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Let $T$ be a transitive permutation group in $S_n$, embedded in $GL_n(F)$ as permutation matrices. Let $D$ be the group of diagonal matrices in $GL_n(F)$. Let $G$ be the group generated by $T$ and $D$. That is, $G$ is a subgroup of the monomial group in $GL_n(F)$.

Question: is $G$ irreducible as a matrix group?

Probably this is a very basic question, but I cannot figure it out or find a reference...

Thank you for your help!

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    $\begingroup$ The answer depends on the size of the field $F$. For example, if $F$ has two elements, the all $1$-vector is fixed by $G$. If $F$ has more than $n$ elements, then the representation is irreducible, because only diagonal matrices commute with all of $D$, and the only diagonal matrices which commute with all of $T$ are scalars. $\endgroup$
    – Geoff Robinson
    Commented Jul 19, 2021 at 12:38
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    $\begingroup$ @GeoffRobinson Do you really need $|F|>n$ for that. Doesn't the argument work whenever $|F|>2$? The only subspaces fixed by $D$ are those spanned by some of the natural basis vectors, and none of those (except for the whole space) are fixed by the permutation matrices. $\endgroup$
    – Derek Holt
    Commented Jul 19, 2021 at 12:44
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    $\begingroup$ Thank you, @GeoffRobinson and @DerekHolt! $\endgroup$
    – Jimmy
    Commented Jul 19, 2021 at 12:52
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    $\begingroup$ To reformulate Derek's point: if $|F|>2$, the (unital) subalgebra generated by the invertible diagonal matrices consists of all diagonal matrices, and the invariant subspaces under this algebra are sum of coordinate subspaces, so transitivity on coordinates implies (absolute) irreducibility. If $|F|=2$ and $n>1$, irreducibility fails since $S_n$ itself is not irreducible. $\endgroup$
    – YCor
    Commented Jul 19, 2021 at 13:26
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    $\begingroup$ @DerekHolt : No, $|F| >n$ is indeed overkill ( I was going to suggest checking the intermediate cases, but cut my comment short). It's a bit harder to see when 2 < $|F| < n$, but for $|F| >n$, there is (as you know) an invertible diagonal matrix with $n$ distinct elements on the main diagonal, and it is obvious that its centralizer is diagonal. $\endgroup$
    – Geoff Robinson
    Commented Jul 19, 2021 at 13:39

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This is just a summary of the answers in the comments.

If $|F|=2$, then the vector $(1,1,\ldots,1)$ spans a subspace invariant under $G$, so the group is reducible (assuming that $n>1$).

Otherwise, if $|F|>2$ then, under the action of $D$, the natural module $V$ is the sum $V_1 \oplus V_2 \cdots \oplus V_n$ of $1$-dimensional irreducible submodules which are mutually nonisomorphic. (The mutual nonisomorphism follows from the fact the actions of $D$ on these submodules have different kernels.) It follows that the only submodules under the action of $D$ are sums of subsets of $\{V_1,\ldots,V_n\}$.

But, the action of $T$ is transitive, so the $V_i$ are permuted transitively by $T$, and none of these subspaces except for $V$ itself is invariant under $T$. So the action of $G$ is irreducible.

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