Isoclinic groups have same proportions of conjugacy class sizes

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Statement

Suppose G_1 and G_2 are finite groups that are isoclinic groups. Suppose c is a positive integer. Denote by m_1 the number of conjugacy classes of G_1 of size c and denote by m_2 the number of conjugacy classes of G_2 of size c. Then, m_1 is nonzero if and only if m_2 is nonzero, and if so, we have:

\frac{m_1}{m_2} = \frac{|G_1|}{|G_2|}

In other words, given the conjugacy class size statistics of G_1, we can obtain the conjugacy class size statistics of G_2 by scaling the number of occurrences of each conjugacy class size by a factor of |G_2|/|G_1|.

In particular, if G_1 and G_2 also have the same order, then they have precisely the same conjugacy class size statistics.

Relation with Schur covering groups

All the Schur covering groups of a given finite group are isoclinic groups, hence have the same conjugacy class size statistics.

Related facts

Facts used

  1. Size of conjugacy class equals index of centralizer

Proof

Proof outline

The idea behind the proof is to show that the size of the conjugacy class of an element depends only on its coset modulo the center, and is completely determined by the information of the commutator map. We use Fact (1).

Proof details

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Given: Two isoclinic groups G_1 and G_2, a positive integer c. m_1 and m_2 are respectively the number of conjugacy classes in G_1 and G_2 of size c. Note that the actual number of elements in G_1 and G_2 with these conjugacy class sizes are m_1c and m_2c respectively.

To prove: m_1 is nonzero if and only if m_2 is nonzero, and if so, \frac{m_1}{m_2} = \frac{|G_1|}{|G_2|}

Proof: Let W be the group \operatorname{Inn}(G_1) \cong \operatorname{Inn}(G_2) and T be the group G_1' \cong G_2'. Denote by \gamma: W \times W \to T the map obtained from the commutator map in either group (we know both maps are equivalent via the isoclinism). Denote by \alpha_1:G_1 \to W and \alpha_2:G_2 \to W the quotient maps modulo the respective centers.

Step no. Asssertion/construction Facts used Given data used (column to be filled) Previous steps used Explanation
1 For w \in W, the centralizer in G_1 of any element in \alpha_1^{-1}(w) is precisely \alpha_1^{-1}(\mathcal{C}(w)) where \mathcal{C}(w) = \{ u \in W \mid \gamma(u,w) \mbox{ is the identity element of } T \}. This follows by definition of centralizer.
2 For any w \in W, the size of the conjugacy class in G_1 of any element in \alpha_1^{-1}(w) is the index of the subgroup \mathcal{C}(w) = \{ u \in W \mid \gamma(u,w) \mbox{ is the identity element of } T \} in W. Fact (1) Step (1) By Fact (1), the conjugacy class size is [G_1:\alpha_1^{-1}(\mathcal{C}(w))], which, by the fourth isomorphism theorem, is [W:\mathcal{C}(w)].
3 The set of elements of G_1 with conjugacy class size c is \alpha_1^{-1}(S) where S is the set of w \in W for which the index of the subgroup \mathcal{C}(w) = \{ u \in W \mid \gamma(u,w) \mbox{ is the identity element of } T \} in W is c. Step (2)
4 m_1c = |S||Z(G_1)| where S is as defined in Step (3). Both sides of the equality represent the number of elements in conjugacy classes of size c in G_1. Step (3) follows directly, since the set is the union of |S| many cosets, each of size |Z(G_1)|.
5 m_2c = |S||Z(G_2)| where S is as defined in Step (3). Both sides of the equality represent the number of elements in conjugacy classes of size c in G_2. Analogous to reasoning for Steps (1)-(4) Replace G_1 and \alpha_1 with G_2 and \alpha_2 throughout. Note that S is defined in the same way in both cases.
6 m_1 is nonzero if and only if m_2 is nonzero, and if so, m_1/m_2 = |Z(G_1)|/|Z(G_2)|. Steps (4), (5) Direct by taking the quotient of the descriptions.
7 m_1 is nonzero if and only if m_2 is nonzero, and if so, m_1/m_2 = |G_1|/|G_2|. Step (6) We have that [G_1:Z(G_1)] = [G_2:Z(G_2)] since the inner automorphism groups are isomorphic, so rearranging gives |G_1|/|G_2| = |Z(G_1)|/|Z(G_2)|. Combine with Step (5) to get the result.