Isoclinic groups have same proportions of degrees of irreducible representations
Suppose and are finite groups that are isoclinic groups. Suppose is a positive integer. Denote by the number of equivalence classes of irreducible linear representations of of degree and denote by the number of equivalence classes of irreducible linear representations of of degree . Then, is nonzero if and only if is nonzero, and if so, we have:
In other words, given the degrees of irreducible representations of , we can obtain the degrees of irreducible representations of by scaling the number of occurrences of each degree by a factor of .
In particular, if and 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.
- Criterion for projective representation to lift to linear representation
- Character determines representation in characteristic zero
- Third isomorphism theorem
- Fundamental theorem of group actions
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 if and only if there exists a lift to a linear representation of .
- The number of such lifts are in the ratio .
Showing the result from that point onwards is straightforward.
Given: Two isoclinic groups and , a positive integer . and are respectively the number of irreducible representations of and of degree over .
To prove: is nonzero if and only if is nonzero, and if so, .
Proof: We prove some subclaims. Let be the group and be the group . Denote by the map obtained from the commutator map in either group (we know both maps are equivalent via the isoclinism). Denote by and the quotient maps modulo the respective centers. In other words, we have the following short exact sequences:
Existence of lift of projective representation
Given: A projective representation . Let be the natural quotient map.
To prove: There exists a linear representation that is a lift of (i.e., ) if and only if there exists a linear representation that is a lift of (i.e., ).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 that lifts to linear representations for both and . Let be the natural quotient map.
To prove: The number of lifts to and the number of lifts to are in the ratio .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||The group of one-dimensional representations of , which is identified as the Pontryagin dual of , acts on the set of all linear lifts of by scalar multiplication, and the action is transitive.||That there is an action is clear. For transitivity, if are two representations, then gives the acting element taking one to the other.|
|2||Under the action of Step (1), the stabilizer of any linear lift is the set of those one-dimensional representations of whose kernel contains all the points where 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 has a nonzero-valued character, then that means that for any , either has trace zero, or is the identity. Thus, in all cases, we have that and have the same trace, so the representations and have equal character. By Fact (2), this means that and are equivalent as representations.
Stabilizer implies kernel contains all nonzero-valued trace points: if is in the stabilizer and is such that has nonzero-valued trace, then forces that and have the same trace. Since has nonzero trace, this forces .
|3||The stabilizer as determined in Step (2) is the set of one-dimensional representations whose kernel contains the inverse image in of the subgroup of , which we denote , defined as follows: it is generated by and the set of for which any representative of 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 , "whether or not the trace is zero" is independent of the choice of representative.|
|4||The size of the stabilizer of in the Pontryagin dual is , where is as determined in Step (3): it is generated by and the set of for which any representative of has a nonzero trace.||Fact (3)||Step (3)||A reinterpretation of Step (3) gives that the stabilizer is the Pontryagin dual of , so its size is , which, by Fact (3), is the same as .|
|5||The number of lifts of to is .||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 to is .||Steps (1) - (5) (analogous reasoning)|
|7||The quotient of the number of lifts to to to is .||Note that 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 of degree . Next, filter this to the set of those that have linear lifts to (and equivalently, to ). For each such, the number of lifts to and to are in the ratio . Hence, the overall number of irreducible linear representations of and of degree are in the ratio .