# Irreducible factors of primitive permutation group representation

Consider a primitive permutation group $$\Gamma\subseteq\mathrm{Sym}(N)$$ on the $$n$$-element set $$N=\{1,...,n\}$$, that is, $$\Gamma$$ does not preserve any non-trivial partition of $$N$$.

Consider the linear representation of $$\Gamma$$ by permutation matrices on $$\Bbb R^n$$ in the obvious way. What can be said about the decomposition of this representation into irreducible factors? I am especially interested in the multiplicities of the factors: is it known whether any two irreducible factors in such a decomposition are non-isomorphic?

• There are primitive permutation groups for which $n$ exceeds the sum of the degrees of all the complex irreducible representations. For example, $M_{24}$ has a primitive permutation representation in which a point stabilizer is $L_2(7)$. Also there are infinitely many $PSL_2(q)$, $q$ odd prime, with a maximal subgroup isomorphic to $A_5$. – Richard Lyons Jul 29 at 14:44

An example in which there are two isomorphic irreducible modules in the decomposition is the group $${\rm PSL}(2,11)$$ in its primitive permutation representation of degree $$55$$ coming from the action of $$G$$ on the cosets of a dihedral subgroup of order $$12$$. The permutation module over the real numbers decomposes into modules of dimensions $$1,10,10,10,12,12$$, where two of the $$10$$-dimensional constituents are isomorphic.

Here is a calculation in Magma that verifies this. I am doing this calculation over the complex field, where the decomposition is $$1+5+5+10+10+12+12$$, but note that the two $$5$$-dimensional constituents are contragredient, and they combine to make a $$10$$- real representation.

> G := PrimitiveGroup(55,1);
> ChiefFactors(G);
G
|  A(1, 11)                   = L(2, 11)
1
> CT := CharacterTable(G);
> CT;

Character Table of Group G
--------------------------

-------------------------------------------
Class |    1  2   3    4    5   6    7    8
Size  |    1 55 110  132  132 110   60   60
Order |    1  2   3    5    5   6   11   11
-------------------------------------------
p  =  2    1  1   3    5    4   3    8    7
p  =  3    1  2   1    5    4   2    7    8
p  =  5    1  2   3    1    1   6    7    8
p  = 11    1  2   3    4    5   6    1    1
-------------------------------------------
X.1   +    1  1   1    1    1   1    1    1
X.2   0    5  1  -1    0    0   1   Z2 Z2#2
X.3   0    5  1  -1    0    0   1 Z2#2   Z2
X.4   +   10 -2   1    0    0   1   -1   -1
X.5   +   10  2   1    0    0  -1   -1   -1
X.6   +   11 -1  -1    1    1  -1    0    0
X.7   +   12  0   0   Z1 Z1#2   0    1    1
X.8   +   12  0   0 Z1#2   Z1   0    1    1

Explanation of Character Value Symbols

# denotes algebraic conjugation, that is,
#k indicates replacing the root of unity w by w^k

Z1     = (CyclotomicField(5: Sparse := true)) ! [ RationalField() | 0, 0, 1, 1 ]
Z2     = (CyclotomicField(11: Sparse := true)) ! [ RationalField() | 0, 1, 0, 1,
1, 1, 0, 0, 0, 1 ]

> K := CyclotomicField(55);
> M := PermutationModule(G,K);
> c := Character(M);
> Decomposition(CT,c);
[ 1, 1, 1, 0, 2, 0, 1, 1 ]

• Thank you for this nice example. I am indeed mainly interested in real irreducible represenations, so this fits perfectly. Do you see any immediate reason why the dimension of such an irreducible representation of multiplicity $\ge 2$ has to be even? I would be highly interested in an example of odd dimension if you know of one. – M. Winter Jul 29 at 15:58
• Yes, the primitive permutation representation of degree $91$ of ${\rm PSL}(2,13)$ has a repeated real constituent of degree $13$. – Derek Holt Jul 30 at 12:18

Peripherally related: In a paper I wrote in 1997 about bases for primitive permutation groups, it is noted that if $$G$$ is a (faithful) primitive permutation group of degree $$n$$, and a complex irreducible character $$\chi$$ of $$G$$ occurs with multiplicity $$m$$ in the associated permutation character of degree $$n$$, then there is a base for $$G$$ of size at most $$\frac{\chi(1)}{m}.$$

Recall that a base for the permutation group $$G$$ acting on $$\Omega$$ is a subset $$\beta$$ of $$\Omega$$ such that only the identity element of $$G$$ fixes every element of $$\beta$$.

Hence we obtain $$|G| \leq n(n-1) \ldots (n+1 - \frac{\chi(1)}{m}) < n^{\frac{\chi(1)}{m}}.$$