2. Preliminaries

In this section, we fix notation and include a few preliminary results that we use throughout this article. We refer the reader to Karpilovsky [Reference Karpilovsky14] for most of the results included in this section.

Let $G$ be a finite group. Recall, for a group $G$ , $Z^2(G, \mathbb C^\times )$ consists of all functions $f\;:\; G \times G \rightarrow \mathbb C^\times$ such that $f(x,1) = f(1,x) = 1$ and $f(x,y)f(xy, z) = f(x, yz) f(y, z)$ for all $x,y, z \in G$ . We shall call elements of $Z^2(G, \mathbb C^\times )$ as 2-cocycles (or sometimes just cocycles when it is clear from the context). Then ${\mathrm{H}}^2(G, \mathbb C^\times ) = Z^2(G, \mathbb C^\times )/B^2(G, \mathbb C^\times )$ , where $B^2(G, \mathbb C^\times )$ is the set of $2$ -coboundaries of $G$ , is called the second cohomology group of $G$ or the Schur multiplier of $G$ . The elements of ${\mathrm{H}}^2(G, \mathbb C^\times )$ are called the cohomology classes of $G$ . For an element $\alpha \in Z^2(G, \mathbb C^\times )$ , the corresponding element of ${\mathrm{H}}^2(G, \mathbb C^\times )$ will be denoted by $[\alpha ]$ . For 2-cocycles $\alpha, \beta \in Z^2(G, \mathbb C^\times )$ we say $\alpha$ is cohomologous to $\beta$ , whenever $[\alpha ] = [\beta ]$ .

A central extension,

(2.1) \begin{equation} 1\to A \to G \to G/A \to 1 \end{equation}

is called a stem extension, if $A \subseteq Z(G) \cap G'$ . For a given stem extension (2.1), the Hochschild-Serre spectral sequence [Reference Hochschild and Serre10, Theorem 2, p. 129] yields the following exact sequence

\begin{equation*}1 \rightarrow \mathrm {Hom}( A,\mathbb C^\times ) \xrightarrow []{{\mathrm {tra}}} {\mathrm {H}}^2(G/A, \mathbb C^\times ) \xrightarrow {\inf } {\mathrm {H}}^2(G, \mathbb C^\times ),\end{equation*}

where ${\mathrm{tra}}\;:\;\mathrm{Hom}( A,\mathbb C^\times ) \to{\mathrm{H}}^2(G/A, \mathbb C^\times )$ given by $f \mapsto [{\mathrm{tra}}(f)]$ , where

\begin{equation*}{\mathrm {tra}}(f)(\overline {x},\overline {y}) = f(\mu (\overline {x})\mu (\bar {y})\mu (\bar {xy})^{-1}),\,\, \overline {x}, \overline {y} \in G/A, \end{equation*}

for a section $\mu \;:\; G/A \rightarrow G$ , denotes the transgression homomorphism and the inflation homomorphism, $\inf \;:\;{\mathrm{H}}^2(G/A, \mathbb C^\times ) \to{\mathrm{H}}^2(G, \mathbb C^\times )$ is given by $[\alpha ] \mapsto [\inf (\alpha )]$ , where $\inf (\alpha )(x,y) = \alpha (xA,yA)$ .

Let $G$ be a finite group and $V$ be a finite-dimensional complex vector space. A mapping $\rho \;:\; G \rightarrow \mathrm{GL}(V)$ is called a (finite-dimensional and complex) projective representation of $G$ if thhere xists a mapping $\alpha \;:\; G \times G \rightarrow \mathbb C^\times$ such that

• $\rho (x) \rho (y) = \alpha (x, y) \rho (xy)$ for all $x,y \in G$ ,

• $\rho (1) = \mathrm{Id}_V$ .

In this case, we say $\rho$ is an $\alpha$ -representation of $G$ on $V$ . For $\alpha \in Z^2(G, \mathbb C^\times )$ , we use $\mathrm{Irr}^\alpha (G)$ to denote the set of equivalence classes of irreducible $\alpha$ -representations of $G$ . For $\alpha = 1$ , we use $\mathrm{Irr}(G)$ instead of $\mathrm{Irr}^\alpha (G)$ and call this the set of ordinary irreducible representations of $G$ . For $\rho \in \mathrm{Irr}(G)$ , at times, we shall also omit the word “ordinary” and call $\rho$ to be a representation of $G$ . For $\alpha \in{\mathrm{H}}^2(G, \mathbb C^\times )$ , the complex group algebra $\mathbb C^\alpha G$ is semisimple and

(2.2) \begin{eqnarray} \mathbb C^\alpha [G] \cong \prod _{\rho \in \mathrm{Irr}^\alpha (G)} M_{\dim (\rho )}(\mathbb C), \end{eqnarray}

where $M_n(\mathbb C)$ denotes the $n \times n$ matrix algebra over $\mathbb C$ . Hence, the information of projective representations of $G$ is also helpful to study the twisted group algebra isomorphism problem. To this end, the representation group (also called a covering group) of a group $G$ will play an important role. We now recall the definition of the representation group.

In [Reference Schur28], Schur proved that representation group of a finite group always exists (see also [Reference Karpilovsky14, Chapter 3 (Corollary 3.3)]). From above, it is clear that to determine the projective representations of a group $G$ , it is enough to determine a representation group of $G$ (there may be many non-isomorphic ones) and its ordinary representations. We now recall a result that elaborates the above correpondence between the projective representations of $G$ and the ordinary representations of $G^\star$ .

Let $N$ be a normal subgroup of $G$ and $\chi \in \mathrm{Irr}(N)$ . Let $\mathrm{Irr}(G \mid \chi )$ denote the set of inequivalent ordinary irreducible representations of $G$ lying above $\chi$ , that is $\rho \in \mathrm{Irr}(G \mid \chi )$ if and only if $\mathrm{Hom}_{N}(\rho |_{N}, \chi )$ is non-trivial. The following well-known result relates the projective representations of $G$ and the ordinary ones of $G^\star$ , see [Reference Karpilovsky14, Chapter 3, Lemma 3.1] and [Reference Hatui and Singla8, Theorem 3.2] for its proof.

