Isoclinic groups have same proportions of degrees of irreducible representations

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

Isoclinic groups have same proportions of conjugacy class sizes

Facts used

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	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 $|G_{1}|/|G_{2}|$ .