TITLE: on the group generated by transitive permutation groups and diagonal groups
QUESTION [5 upvotes]: 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!

REPLY [6 votes]: 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.