Theorem 2.2. Let $\alpha$ be a $2$ -cocycle of $G$ . Let $\chi \in \mathrm{Hom}(A,\mathbb C^\times )$ be such that $\mathrm{tra}(\chi ) = [\alpha ]$ . There is a bijective correspondence between

\begin{equation*}\mathrm {Irr}^{\alpha }(G)\leftrightarrow \mathrm {Irr}{(G^\star \mid \chi )}\end{equation*}

obtained via lifting a projective representation of $G$ to an ordinary representation of $G^\star$ . In particular, we obtain the following:

\begin{equation*}\bigcup _{[\alpha ] \in {\mathrm {H}}^2(G,\mathbb C^\times )} \mathrm {Irr}^{\alpha }(G)\leftrightarrow \mathrm {Irr}{(G^\star )}.\end{equation*}

To understand the ordinary representations of a representation group, we also require a few general results from the theory of the ordinary characters of a finite group. These results are usually known by the name of Clifford’s theory and they provide an important connection between the complex representations of a finite group $G$ and its normal subgroups. Recall that $\mathrm{Irr}(G)$ , for a finite group $G$ , denotes the set of all inequivalent irreducible representations of $G$ . For an abelian group $A$ , we also use $\widehat{A}$ to denote $\mathrm{Irr}(A)$ and call this to be the set of characters (or one-dimensional representations) of $A$ . Let $N$ be a normal subgroup of $G$ and $\rho \in \mathrm{Irr}(N)$ be an irreducible representation of $N$ . The representation obtained by inducing $\rho$ from $N$ to $G$ will be denoted by $\mathrm{Ind}_N^G(\rho )$ . We say $\rho \in \mathrm{Irr}(N)$ has an extension to $G$ if thhere xists $\tilde{\rho } \in \mathrm{Irr}(G)$ such that $\tilde{\rho }|_{N} = \rho .$

The group $G$ acts on $\mathrm{Irr}(N)$ via conjugation action of $G$ on $N$ . For $\rho \in \mathrm{Irr}(N)$ and $g\in G$ , define $\rho ^g \in \mathrm{Irr}(N)$ by $\rho ^g(x) = \rho (gxg^{-1})$ for all $x \in N$ . For $\rho, \rho ' \in \mathrm{Irr}(N)$ , we use $\rho \cong \rho '$ to denote that $\rho$ and $\rho '$ are equivalent representations of $N$ . For $\rho \in \mathrm{Irr}(N)$ , let $I_G(\rho ) = \{ g \in G |\,\, \rho ^{g} \cong \rho \}$ denote the stabilizer (or the inertia) group of $\rho$ in $G$ . We will use the following results of Clifford’s theory:

We conclude this section by including a proof of Theorem 1.5.

Proof of Theorem 1.5. Our goal is to define an isomorphism $\gamma \;:\;{\mathrm{H}}^2(G_2, \mathbb C^\times ) \rightarrow{\mathrm{H}}^2(G_1, \mathbb C^\times )$ that gives $\mathbb C$ -twist isomorphism between $G_2$ and $G_1$ . It follows from [Reference Karpilovsky14, Theorem 2.9] and [Reference Hatui and Singla8, Theorem 3.2] that the projective representations of $G_{i}$ are obtained from those of $G_{i}/G'_{\!\!i}$ via inflation and

(2.3) \begin{eqnarray} \mathbb C^\alpha [G_{i}] \cong \prod _{\mathrm{inf_i}([\beta ]) = [\alpha ]} \mathbb{C}^\beta [G_{i}/G'_{\!\!i}]. \end{eqnarray}

The map $\tilde{\sigma }$ is an induced isomorphism obtained from $\sigma$ . Hence, $\mathbb{C}^\beta [G_1/G'_{\!\!1}] \cong \mathbb{C}^{\tilde{\sigma }(\beta )}[G_2/G'_{\!\!2}].$ Therefore, it is sufficient to define an isomorphism $\gamma \;:\;{\mathrm{H}}^2(G_2, \mathbb C^\times ) \rightarrow{\mathrm{H}}^2(G_1, \mathbb C^\times )$ such that Figure 1 is commutative. Indeed, such a $\gamma$ is obtained by defining

\begin{equation*} \gamma ([\alpha ]) = \mathrm {inf}_1 (\tilde {\sigma }([\alpha _0])) \,\, \mathrm {for}\,\, [\alpha ] \in H^2(G_2, \mathbb C^\times ), \end{equation*}

where $[\alpha _0] \in H^2(G_2/G'_{\!\!2}, \mathbb C^\times )$ is any element such that $\inf _2([\alpha _0]) = [\alpha ]$ .

3. $p$ -groups with $s(G) \leq 2$

In this section, we study the $\mathbb{C}$ -twist isomorphism classes of $p$ -groups with $s(G) \leq 2$ . We first deal with the case of $s(G) \in \{0,1\}$ .

Proof of Lemma 1.1. Niroomand [Reference Niroomand22, Theorem 21, Corollary 23] proved the following classification of finite non-abelian $p$ -groups $G$ with $s(G) \in \{0,1\}$ :

1. (a) $s(G) = 0$ if and only if $G$ is isomorphic to $H_{1}^{1} \times C_{p}^{(n-3)}$ .

2. (b) $s(G) = 1$ if and only if $G$ is isomorphic to $D_{8} \times C_{2}^{(n-2)}$ or $C_{p}^{(4)} \rtimes C_{p}\, (p \neq 2)$ .

We remark that in [Reference Niroomand22], Niroomand uses the corank of a group $G$ (denoted $t(G)$ ) instead of the generalized corank of $G$ . We have used the well-known relation $t(G) = s(G) + (n-2)$ for any non-abelian $p$ -group $G$ to use the results of [Reference Niroomand22]. We observe that among the groups mentioned in (a) and (b), thhere xists at most one group for any given order and fixed $s(G)$ . Since both order and $s(G)$ are invariant of $\mathbb C$ -twist isomorphism, our result follows.

The rest of this section is devoted to the $s(G) = 2$ case.

Proof. We proceed to prove (i). The proof of (ii) is similar so we only give essential ingredients there and omit the details.

(i)(a) For simplification of notations, we denote $E(2) \times C_p^{(n-4)}$ and $H_1^2 \times C_p^{(n-3)}$ by $G_1$ and $G_2$ , respectively. The groups $G_1$ and $G_2$ have following presentations:

\begin{equation*} G_1 = \lt x_1,y_1,z_1, \gamma _1, a_1, a_2, \cdots a_{n-4} \mid [x_1,y_1] = z_1 = \gamma _1^p, x_1^p = y_1^p = a_i^p = 1, \gamma _1^{p^2} = 1\gt \end{equation*}

\begin{equation*} G_2 = \lt x_2,y_2,z_2, b_1, \cdots, b_{n-3} \mid [x_2,y_2] = x_2^p = z_2, x_2^{p^2} = y_2^p = z_2^p = b_i^p = 1\gt. \end{equation*}

Therefore, $G'_{\!\!i} \cong C_{p}$ and $G_{i}/G'_{\!\!i} \cong C_{p}^{ n-1}$ for $i \in \{1, 2\}$ . Also, in view of Proposition 1.3 of [Reference Karpilovsky14],

\begin{equation*} {\mathrm {H}}^2(G_{i}/G'_{\!\!i}, \mathbb C^\times ) \cong C_{p}^{ \frac {(n-1)(n-2)}{2}}, \,\, {\mathrm {H}}^2(G_{i}, \mathbb C^\times ) \cong C_{p}^{ \frac {(n-1)(n-2)}{2}-1}. \end{equation*}

