5
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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 $A_5$. $\endgroup$ – Richard Lyons Jul 29 at 14:44
4
$\begingroup$

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 ]
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – M. Winter Jul 29 at 15:58
  • 2
    $\begingroup$ Yes, the primitive permutation representation of degree $91$ of ${\rm PSL}(2,13)$ has a repeated real constituent of degree $13$. $\endgroup$ – Derek Holt Jul 30 at 12:18
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.