Isoclinic groups have same proportions of degrees of irreducible representations

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Statement

Suppose G_1 and G_2 are finite groups that are isoclinic groups. Suppose d is a positive integer. Denote by m_1 the number of equivalence classes of irreducible linear representations of G_1 of degree d and denote by m_2 the number of equivalence classes of irreducible linear representations of G_2 of degree d. 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|}

All this is over a field that is a splitting field for both G_1 and G_2. Note that the minimal splitting field for G_1 may be different from that of G_2.

In other words, given the degrees of irreducible representations of G_1, we can obtain the degrees of irreducible representations of G_2 by scaling the number of occurrences of each degree 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 degrees of irreducible representations.

Relation with Schur covering groups

All the Schur covering groups of a given finite group have the same collection of degrees of irreducible representations and these correspond, with some adjustment for multiplicity, to the degrees of the irreducible projective representations of the original finite group.

Related facts

Facts used

  1. Criterion for projective representation to lift to linear representation
  2. Character determines representation in characteristic zero
  3. Third isomorphism theorem
  4. Fundamental theorem of group actions

Proof

Proof idea

The idea will be to show the following two facts:

  • For any irreducible projective representation of the inner automorphism lift, there exists a lift to a linear representation of G_1 if and only if there exists a lift to a linear representation of G_2.
  • The number of such lifts are in the ratio |G_1|/|G_2|.

Showing the result from that point onwards is straightforward.

Proof details

Given: Two isoclinic groups G_1 and G_2, a positive integer d. m_1 and m_2 are respectively the number of irreducible representations of G_1 and G_2 of degree d over \mathbb{C}.

To prove: m_1 is nonzero if and only if m_2 is nonzero, and if so, m_1/m_2 = |G_1|/|G_2|.

Proof: We prove some subclaims. 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. In other words, we have the following short exact sequences:

\! 1 \to Z(G_1) \to G_1 \stackrel{\alpha_1}{\to} W \to 1

\! 1 \to Z(G_2) \to G_2 \stackrel{\alpha_2}{\to} W \to 1

Existence of lift of projective representation

Given: A projective representation \rho: W \to PGL_d(\mathbb{C}). Let \pi_d: GL_d(\mathbb{C}) \to PGL_d(\mathbb{C}) be the natural quotient map.

To prove: There exists a linear representation \theta_1: G_1 \to GL_d(\mathbb{C}) that is a lift of \rho (i.e., \pi_d \circ \theta_1 = \rho \circ \alpha_1) if and only if there exists a linear representation \theta_2: G_2 \to GL_d(\mathbb{C}) that is a lift of \rho (i.e., \pi_d \circ \theta_2 = \rho \circ \alpha_2).

Proof: By Fact (1), checking that the projective representation lifts is equivalent to checking a certain condition, that turns out to be invariant under isoclinism. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

Proportionality of number of lifts

Given: A projective representation \rho: W \to PGL_d(\mathbb{C}) that lifts to linear representations for both G_1 and G_2. Let \pi_d: GL_d(\mathbb{C}) \to PGL_d(\mathbb{C}) be the natural quotient map.

To prove: The number of lifts to G_1 and the number of lifts to G_2 are in the ratio |G_1|/|G_2|.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The group of one-dimensional representations of G_1, which is identified as the Pontryagin dual of G_1/G_1', acts on the set of all linear lifts of \rho by scalar multiplication, and the action is transitive. That there is an action is clear. For transitivity, if \theta_1,\mu_1 are two representations, then g \mapsto \theta_1(g)\mu_1(g)^{-1} gives the acting element taking one to the other.
2 Under the action of Step (1), the stabilizer of any linear lift \theta_1 is the set of those one-dimensional representations \beta of G_1/G_1' whose kernel contains all the points where \theta_1 has a nonzero-valued trace. Fact (2) Step (1) (for jargon) We need to show both directions.
Kernel contains all nonzero-valued trace points implies in stabilizer: If a one-dimensional representation has a kernel containing all the points where \theta_1 has a nonzero-valued character, then that means that for any g \in G_1, either \theta_1(g) has trace zero, or \beta(g) is the identity. Thus, in all cases, we have that \beta(g)\theta_1(g) and \theta_1(g) have the same trace, so the representations \beta \theta_1 and \theta_1 have equal character. By Fact (2), this means that \beta\theta_1 and \theta_1 are equivalent as representations.
Stabilizer implies kernel contains all nonzero-valued trace points: if \beta is in the stabilizer and g is such that \theta_1(g) has nonzero-valued trace, then \beta\theta_1 \simeq \theta_1 forces that \beta(g)\theta_1(g) and \theta_1(g) have the same trace. Since \theta_1(g) has nonzero trace, this forces \beta(g) = 1.
3 The stabilizer as determined in Step (2) is the set of one-dimensional representations whose kernel contains the inverse image in G_1 of the subgroup of W, which we denote \mathcal{N}(\rho), defined as follows: it is generated by W' and the set of g \in W for which any representative of \rho(g) has a nonzero trace. Step (2) Direct from Step (2). Note that although the trace is not the same for different representatives of an element of PGL_d(\mathbb{C}), "whether or not the trace is zero" is independent of the choice of representative.
4 The size of the stabilizer of \rho in the Pontryagin dual is [W:\mathcal{N}(\rho)], where \mathcal{N}(\rho) is as determined in Step (3): it is generated by W' and the set of g \in W for which any representative of \rho(g) has a nonzero trace. Fact (3) Step (3) A reinterpretation of Step (3) gives that the stabilizer is the Pontryagin dual of G_1/\alpha_1^{-1}(\mathcal{N}(\rho)), so its size is [G_1:\alpha_1^{-1}(\mathcal{N}(\rho))], which, by Fact (3), is the same as [W:\mathcal{N}(\rho)].
5 The number of lifts of \rho to G_1 is |G_1/G_1'|/[W:\mathcal{N}(\rho)] = \frac{|G_1||\mathcal{N}(\rho)|}{|G_1'||W|}. Fact (4) Steps (1), (4) We use Fact (4) and get that the size of the orbit is the index of the stabilizer. Simplifying, we get the expression indicated.
6 The number of lifts of \rho to G_1 is |G_2/G_2'|/[W:\mathcal{N}(\rho)] = \frac{|G_2||\mathcal{N}(\rho)|}{|G_2'||W|}. Steps (1) - (5) (analogous reasoning)
7 The quotient of the number of lifts to G_1 to to G_2 is |G_1|/|G_2|. Note that |G_1'| = |G_2'| cancels, and we get the result.

Proportionality of the overall number of representations

We can now finish off the proof: first, list the set of all irreducible projective representations of W of degree d. Next, filter this to the set of those that have linear lifts to G_1 (and equivalently, to G_2). For each such, the number of lifts to G_1 and to G_2 are in the ratio |G_1|/|G_2|. Hence, the overall number of irreducible linear representations of G_1 and G_2 of degree d are in the ratio |G_1|/|G_2|.