This yields the following short exact sequences for $i \in \{1, 2\}$ :

\begin{equation*} 1 \longrightarrow \mathrm {Hom}(G'_{\!\!i}, \mathbb C^\times ) \stackrel {\mathrm {tra}_i}{\longrightarrow } H^2(G_{i}/G'_{\!\!i}, \mathbb C^\times ) \stackrel {\mathrm {inf_i}}{\longrightarrow } H^2(G_{i}, \mathbb C^\times ) \longrightarrow 1. \end{equation*}

We now define $\delta \;:\; G'_{\!\!1} \rightarrow G'_{\!\!2}$ and $\sigma \;:\; G_1/G'_{\!\!1} \rightarrow G_2/G'_{\!\!2}$ such that the Figure 1 is commutative. Define $\delta$ by $\delta (z_1) = z_2$ and $\sigma$ by

\begin{equation*} \sigma (x_1 G'_{\!\!1}) = x_2 G'_{\!\!2}, \sigma (y_1 G'_{\!\!1}) = y_2 G'_{\!\!2}, \sigma (\gamma _1 G'_{\!\!1}) = b_{n-3} G'_{\!\!2}, \sigma (a_i G'_{\!\!1}) = b_i G'_{\!\!2}, \end{equation*}

for all $i \in \{1,\ldots, n-4\}$ . We now describe transgression maps for these groups.

Define a section $s_1\;:\;G_1/G'_{\!\!1} \rightarrow G_1$ by

\begin{equation*} s_1(x_1^i y_1^j \gamma _1^k a_1^{r_1} \cdots a_{n-4}^{r_{n-4}}G'_{\!\!1} ) = x_1^i y_1^j \gamma _1^k a_1^{r_1} \cdots a_{n-4}^{r_{n-4}} \end{equation*}

For $u= x_1^i y_1^j \gamma _1^k a_1^{r_1} \cdots a_{n-4}^{r_{n-4}} G'_{\!\!1}$ and $v = x_1^{i^{\prime}} y_1^{j^{\prime}} \gamma _1^{k^{\prime}} a_1^{r^{\prime}_1} \cdots a_{n-4}^{r^{\prime}_{\!n-4}} G'_{\!\!1},$ we have

\begin{equation*} s_1(u)s_1(v)s_1(uv)^{-1} = \gamma _1^{-p j i^{\prime}}. \end{equation*}

Hence, a representative of $[\mathrm{tra_1}(\chi )]$ is given by ${\mathrm{tra}}_1(\chi )(u, v)= \chi (z_1^{-ji^{\prime}})$ for $\chi \in \mathrm{Hom}(G'_{\!\!1}, \mathbb C^\times )$ . Define a section $s_2\;:\;G_2/G'_{\!\!2} \rightarrow G_2$ by

\begin{equation*} s_2(x_2^i y_2^j b_1^{r_1} \cdots b_{n-3}^{r_{n-3}}G'_{\!\!2} ) = x_2^i y_2^j b_1^{r_1} \cdots b_{n-3}^{r_{n-3}} \end{equation*}

For $u= x_2^i y_2^j b_1^{r_1} \cdots b_{n-3}^{r_{n-3}} G'_{\!\!2}$ and $v = x_2^{i^{\prime}} y_2^{j^{\prime}} b_1^{r^{\prime}_1} \cdots b_{n-3}^{r^{\prime}_{\!n-3}} G'_{\!\!2},$ we have

\begin{equation*} s_2(u)s_2(v)s_2(uv)^{-1} = z_2^{-ji^{\prime}}. \end{equation*}

Therefore, a representative of $[\mathrm{tra_2}(\chi )]$ is given by $\mathrm{tra_2}(\chi ) (u,v) = \chi (z_2^{-ji^{\prime}})$ for $\chi \in \mathrm{Hom}(G'_{\!\!2}, \mathbb C^\times )$ . This combined with the given isomorphisms $\delta$ and $\sigma$ gives the commutativity of Figure 1. Now, the result follows as a direct consequence of Theorem 1.5.

For $(i)(b)$ , proof is along the same lines as that of $(i)(a)$ with only difference that $Q_8 \times C_2^{(n-3)}$ has the following presentation:

\begin{equation*} \langle a, b, c, b_1, \cdots, b_{n-3} \ | \ a^4=1, a^2=b^2=c, b^{-1}ab = ca, b_i^{2} = 1 \ \forall \ 1 \leq i \leq (n-3) \rangle. \end{equation*}

We leave the rest of the details for the reader.

(ii) We denote the groups $E(2) \times C_p^{(n-2m-2)}$ , $H_m^1 \times C_p^{(n-2m-1)}$ and $H_m^2 \times C_p^{(n-2m-1)}$ by $G^m_{1}$ , $G^m_{2}$ and $G^m_{3}$ , respectively. Note that if $m \neq m'$ , then for any $i,j \in \{ 1, 2,3\},$ the complex group algebras of $G_{i}^m$ and $G_j^{m^{\prime}}$ are not isomorphic. Further, observe that for any $m \geq 2$ , the order of $G_{1}^m$ is $p^n$ such that $n \geq 6.$ Therefore, for some $n \geq 6,$ if $G_{i}^{m}$ is $\mathbb{C}$ -twist isomorphic to $G_j^{m^{\prime}},$ then it implies that $m = m'$ . Similarly, when $n = 5,$ a necessary condition for the $\mathbb{C}$ -twist isomorphism of $G_1^{m}$ and $G_2^{m^{\prime}}$ is that $ m = m'$ . Therefore, from now onwards, we fix $m$ and prove the result.

The commutator subgroup of $G_{i}^m$ is central and is isomorphic to $C_{p};$ and $G_{i}^m/(G_{i}^m)' \cong C_p^{(n-1)}.$ Further, since for any $i \in \{1, 2, 3\},$ ${\mathrm{H}}^2(G_{i}^m, \mathbb C^\times ) \cong C_p^{ \frac{n^2-3n}{2} }$ , we get the following short exact sequences:

\begin{equation*} 1 \longrightarrow \mathrm {Hom}(G'_{\!\!i}, \mathbb C^\times ) \stackrel {\mathrm {tra}_i}{\longrightarrow } {\mathrm {H}}^2(G_{i}/G'_{\!\!i}, \mathbb C^\times ) \stackrel {\mathrm {inf_i}}{\longrightarrow } {\mathrm {H}}^2(G_{i}, \mathbb C^\times ) \longrightarrow 1. \end{equation*}

