Isoclinic groups have same proportions of conjugacy class sizes
Contents
Statement
Suppose and
are finite groups that are isoclinic groups. Suppose
is a positive integer. Denote by
the number of conjugacy classes of
of size
and denote by
the number of conjugacy classes of
of size
. Then,
is nonzero if and only if
is nonzero, and if so, we have:
In other words, given the conjugacy class size statistics of , we can obtain the conjugacy class size statistics of
by scaling the number of occurrences of each conjugacy class size by a factor of
.
In particular, if and
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
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 and
, a positive integer
.
and
are respectively the number of conjugacy classes in
and
of size
. Note that the actual number of elements in
and
with these conjugacy class sizes are
and
respectively.
To prove: is nonzero if and only if
is nonzero, and if so,
Proof: 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.
Step no. | Asssertion/construction | Facts used | Given data used (column to be filled) | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | For ![]() ![]() ![]() ![]() ![]() |
This follows by definition of centralizer. | |||
2 | For any ![]() ![]() ![]() ![]() ![]() |
Fact (1) | Step (1) | By Fact (1), the conjugacy class size is ![]() ![]() | |
3 | The set of elements of ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
Step (2) | |||
4 | ![]() ![]() ![]() ![]() |
Step (3) | follows directly, since the set is the union of ![]() ![]() | ||
5 | ![]() ![]() ![]() ![]() |
Analogous to reasoning for Steps (1)-(4) | Replace ![]() ![]() ![]() ![]() ![]() | ||
6 | ![]() ![]() ![]() |
Steps (4), (5) | Direct by taking the quotient of the descriptions. | ||
7 | ![]() ![]() ![]() |
Step (6) | We have that ![]() ![]() |