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A transitive permutation group $G \subset S_n$ is primitive if $G_1 \subset G$ is a maximal subgroup.
A finite group $G$ is linearly primitive if it has a faithful complex irreducible representation.

Question: Are the primitive finite groups linearly primitive?

Remark: I've checked by a GAP computation that it's true for $n=[G:G_1] \le 200$ and $\vert G \vert \le 10^4$.

Sebastien Palcoux
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2 Answers2

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Yes. The O'Nan-Scott Theorem classifies finite primitive groups into a number of different types, such as affine type, almost simple type, product type, etc. The groups in all but one of these types have a unique minimal normal subgroup $N$, which implies that they are linearly primitive, since the kernel of any unfaithful representation must contain $N$, and so any faithful representation must have an irreducible constituent whose kernel does not contain $N$ and is consequently faithful.

The groups $G$ in the remaining type have two isomorphic minimal normal subgroups $N_1$ and $N_2$, both of which are direct products of one or more isomorphic nonabelian simple groups. The smallest such example is $A_5 \times A_5$ acting on the cosets of a diagonal subgroup.

But direct products of nonabelian simple groups are linearly primitive, because the tensor product of nontrivial irreducible representations of their direct factors is a faithful irreducible representation. So $N_1 \times N_2$ has a faithful irreducible representation $\rho$ and any irreducible constituent of the induced representation $\rho^G$ is faithful for $G$.

Derek Holt
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  • Very nice, thank you! I believed there is a direct formal proof using on one hand the fact that a transitive group $G$ is primitive iff $G_1$ is maximal in $G$, and on the other hand the fact that an irreducible representation is faithful iff its tensor powers generate all the others irreducible representations. Anyway, I guess it's easy to find a linearly primitive group without a faithful primitive action (for example the cyclic groups of non-prime order), so the converse is not true. – Sebastien Palcoux Jan 03 '15 at 12:11
  • I've given (as an other answer) a short self-contained proof of the fact that a primitive permutation group is linearly primitive. Do you think it can simplify the proof of the O'Nan-Scott Theorem (for example by excluding some cases) ? – Sebastien Palcoux Jan 27 '16 at 13:15
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Yes, the following is a short self-contained proof:

Let $G \subset S_n$ be a primitive permutation group; by definition $G_1$ is a core-free maximal subgroup of $G$. Let $V$ be a non-trivial irreducible complex representation of $G$ such that the fixed point subspace $V^{G_1} \neq 0$ (such a $V$ exists, see the comment). By construction $G_1 \subset G_{(V^{G_1})} $ but by maximality $G_{(V^{G_1})} = G_1$ or $G$, if it is $G$ then $V^{G_1} = 0$ contradiction, so it is $G_1$. Now $K=ker(\pi_V) \subset G_{(V^{G_1})} = G_1$, but $G_1$ is core-free in $G$ and $K$ is a normal subgroup of $G$. It follows that $K=\{ e \}$ and so $G$ is linearly primitive. $\square$

Sebastien Palcoux
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  • I am struggling a little with your notation. What does $V^{G_1}$ mean? (I would expect that to denote an induced representation, but that doesn't make sense here.) And how do you know that such a $V$ exists? – Derek Holt Jan 27 '16 at 13:48
  • @DerekHolt: it is not the induced representation, it is just the fixed point subspace: $V^{G_1} = { v \in V ; g.v = v \ \forall g \in G_1 }$ – Sebastien Palcoux Jan 27 '16 at 13:53
  • @DerekHolt: sorry, I have to suppose $V$ non-trivial. So if for every non-trivial irreducible complex representation $V$ we have $V^{G_1} = 0$ , then for $H = \mathbb{C} G$ the left regular representation, we have $H^{G_1} = trivial $, but $H^{G_1} \neq trivial $ because it contains the vector $v = \sum_{g \in G_1} e_g$, and $ \mathbb{C} v$ is not the trivial rep. So such a $V$ exists. – Sebastien Palcoux Jan 27 '16 at 14:15