As in (i), the proof of $\mathbb C$ -twist isomorphism follows by considering the image of the $\mathrm{tra_i}$ for $i \in \{1, 2, 3\}$ and by proving the commutativity of Figure 1. This is obtained by using the following presentation of groups $G_{i}^m.$

\begin{eqnarray*} G_1^m & = & \lt x_1, \cdots, x_m,y_1, \cdots y_m,z, \gamma, a_1, a_2, \cdots a_{n-2m-2} \mid [x_i,y_i] = z = \gamma ^p, \\ & & x_i^p = y_i^p = a_i^p = 1, \gamma ^{p^2} = 1\gt. \\ G_2^m & = & \lt x_1, \cdots, x_m,y_1, \cdots y_m,z, a_1, a_2, \cdots, a_{n-2m-1} \mid [x_i,y_i] = z, \\ & & x_i^p = y_i^p = a_i^p = 1\gt. \\ G_3^m & = & \lt x_1, \cdots, x_m,y_1, \cdots y_m,z, a_1, a_2, \cdots, a_{n-2m-1} \mid [x_i,y_i] = z = x_m^p = y_m^p, \\ & & x_i^p = y_i^p = z^p = 1 (1 \leq i \leq m-1), a_i^p = 1\gt. \end{eqnarray*}

Below we calculate $\mathrm{tra}_1$ explicitly and leave the details for ${\mathrm{tra}}_2$ and ${\mathrm{tra}}_3$ as those are similar. By the given presentation of $G_1^m$ , we have $(G_1^m)' = \lt \gamma ^p\gt$ . Define a section $s\;:\;G_1^m/(G_1^m)' \rightarrow G_1^m$ by

\begin{equation*} s(x_1^{i_1} \cdots x_m^{i_m} y_1^{j_1} \cdots y_m^{j_m} \gamma ^k a_1^{r_1} \cdots a_{n-2m-2}^{r_{n-2m-2}} (G_1^m)' ) = x_1^{i_1} \cdots x_m^{i_m} y_1^{j_1} \cdots y_m^{j_m} \gamma ^k a_1^{r_1} \cdots a_{n-2m-2}^{r_{n-2m-2}} \end{equation*}

Note that for any two elements

\begin{equation*}u= x_1^{i_1} \cdots x_m^{i_m} y_1^{j_1} \cdots y_m^{j_m} \gamma ^k a_1^{r_1} \cdots a_{n-2m-2}^{r_{n-2m-2}}(G_1^m)^{\prime}, v = x_1^{i^{\prime}_1} \cdots x_m^{i^{\prime}_m} y_1^{j^{\prime}_1} \cdots y_m^{j^{\prime}_m} \gamma ^{k^{\prime}} a_1^{r^{\prime}_1} \cdots a_{n-2m-2}^{r^{\prime}_{\!n-2m-2}}(G_1^m)^{\prime}\end{equation*}

of $G_1^m/(G_1^m)^{\prime},$ we have $ s(u)s(v)s(uv)^{-1} = \gamma ^{-p \sum _{l=1}^mj_l i^{\prime}_l}.$ Therefore, for any $\chi \in \mathrm{Hom}((G_1^m)', \mathbb C^\times )$ , a representative of $[\mathrm{tra_1}(\chi )]$ is given by $\mathrm{tra_1}(\chi )(u,v) = \chi (z^{-\sum _{l=1}^mj_l i^{\prime}_l}).$ By a similar computation of ${\mathrm{tra}}_2$ and ${\mathrm{tra}}_3$ , we obtain that for all $i \in \{1, 2, 3\},$ the groups $G_{i}^m$ pairwise satisfy the hypothesis of Theorem 1.5 and hence are $\mathbb C$ -twist isomorphic.

We now complete the details regarding the $\mathbb{C}$ -twist isomorphism classes for $p$ -groups with $s(G)=2$ .

Proof of Theorem 1.2. It follows from [Reference Niroomand24, Theorem 11] that for any fixed $p$ , thhere xists only one $p$ -group of order $p^3$ with $s(G) = 2$ and so it forms a singleton $\mathbb C$ -twist class. We now consider cases when $n \geq 4.$

$n=4$ : Any group of order $p^4$ with $s(G) = 2$ is isomorphic to one of the following:

• $E(2)$

• $H_1^{2} \times C_p, p \neq 2$

• $Q_8 \times C_2$

• $\langle a,b\,|\,a^4=1,b^4=1,[a,b,a]=[a,b,b]=1, [a,b]=a^2b^2 \rangle$

• $ \langle a,b,c\,|\,a^2=b^2=c^2=1, abc=bca=cab \rangle$

• $\langle a,b\,|\,a^{p^2}=1,b^{p}=1, [a,b,a]=[a,b,b]=1 \rangle$

• $\langle a,b\,|\, a^9=b^3=1, [a,b,a] =1, [a,b,b]=a^6, [a,b,b,b]=1 \rangle$

• $\langle a,b\,|\, a^p=1, b^p=1, [a,b,a]=[a,b,b,a]=[a,b,b,b]=1 \rangle$ $(p \neq 3).$

Here, the group $\langle a,b,c\,|\,a^{2}=b^{2}=c^{2} = 1,abc=bca=cab \rangle$ is isomorphic to $E(2).$ As mentioned in Theorem $4.3$ in [Reference Margolis and Schnabel17], any non-singleton $\mathbb{C}$ -twist isomorphism class of groups of order $p^4$ consists of two groups, when $p=2;$ and of three groups, when $p$ is an odd prime. Thus, comparing with the groups given in Tables 3 and 4 in [Reference Margolis and Schnabel17], we obtain the following non-singleton $\mathbb C$ -twist isomorphism classes of groups of order $p^4$ with generalized corank $2$ :

• $\mathbb{Q}_{8} \times C_{2} \sim _{\mathbb{C}} E(2),$ when $p=2$

• $E(2) \sim _{\mathbb{C}} H_1^1 \times C_p \sim _{\mathbb{C}} \langle a,b\,|\,a^{p^2}=1,b^{p}=1, [a,b,a]=[a,b,b]=1 \rangle$ , when $p$ is odd.

Thus, each of the remaining groups of order $p^4$ in the above list constitutes a $\mathbb C$ -twist isomorphism class of size $1.$

$n \geq 5:$ Any group of order $p^n$ with $n \geq 5$ and $s(G) = 2$ is isomorphic to one of the following:

• $E(2) \times C_p^{(n-2m-2)}$

• $H_1^{2} \times C_p^{(n-3)}, p \neq 2$

• $Q_8 \times C_2^{(n-3)}$

• $H_m^1 \times C_p^{(n-2m-1)}$

• $H_m^2 \times C_p^{(n-2m-1)}$

• $C_{p} \times (C_{p}^4 \rtimes _{\theta } C_{p}),\, p \neq 2.$

The derived subgroup of $C_{p} \times (C_{p}^4 \rtimes _{\theta } C_{p}),\,p \neq 2$ is of order $p^2$ ; whereas the derived subgroup of the rest of the groups in the above list is of order $p$ . Therefore, by comparing the complex group algebras, we obtain that for a fixed odd prime $p,$ the group $C_{p} \times (C_{p}^4 \rtimes _{\theta } C_{p})$ forms a singleton $\mathbb C$ -twist isomorphism class. Finally, Proposition 3.1 completes the classification of the rest of the groups into $\mathbb C$ -twist isomorphism classes.

