[Paper Review] Finite $p$-groups of conjugate type $\{ 1, p^3 \}$
This paper classifies finite $p$-groups of conjugate type $\{1, p^3\}$ up to isoclinism, proving they must have nilpotency class 2. It identifies four isoclinism types: Camina $p$-groups with commutator subgroup of order $p^3$, the group $G_3$ from Ito's construction, and two central quotients of $G_3$ for odd $p$. For $p=2$, a similar classification holds, including a family of 2-groups of order $2^{10}$ and exponent 4.
We classify finite $p$-groups, upto isoclinism, which have only two conjugacy class sizes $1$ and $p^3$. It turns out that the nilpotency class of such groups is $2$.
Motivation & Objective
- To classify finite $p$-groups with exactly two conjugacy class sizes: 1 and $p^3$, up to isoclinism.
- To determine the nilpotency class of such groups, resolving a question about whether class 3 groups can exist for this conjugate type.
- To extend previous classifications of $p$-groups of conjugate type $\{1,p\}$ and $\{1,p^2\}$ to the case $\{1,p^3\}$.
- To include the case $p=2$, where the structure of such groups differs from odd $p$, particularly in exponent and isoclinism types.
- To provide a complete isoclinism classification for $2$-groups of conjugate type $\{1,8\}$, including central quotients of a specific group $\mathcal{G}$.
Proposed method
- Use isoclinism as the equivalence relation to classify $p$-groups with conjugate type $\{1,p^3\}$, reducing the problem to classifying groups up to this invariant.
- Apply the theory of Camina groups and special $p$-groups to analyze groups of nilpotency class 2 with $|G'| = p^3$.
- Utilize Ito's construction of the group $G_r$ to define $G_3$, a special $p$-group of exponent $p$ and order $p^{10}$ with conjugacy class sizes $1$ and $p^3$.
- Analyze central quotients of $G_3$ by subgroups of order $p$ and $p^2$ to identify additional isoclinism types, particularly for $p>2$.
- For $p=2$, use Wilkens' classification result to eliminate certain isoclinism types and show only Camina groups and quotients of $\mathcal{G}_3$ (a 2-group of order $2^{10}$) remain.
- Employ group-theoretic tools including the Frattini subgroup $\Phi(G)$, the center $Z(G)$, and the commutator subgroup $G'$, with the condition $Z(G) \leq G'$ to constrain structure.
Experimental results
Research questions
- RQ1Can finite $p$-groups of conjugate type $\{1, p^3\}$ have nilpotency class greater than 2?
- RQ2What are the isoclinism classes of finite $p$-groups with conjugacy class sizes $1$ and $p^3$?
- RQ3How do the isoclinism types differ between odd primes $p$ and $p=2$?
- RQ4Are there $p$-groups of conjugate type $\{1, p^3\}$ that are not quotients of Ito's group $G_3$?
- RQ5What is the structure of $2$-groups of conjugate type $\{1,8\}$, and how do they relate to the family $\hat{\mathcal{G}}_3$?
Key findings
- Finite $p$-groups of conjugate type $\{1, p^3\}$ for $p > 2$ have nilpotency class exactly 2.
- For $p > 2$, such groups are isoclinic to one of four types: Camina groups with $|G'| = p^3$, $G_3$, $G_3/M$ with $|M| = p$, or $G_3/N$ with $|N| = p^2$ and $N$ generated by two commutators with a non-square parameter.
- The group $G_3$ from Ito's construction is a special $p$-group of exponent $p$, order $p^{10}$, and has conjugacy class sizes $1$ and $p^3$.
- For $p = 2$, the classification includes Camina $2$-groups with $|G'| = 8$, the group $\mathcal{G} \in \hat{\mathcal{G}}_3$ of order $2^{10}$, and two central quotients $\mathcal{G}/M$ and $\mathcal{G}/N$ with $|M| = 2$, $|N| = 4$.
- The family $\hat{\mathcal{G}}_3$ of $2$-groups of class 2 and order $2^{10}$ contains exactly 989 non-isomorphic groups, all of which are of conjugate type $\{1,8\}$.
- Groups of conjugate type $\{1, p^3\}$ cannot be of nilpotency class 3, as shown by contradiction using the structure of $Z(G)$ and $G'$, and the fact that $|G:Z(G)| = p^4$ forces $G$ to be minimally generated by 4 elements and isoclinic to $G_3$ or its quotients.
Better researchstarts right now
From reading papers to final review, dramatically reduce your research time.
No credit card · Free plan available
This review was created by AI and reviewed by human editors.