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Let $D_n$ be the Dihedral group (of order $2n$).

For $p>2$ a prime number, $\mathbb{Z}/2$ is a core-free maximal subgroup of $D_p$, then $D_p$ is a primitive permutation group, and so linearly primitive (see here).

Question: Is $D_n$ linearly primitive for any $n>2$?

Sebastien Palcoux
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2 Answers2

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This is just writing up the comment by Tobias Kildetoft. Recall that by "linearly primitive" OP means "has a faithful complex irreducible representation" (I actually think this is nice terminology I just haven't heard it before). Then as claimed by Tobias Kildetoft there are lots of explicit examples of such representations for dihedral groups $ n\geq 3 $ and finite Coxeter groups in general.

A finite coxeter group is a direct product of irreducible finite Coxeter groups. The full list of irreducible finite Coxeter groups is given here

https://en.wikipedia.org/wiki/Coxeter_group

the irreducible finite Coxeter groups are linearly primitive since they are exactly equivalent to the irreducible real reflection groups. For an explicit list that includes all the irreducible finite Coxeter groups as well as some other interesting linearly primitive groups see

https://en.wikipedia.org/wiki/Complex_reflection_group#List_of_irreducible_complex_reflection_groups

Summary: Wikipedia has the full list of irreducible complex reflection groups. This list of course contains all the irreducible real reflection groups, which are equivalent to the irreducible finite Coxeter groups. So linearly primitive groups includes all finite irreducible Coxeter groups: the dihedral groups but also the symmetric groups, and indeed all Weyl groups including those for the exceptional cases like $ W(E_6)\cong SO_5(3) $.

The list of complex reflection groups also furnishes many examples of linearly primitive groups which are not Coxeter groups. These include all cyclic groups as well as all the $ G(m,p,n) $ for $ m \geq 3 $ (the $ G(2,1,n) $ and $ G(2,2,n) $ are Coxeter) as well as some cool exceptional cases with simple composition factors like $ PSL_2(7) $ and $ PSU_4(3) $.

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Yes

A theorem of W. Gaschütz states that a finite group $G$ is linearly primitive iff there exists $g \in S(G)$ the socle of $G$ [the subgroup generated by all the minimal normal subgroups] such that the conjugate class of $g$ in $G$ generates $S(G)$ [see the theorem 42.7 p558 of the book "Character Theory of Finite Groups" by Bertram Huppert, and see my source here].

Now for $G=D_n = \mathbb{Z}/n \rtimes \mathbb{Z}/2$ and using this answer, we get that for $n>2$, $S(D_n) = S(\mathbb{Z}/n)$ which is cyclic, so its generator $g$ (and a fortiori the conjugate class of $g$) generates $S(D_n)$, then by Gaschütz theorem, $D_n$ is linearly primitive.

Sebastien Palcoux
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