4. $p$ -groups with $s(G)=3$

In this section, we proceed with the determination of the $\mathbb{C}$ -twist isomorphism classes of the $p$ -groups with $s(G) = 3.$ The following result, using the notations of [Reference James12], gives a complete list of $p$ -groups with $s(G) = 3$ .

As mentioned earlier, the $\mathbb{C}$ -twist isomorphism classes of the groups of order $p^4$ were described by Margolis-Schnabel [Reference Margolis and Schnabel17]. We now consider the groups of order $p^5$ with $s(G) = 3$ . Let $H_1$ and $H_2$ denote the groups $\phi _2(2111)d$ and $\phi _2(2111)c,$ respectively. We proceed to prove that $H_1$ and $H_2$ are $\mathbb{C}$ -twist isomorphic. We use the following general result to prove this.

Lemma 4.2. Let $G_1$ and $G_2$ be two finite groups with $\widetilde{G_1}$ and $\widetilde{G_2}$ as their representation groups, respectively. For $i \in \{1, 2\}$ , let $A_i$ be a central subgroup of $\widetilde{G_{i}}$ such that $\widetilde{G_{i}}/A_i \cong G_{i}$ and the transgression maps $\mathrm{tra_i}\;:\; \mathrm{Hom}(A_i, \mathbb C^\times ) \rightarrow{\mathrm{H}}^2(G_{i}, \mathbb C^\times )$ are isomorphisms. Let $\sigma \;:\; \mathrm{Hom}(A_1, \mathbb C^\times ) \rightarrow \mathrm{Hom}(A_2, \mathbb C^\times )$ be an isomorphism such that the following sets are in a dimension preserving bijection for every $\chi \in \mathrm{Hom}(A_1, \mathbb C^\times )$ :

\begin{equation*} \mathrm {Irr}(\widetilde {G_1} \mid \chi ) \leftrightarrow \mathrm {Irr}(\widetilde {G_2} \mid \sigma (\chi )). \end{equation*}

Then $G_1$ and $G_2$ are $\mathbb C$ -twist isomorphic.

Proof. Consider the following diagram:

Define $\tilde{\sigma }\;:\;{\mathrm{H}}^2(G_1, \mathbb C^\times ) \rightarrow{\mathrm{H}}^2(G_2, \mathbb C^\times )$ by $ \tilde{\sigma }({\mathrm{tra}}_1(\chi )) = \mathrm{tra}_2(\sigma (\chi ))$ for $\chi \in \mathrm{Hom}(A_1, \mathbb C^\times )$ . It is easy to see that $\tilde{\sigma }$ is a group isomorphism. By Theorem 2.2, the dimension preserving bijection

\begin{equation*} \mathrm {Irr}(\widetilde {G_1} \mid \chi ) \leftrightarrow \mathrm {Irr}(\widetilde {G_2} \mid \sigma (\chi )) \end{equation*}

for any $\chi \in \mathrm{Hom}(A_1, \mathbb C^\times )$ gives

\begin{equation*} \mathbb C^{\alpha _1}[ G_1] \cong \mathbb C^{\alpha _2}[ G_2 ], \end{equation*}

where $[\alpha _1] = \mathrm{tra}_1(\chi )$ and $[\alpha _2] = \mathrm{tra}_2(\sigma (\chi ))$ . Therefore, $\tilde{\sigma }$ gives the required $\mathbb C$ -twist isomorphism between $G_1$ and $G_2$ .

Note that $H_1 = E_1 \times C_{p^{2}}$ and $H_2 = \langle \alpha, \alpha _1, \alpha _2 \mid [\alpha _1, \alpha ] = \alpha _2, \alpha ^{p^2} = \alpha _1^p = \alpha _2^p = 1 \rangle \times \langle \alpha _3 \rangle$ . Define the following groups:

\begin{eqnarray*} \widetilde{H_1} & = & \langle \alpha _1, \alpha _2, \alpha _3, \alpha _4, x,y,z,\alpha \mid [x,y] = z, [x,z] = \alpha _1, [y,z] = \alpha _2, \\ & & [x,\alpha ] = \alpha _3, [y,\alpha ] = \alpha _4, x^p = y^p = z^p = \alpha _{i}^{p}=\alpha ^{p^2} = 1 \rangle. \end{eqnarray*}

and

\begin{eqnarray*} \widetilde{H_2} & = & \langle x,y,z,w, \alpha, \alpha _1, \alpha _2,\alpha _{3} \mid [\alpha _1, \alpha ] = \alpha _2, [\alpha _1, \alpha _2] = x, [\alpha, \alpha _2] = y, \\ & & [\alpha _3, \alpha _{1}] = z, [\alpha _3,\alpha ] = w, \alpha ^{p^{2}}=\alpha _i^p = x^{p}=y^{p}=z^{p}=w^{p} = 1 \rangle. \end{eqnarray*}

Proof. Here, we give a proof for $H_1$ . The proof for $H_2$ is along the same lines so we omit that part. The group $H_1$ has the following presentation:

\begin{equation*} H_1 = \langle x,y,z, \alpha \mid [x,y] = z, x^p = y^p = \alpha ^{p^2} = 1 \rangle. \end{equation*}

Consider the projection map from $\widetilde{H_1}$ onto $H_1$ obtained by mapping $x,y,z,\alpha$ to $x,y,z,\alpha$ , respectively, and all $\alpha _i$ to $1$ . Let $K_1$ be the kernel of this projection map. Then $|K_1| = |{\mathrm{H}}^2(H_1, \mathbb C^\times )| = p^4$ and $K_1 \subseteq Z(\widetilde{H_1}) \cap [\widetilde{H_1}, \widetilde{H_1}]$ . Therefore, by [Reference Karpilovsky14, Theorem 3.7 (Chapter 3)], $\widetilde{H_1}$ is a representation group of $H_1$ .

