nilpotency class 2 or $|G/\Phi (G)\textrm{Z}(G)|=p^2$.

Proof

Examination of the upper central series yields that the nilpotency class of G is at most 3. The quotient $G/\textrm{Z}(G)$ is isomorphic to one of the five groups of order $p^3$, namely to one of $C_{p^3}$, $C_p^3$, $C_{p^2}\times C_p$, $(C_p\times C_p)\rtimes C_p$, or $C_{p^2}\rtimes C_p$. Observe that G/Z(G) is not cyclic as G is abelian otherwise.

Next, we show that $\gamma _2(G)$ is elementary abelian. It is clear that $\gamma _2(G)$ is abelian as $[\gamma _2(G), \gamma _2(G)]\subseteq \gamma _4(G)=1$. Thus it suffices to show that the generators of $\gamma _2(G)$ have order p. Note that $\gamma _3(G)$ is elementary abelian. Indeed, it is generated by elements of the form [x, y] with $x\in G$ and $y\in \gamma _2(G)$; since $\gamma _2(G)\textrm{Z}(G)/\textrm{Z}(G)$ has order at most p, we have $y^p\in \textrm{Z}(G)$, so Lemma 2.1 shows that $1=[x,y^p]=[x,y]^p$. Let $a,b_1, b_2\in G$ be such that $G/\textrm{Z}(G)=\left\langle a\textrm{Z}(G), b_1 \textrm{Z}(G), b_2\textrm{Z}(G) \right\rangle$. By the structure of $G/\textrm{Z}(G)$, we may assume that $b_i^p\in \textrm{Z}(G)$ for $i=1,2$. Then for every $x\in G$ and $i \in \{1,2\}$, Lemma 2.1 yields that

$$\begin{aligned} 1=[x, b_i^p]= [x,b_i]^p [[x,b_i],b_i]^{{p\atopwithdelims ()2}} =[x,b_i]^p. \end{aligned}$$

This shows that $\gamma _2(G)$ is elementary abelian.

Finally, if G has nilpotency class 3, then $G/\textrm{Z}(G)$ is isomorphic to one of the two non-abelian p-groups of order $p^3$. Then it is clear that $|\gamma _2(G)\textrm{Z}(G)/\textrm{Z}(G)|=p$, and hence $|G/\Phi (G)\textrm{Z}(G)|=p^2$. $\square$

We can now prove our main result:

Theorem 4.2

Let p be an odd prime number, let F be the field with p elements, and let G and H be finite p-groups. Suppose that $|G:\textrm{Z}(G)|=p^3$. If $FG\cong FH$, then $G\cong H$.

Proof

By Lemma 4.1, we know that $\gamma _2(G)$ is elementary abelian. If the nilpotency class of G is 2, then $G\! \cong \! H/\gamma _2(H)^p \! \gamma _3(H)$ by [22], so $G \! \cong \! H$ due to $|G|\! =\! |H|$. Thus, by Lemma 4.1, we can assume that the nilpotency class of G is 3, and that $|G/\Phi (G)\textrm{Z}(G)|\! =\! p^2$. By Corollary 3.5, we then have $G\cong H$. $\square$

Remark 4.3

We point out that there cannot be an analogue of Theorem 4.2 for $p=2$, as in [9] non-isomorphic finite 2-groups with centers of index 8 and isomorphic group algebras over every field of characteristic 2 are presented. Hence our result underlines the difference between the cases $p=2$ and $p>2$ for this problem (we refer to [8] for other contrasts).

Remark 4.4

Observe that [4, Corollary 8.2] yields that the finite p-groups (p odd) with extraspecial central quotient are exactly the groups with center of index $p^3$ and nilpotency class 3. Thus Theorem 4.2 gives a positive answer to the modular isomorphism problem for this class of groups.

We conclude this section with examples illustrating that the class of p-groups with a center of index $p^3$ is not contained in one of the classes for which the modular isomorphism problem was previously known to have a positive answer.

Example 4.5

   

• (1) Consider the following groups of order $5^7$:

  $$\begin{aligned}&G=\left\langle a,b,c,d \left| \begin{array}{c} a^{625} = b^5=c^{25}=[b,a]^5=[c,a]=[c,b] =[d,a]=[d,b] =[d,c] =1, \\ a^5=d, \left[ [b,a], a \right] = d^5, \left[ [b,a],b\right] = c^5 \end{array} \right. \right\rangle , \\&H= \left\langle a,b,c,d \left| \begin{array}{c} a^{625} = b^5=c^{25}=[b,a]^5=[c,a]=[c,b] =[d,a]=[d,b] =[d,c] =1, \\ a^5=d, \left[ [b,a], a \right] = c^5, \left[ [b,a],b\right] = d^5 \end{array} \right. \right\rangle , \\&K= \left\langle a,b,c,d \left| \begin{array}{c} a^{625} = b^5=c^{25}=[b,a]^5=[c,a]=[c,b] =[d,a]=[d,b] =[d,c] =1, \\ a^5=d^2, \left[ [b,a], a \right] = c^5, \left[ [b,a],b\right] = d^5 \end{array} \right. \right\rangle . \end{aligned}$$

  In GAP, they are listed as $\texttt {SmallGroup}(5^7, 1599)$, $\texttt {SmallGroup}(5^7,1734)$, and $\texttt{SmallGroup}(5^7,1766)$, respectively. Each of these groups is 3-generated with minimal set of generators $\{a,b,c\}$. The center is given by $\langle c,d \rangle \cong C_{25} \times C_{25}$, and hence it has index $5^3$, so Theorem A applies to G, H, and K. On the other hand, a quick verification shows that none of these groups is covered by any of the cases for which MIP is known (according to the survey in [12]). Moreover, these three groups agree in all the group theoretical invariants implemented in the GAP package ModIsomExt [13, 25].

• (2) Let $p \ge 7$. In the list of groups of order $p^6$ given in [16], consider the groups $G_{6,3}$ as well as $G_{6,5r}$ for $r \in \{1, \nu \}$. These groups have centers of index $p^3$ and do not lie in the union of classes of p-groups for which the modular isomorphism problem is known to have a positive answer (according to [12]). For small values of p, one can verify computationally that these groups agree in all group theoretical invariants implemented in ModIsomExt.