Much as
character theory simplifies working with
group representations,
marks simplify working with
permutation representations and the Burnside ring. If
G acts on
X, and
H ≤
G (
H is a
subgroup of
G), then the
mark of
H on
X is the number of elements of
X that are fixed by every element of
H: m_X(H) = \left|X^H\right|, where :X^H = \{ x\in X \mid h\cdot x = x, \forall h\in H\}. If
H and
K are conjugate subgroups, then
mX(
H) =
mX(
K) for any finite
G-set
X; indeed, if
K =
gHg−1 then
XK =
g ·
XH. It is also easy to see that for each
H ≤
G, the map
Ω(
G) →
Z :
X ↦
mX(
H) is a homomorphism. This means that to know the marks of
G, it is sufficient to evaluate them on the generators of
Ω(
G),
viz. the orbits
G/
H. For each pair of subgroups
H,
K ≤
G define :m(K, H) = \left|[G/K]^H\right| = \# \left\{ gK \in G/K \mid HgK=gK \right\}. This is
mX(
H) for
X =
G/
K. The condition
HgK =
gK is equivalent to
g−1
Hg ≤
K, so if
H is not conjugate to a subgroup of
K then
m(
K,
H) = 0. To record all possible marks, one forms a table, Burnside's
Table of Marks, as follows: Let
G1 (= trivial subgroup),
G2, ...,
GN =
G be representatives of the
N conjugacy classes of subgroups of
G, ordered in such a way that whenever
Gi is conjugate to a subgroup of
Gj, then
i ≤
j. Now define the
N ×
N table (
square matrix) whose (
i,
j)th entry is
m(
Gi,
Gj). This matrix is lower triangular, and the elements on the diagonal are non-zero so it is invertible. It follows that if
X is a
G-set, and
u its
row vector of marks, so
ui =
mX(
Gi), then
X decomposes as a
disjoint union of
ai copies of the orbit of type
Gi, where the vector
a satisfies, :
aM =
u, where
M is the matrix of the table of marks. This theorem is due to . ==Examples==