Proposition 4.4. The groups $\widetilde{H_1}$ and $\widetilde{H_2}$ satisfy the following:

\begin{equation*} \mathbb {C}[\widetilde {H_1}] \cong \mathbb {C}[\widetilde {H_2}] \cong \mathbb C^{\oplus p^4 } \oplus (\mathrm {M}_p(\mathbb C))^{\oplus (p^3+4p^2-p+1)p^2(p-1)} \oplus (\mathrm {M}_{p^2}(\mathbb C))^{\oplus p^3(p-1)^3(p+2)}. \end{equation*}

Furthermore, for the subgroups $A = \langle \alpha _1, \alpha _2, \alpha _3, \alpha _4 \rangle$ and $B = \langle x,y,z,w \rangle$ of $\widetilde{H_1}$ and $\widetilde{H_2}$ respectively, thehere ists an isomorphism $\sigma \;:\; \widehat{A} \rightarrow \widehat{B}$ such that the following sets are in a dimension preserving bijection for every $\chi \in \hat{A}$ :

\begin{equation*} \mathrm {Irr}(\widetilde {H_1} \mid \chi ) \leftrightarrow \mathrm {Irr}(\widetilde {H_2} \mid \sigma (\chi )). \end{equation*}

Proof. Representations of $\widetilde{H_1}$ : We first justify the representations of $\widetilde{H_1}$ . By the definition of $\widetilde{H_1}$ , the derived subgroup of $\widetilde{H_{1}}$ (denoted $\widetilde{H}^{'}_{1}$ ) is $\langle \alpha _{1},\alpha _{2},\alpha _{3},\alpha _{4}, z\rangle$ and the center of $\widetilde{H_{1}}$ is $\langle \alpha _{1},\alpha _{2},\alpha _{3},\alpha _{4}, \alpha ^p \rangle$ . By considering the quotient group $\widetilde{H_{1}}$ / $\widetilde{H}^{'}_{1}$ , we obtain that $\widetilde{H_1}$ has exactly $p^{4}$ one-dimensional representations.

We next consider the abelian normal subgroup $N = \langle \alpha _{1},\alpha _{2},\alpha _{3},\alpha _{4},\alpha,z \rangle$ of $\widetilde{H_{1}}$ . The group $N$ has order $p^7$ . By Frobenius reciprocity and Theorem 2.3, every irreducible representation of $\widetilde{H_1}$ has dimension either $1$ , $p$ or $p^2$ . We have already justified all one-dimensional representations of $\widetilde{H_{1}}$ . Our next goal is to determine all $p$ and $p^2$ dimensional representations of $\widetilde{H_1}$ .

Let $\chi \in \operatorname{Irr}(N)$ such that $\chi (\alpha _{1})= \xi ^{i_{1}}$ , $\chi (\alpha _{2})= \xi ^{i_{2}}$ , $\chi (\alpha _{3})= \xi ^{i_{3}}$ and $\chi (\alpha _{4})= \xi ^{i_{4}}$ , where $\xi$ is a primitive $p^{th}$ root of unity and $0 \leq i_{1},i_{2},i_{3},i_{4} \leq (p-1)$ . We determine the stabilizer (or inertia group) of $\chi$ in $\widetilde{H_1}$ , denoted by $I_{\widetilde{H_{1}}}(\chi )$ . Recall from Section 2, $I_{\widetilde{H_{1}}}(\chi ) = \{g \in \widetilde{H_{1}} \mid \chi ^g = \chi \}.$ By definition of $I_{\widetilde{H_{1}}}(\chi )$ , $N \leq I_{\widetilde{H_{1}}}(\chi )$ . We will obtain the following information from $|I_{\widetilde{H_{1}}}(\chi )|$ and by the virtue of Theorem 2.3.

1. 1. $\Big{|}{\frac{I_{\widetilde{H_{1}}}(\chi )}{N}}\Big{|} = 1$ implies every $\rho \in \mathrm{Irr}(\widetilde{H_{1}} \mid \chi )$ satisfies $\dim (\rho ) = p^2$ .

2. 2. $\Big{|}{\frac{I_{\widetilde{H_{1}}}(\chi )}{N}}\Big{|} = p$ implies every $\rho \in \mathrm{Irr}(\widetilde{H_{1}} \mid \chi )$ satisfies $\dim (\rho ) = p$ .

3. 3. $\Big{|}{\frac{I_{\widetilde{H_{1}}}(\chi )}{N}}\Big{|} = p^2$ implies every $\rho \in \mathrm{Irr}(\widetilde{H_{1}} \mid \chi )$ satisfies $\dim (\rho ) \in \{1, p\}$ .

As we already have information regarding dimension one representations, we will easily obtain the rest of the information regarding $\mathbb C[\widetilde{H_{1}}]$ from the description of $I_{\widetilde{H_{1}}}(\chi )$ . Consider $g = x^{i}y^{j}n$ , with $n \in N$ and $i,j \in \{0, 1, \ldots, p-1\}$ . Recall that every $m \in N$ satisfies $m = \alpha ^{e}z^{f}h$ for some $h \in Z(\widetilde{H_1}).$ Therefore, we have

\begin{equation*}\chi ^{x^{i}y^{j}}(\alpha ^{e}z^{f}h) = \chi (\alpha ^{e}z^{f}h)\end{equation*}

if and only if $\chi (x^{i}y^{j}\alpha ^{e}z^{f} y^{-j}x^{-i}) = \chi (\alpha ^{e}z^{f}).$ Since

\begin{eqnarray*} x^{i}(y^{j}\alpha ^{e})z^{f} y^{-j}x^{-i} & = & x^{i}(\alpha ^e y^{j}\alpha _4^{je})z^{f}y^{-j}x^{-i} = \alpha ^e x^{i} \alpha _3^{ie}y^{j}\alpha _4^{je}z^{f}y^{-j}x^{-i} \\ & = & \alpha ^e x^{i} y^{j}z^{f}y^{-j}x^{-i}\alpha _3^{ie}\alpha _4^{je} = \alpha ^e x^{i} z^f y^{j} \alpha _2^{jf}y^{-j}x^{-i}\alpha _3^{ie}\alpha _4^{je} \\ & = & \alpha ^e z^{f} \alpha _1^{if} \alpha _2^{jf} \alpha _3^{ie}\alpha _4^{je}, \end{eqnarray*}

we obtain that $x^i y^j n \in I_{\widetilde{H_{1}}}(\chi )$ if and only if $\chi (\alpha _1^{if} \alpha _2^{jf} \alpha _3^{ie}\alpha _4^{je}) = 1$ . This is equivalent to saying that $\xi ^{i_1if+i_2jf+i_3ie+i_4je} = 1.$

We now consider various cases of irreducible representations of $N$ . Note that in each of the cases discussed below, all the computations for $i$ and $j$ are done modulo $p$ .

• Case I: Consider $\chi \in \mathrm{Irr}(N)$ such that $i_1 = i_2 = i_3 = i_4 = 0$ . These are exactly $p^3$ in number. Among these there are $p^2$ characters which act trivially on $z$ and hence on $\widetilde{H}^{'}_{1}$ . These give $p^4$ one-dimensional characters of $\widetilde{H_{1}}$ . The other $(p^3-p^2)$ characters of $N$ , by Theorem 2.3, of this case give irreducible characters of $\widetilde{H_{1}}$ of dimension $p$ and therefore we obtain $p^4-p^3$ many characters of $\widetilde{H_{1}}$ of dimension $p$ .

• Case II: When any three of $i_1, i_2, i_3, i_4$ are $0$ and the fourth one is non-zero, then $g = x^i y^j n \in I_{\widetilde{H_{1}}}(\chi )$ if and only if either $i$ or $j$ is $0$ . Thus, $|I_{\widetilde{H_{1}}}(\chi )| = p^8.$

• Case III: When any two of $i_1, i_2, i_3, i_4$ are $0$ and the other two are non-zero, then $i$ is a non-zero multiple of $j$ or one of them is zero and other can take any value. Hence, $|I_{\widetilde{H_{1}}}(\chi )| = p^8.$

• Case IV: When any three of $i_1, i_2, i_3, i_4$ are non-zero and the fourth one is $0$ , then $i = j = 0$ and hence $I_{\widetilde{H_{1}}}(\chi ) = N.$

• Case V: Assume that each $i_t,$ where $t \in \{1, 2, 3, 4\},$ is non-zero. When $f = 0$ and $e = 1, i = \frac{-i_4j}{i_3};$ and when $e = 0$ and $f = 1, i = \frac{-i_2j}{i_1}.$ Therefore, $\frac{-i_4j}{i_3} = \frac{-i_2j}{i_1},$ which holds if and only if $(i_4i_1-i_2i_3)j = 0.$ Now if $i_4i_1 \neq i_2i_3,$ then $i = j =0$ and hence $I_{\widetilde{H_{1}}}(\chi ) = N.$

  Here, note that $(p-1)^3$ many characters of $N$ satisfy $i_4i_1 = i_2i_3$ and their inertia group is of order $p^8.$ Thus, the remaining $((p-1)^4-(p-1)^3)$ characters have inertia group $N.$

Considering the case of $(p^7-p^3)$ many characters of $N,$ discussed in Cases II-V, the inertia group of $ p^3(p-1)^3(p+2)$ characters is $N,$ and for the other $ p^3(p-1)(p^2+4p-1)$ characters, it is of order $p^8.$ Hence, we obtain the following description of the group algebra of $\widetilde{H_{1}}$ .

\begin{equation*} \mathbb C[\widetilde {H_{1}}] \cong \mathbb C^{\oplus p^4 } \oplus (\mathrm {M}_p(\mathbb C))^{\oplus (p^3+4p^2-p+1)p^2(p-1)} \oplus (\mathrm {M}_{p^2}(\mathbb C))^{\oplus p^3(p-1)^3(p+2)}. \end{equation*}

Representations of $\widetilde{H_{2}}$ : We have $\widetilde{H}^{'}_{2} = \langle x, y, z, w, \alpha _{2} \rangle$ and $Z(\widetilde{H_{2}}) = \langle \alpha ^p, x, y, z, w \rangle$ . Clearly, there are exactly $p^{4}$ number of linear characters of $\widetilde{H_{2}}$ . Consider the subgroup

\begin{equation*}M = \langle \alpha _{2},\alpha _{3}, x, y, z, w, \alpha ^p \rangle \end{equation*}

of $\widetilde{H_{2}}$ . This is an abelian normal group of order $p^7$ . Let $\chi \in \mathrm{Irr}(M)$ such that $\chi (x)= \xi ^{i_{1}}$ , $\chi (y)= \xi ^{i_{2}}$ , $\chi (z)= \xi ^{i_{3}}$ , and $\chi (w)= \xi ^{i_{4}},$ where $\xi$ is a primitive $p^{th}$ root of unity and $0 \leq i_{1},i_{2},i_{3},i_{4} \leq (p-1)$ . Let $g = \alpha ^{i} \alpha _1^{j}m$ , where $m \in M,$ be an element of $I_{G}(\chi ).$ Therefore, for any $m^{\prime} = \alpha _2^{k}\alpha _3^{l}h \in M,$ where $h \in Z(\widetilde{H_1}),$ we have $\chi ^{\alpha ^{i} \alpha _1^{j}}(\alpha _2^{k}\alpha _3^{l}h) = \chi (\alpha _2^{k}\alpha _3^{l}h).$ Computations, similar to $\widetilde{H_{1}}$ , yield that $g \in I_{\widetilde{H_{2}}}(\chi )$ if and only if $\xi ^{i_1jk+i_2ik+i_3(-jl)+i_4(-il)} = 1$ and hence the following cases arise:

• Case I: When any three of $i_1, i_2, i_3, i_4$ are $0$ and the fourth one is non-zero, then either $i$ or $j$ is $0.$ Thus, $|I_{\widetilde{H_{2}}}(\chi )| = p^8.$

• Case II: When any two of $i_1, i_2, i_3, i_4$ are $0$ and the other two are non-zero, then $i$ is a non-zero multiple of $j$ or one of them is zero and other can take any value. Hence, $|I_{\widetilde{H_{2}}}(\chi )| = p^8.$

• Case III: When any three of $i_1, i_2, i_3, i_4$ are non-zero and the fourth one is $0$ , then $i = j = 0$ and hence $I_{\widetilde{H_{2}}}(\chi ) = M.$

• Case IV: Assume that each $i_t,$ where $t \in \{1, 2, 3, 4\},$ is non-zero. When $l = 0$ and $k = 1, i = \frac{-i_1j}{i_2};$ and when $k = 0$ and $l = 1, i = \frac{-i_3j}{i_4}.$ Therefore, $\frac{-i_3j}{i_4} = \frac{-i_1j}{i_2},$ which holds if and only if $(i_4i_1-i_2i_3)j = 0.$ Now if $i_4i_1 \neq i_2i_3,$ then $i = j =0$ and hence $I_G(\chi ) = M.$ On the other hand, if $i_4 = \frac{i_2i_3}{i_1},$ then $|I_G(\chi )| = p^8.$

  Therefore, in this case the inertia group of $(p-1)^3$ many characters is of order $p^8$ and for the remaining $((p-1)^4-(p-1)^3)$ characters it is $M.$

Similar to the proof of $\widetilde{H_{1}}$ , all the above cases along with Theorem 2.3 give

\begin{equation*} \mathbb C[\widetilde {H_{2}}] \cong \mathbb C^{\oplus p^4 } \oplus (\mathrm {M}_p(\mathbb C))^{\oplus (p^3+4p^2-p+1)p^2(p-1)} \oplus (\mathrm {M}_{p^2}(\mathbb C))^{\oplus p^3(p-1)^3(p+2)}. \end{equation*}

We now proceed to define required isomorphism $\sigma \;:\;\widehat{A} \rightarrow \widehat{B}$ . For this, define an isomorphism $\sigma '\;:\; A \rightarrow B$ by $\sigma '(\alpha _1) = x, \sigma '(\alpha _2) = y, \sigma '(\alpha _3) = z, \sigma '(\alpha _4) = w.$ This defines an isomorphism, denoted by $\sigma$ , between $\widehat{A}$ and $\widehat{B}$ . From above discussion, by considering various cases of $i_j$ for $j \in \{1, \ldots, 4\}$ , we obtain a dimension preserving bijection between the following sets for every $\chi \in \widehat{A}$ :

\begin{equation*} \mathrm {Irr}(\widetilde {H_1} \mid \chi ) \leftrightarrow \mathrm {Irr}(\widetilde {H_2} \mid \sigma (\chi )). \end{equation*}

Proof. This follows from Lemmas 4.2, 4.3 and Proposition 4.4.

Proof. It follows from the presentations of $\phi _{3}(1^{5})$ and $\phi _{7}(1^{5})$ that the nilpotency class of both the groups is $3.$ Now, consider the abelian normal subgroup $N_1 = \langle \alpha _1, \alpha _2, \alpha _3, \beta \rangle$ of $\phi _3(1^5).$ Since it is of index $p,$ each irreducible representation of $\phi _3(1^5)$ is of dimension at most $p.$

Now note that the derived subgroup of $\phi _{7}(1^5)$ is $\langle \alpha _2 \rangle \times \langle \alpha _3 \rangle$ and its center is $\langle \alpha _3 \rangle$ . Consider the abelian normal subgroup $N_2 = \langle \alpha _1, \alpha _2, \alpha _3 \rangle$ of $\phi _{7}(1^5)$ . Let $\chi \in \mathrm{Irr}(N_2)$ such that $\chi (\alpha _1) = \zeta ^{i_1}, \chi (\alpha _2) = \zeta ^{i_2}$ and $\chi (\alpha _3) = \zeta ^{i_3}$ , where $\zeta$ is a primitive $p$ -th root of unity and $0 \leq i_1, i_2, i_3 \leq (p-1).$ Assume that for some $0 \leq i, j \leq p-1,$ $\alpha ^i \beta ^j$ stabilizes $\chi .$ Let $\alpha _{1}^{a}\alpha _{2}^{b}\alpha _{3}^{c}\in N_{2}$ .

Since the group $\phi _{7}(1^5)$ is of nilpotency class $3$ ,

\begin{eqnarray*} \alpha ^i (\beta ^j \alpha _1^a) \alpha _2^b \alpha _3^c \beta ^{-j} \alpha ^{-i} & = & \alpha ^i \alpha _1^a \beta ^j \alpha _2^b \beta ^{-j} \alpha ^{-i} \alpha _3^{c-aj} \\ & = & \alpha _2^{-ai} \alpha _1^a \alpha ^i \beta ^j \alpha _2^b \beta ^{-j} \alpha ^{-i} \alpha _3^{c-aj-a\binom{i}{2}} \\ & = & \alpha _1^a \alpha _2^{b-ai} \alpha _3^{-ib+c-aj-a\binom{i}{2}} =\alpha _{1}^{a}\alpha _{2}^{b}\alpha _{3}^{c}\alpha _{2}^{-ai}\alpha _{3}^{-ib-aj-a\binom{i}{2}}. \end{eqnarray*}

Thus, $\alpha ^{i}\beta ^{j}$ stabilizes $\chi$ if, and only if, $\zeta ^{-aii_2-(ib+aj+a\binom{i}{2})i_3} = 1.$ When $i_2 = 0$ and $i_3 \neq 0,$ then for $a=0$ and $b = \frac{-1}{i_3},$ we have $\zeta ^{i}=1;$ which implies that $i=0.$ Further, $a=\frac{-1}{i_3}$ (note that $i_{3}$ is invertible modulo $p$ ) gives $\zeta ^{j} =1$ . Hence, $j=0$ and it follows that the inertia group of $\chi$ in $\phi _{7}(1^{5})$ is $N_2.$ Therefore, by Theorem 2.3, the character of $\phi _7(1^5)$ induced from $\chi$ is irreducible of degree $p^2.$ Since $\phi _3(1^5)$ has no irreducible representations of dimension $p^2,$ the complex group algebras of $\phi _3(1^5)$ and $\phi _7(1^5)$ are not isomorphic.

Proof of Theorem 1.3. From Theorem 4.1, it is clear that every $p$ -group with $s(G) = 3$ has order $p^n$ where $n \in \{4, \ldots, 7\}$ .

In the following, we separately consider the cases of $n$ with $4 \leq n \leq 7$ :

n = 4. For the proof of this case, one can refer to [Reference Margolis and Schnabel17, Theorem 4.3].

n = 5. When $p$ is an odd prime, it follows from Theorem 4.1 that the only groups of order $p^5$ with $s(G) = 3$ are $\phi _3(1^5)$ , $\phi _7(1^5), H_1$ and $H_2$ . The derived subgroups of $\phi _3(1^5)$ and $\phi _7(1^5)$ are elementary abelian of order $p^2$ and of $H_1$ and $H_2$ are of order $p.$ Thus, the groups $\phi _3(1^5)$ and $\phi _7(1^5)$ have $p^3$ linear characters; whehere $H_1$ and $H_2$ have $p^4$ linear characters. Therefore, no group in the set $\{\phi _3(1^5), \phi _7(1^5)\}$ can be $\mathbb{C}$ -twist isomorphic to any group in $\{H_1, H_2\}.$ Now it follows from Lemma 4.6 and Proposition 4.5 that in this case the only non-singleton $\mathbb{C}$ -twist isomorphism class is constituted by $H_1$ and $H_2.$

When $p = 2$ , $T_1 = C_{2}^{4} \rtimes C_{2}$ and $T_2 = C_{2} \times ((C_{4}\times C_{2}) \rtimes C_{2})$ are the only two $2$ -groups of order $32$ that have generalized corank $3$ . The GAP ID of these groups is [32,27] and [32,22] and it can be checked using GAP [5] that the size of the derived subgroup of $T_1$ is $4$ and that of $T_2$ is $2.$ Thus, the complex group algebras of these groups are not isomorphic and hence each of these groups constitutes a singleton $\mathbb{C}$ -twist isomorphism class.

n = 6. The groups of order $p^6$ with $s(G) = 3$ are $\phi _{11}(1^6), \phi _{12}(1^6), \phi _{13}(1^6)$ and $\phi _{15}(1^6).$ Note that the size of the commutator subgroup of $\phi _{11}(1^6)$ is $p^3$ and of the rest of the groups is $p^2.$

It can be checked that $\phi _{12}(1^6) = ES_p(p^3) \times ES_p(p^3)$ has $2p^2(p-1)$ inequivalent irreducible representations of dimension $p$ and $(p-1)^2$ of dimension $p^2.$ Further, it follows from [Reference James12, Table 4.1] that $\phi _{13}(1^6)$ has $(p^3-p^2)$ representations of dimension $p$ ; whereas $\phi _{15}(1^6)$ has no representation of dimension $p$ . Thus, the complex group algebras of the groups of order $p^6$ with generalized corank $3$ are not isomorphic. It establishes the desired result.

n = 7. There is a unique group of order $p^7$ with $s(G) = 3$ which is isomorphic to $C_{p}^{(4)} \rtimes C_{p} \times C_{p}^{2}$ . Therefore, it constitutes a singleton $\mathbb C$ -twist isomorphism class.