I would like an answer to the following question. A proof or a counterexample would be enormously appreciated.
Let $G$ be a finite group and let $V$ be a finite-dimensional complex irreducible $G$-module. Does there exist a subgroup $K$ of $G$ such that $ V$ decomposes as a direct sum of inequivalent, one-dimensional $K$-submodules?
Example: Suppose $G = S_3 = \langle r, s \mid r^3 = s^2 = s r s^{-1} r = e \rangle$ and $V = \mathbb C^2$. Consider the representation $\rho : G \to \text{GL}(V, \mathbb C)$, given by
$$ \rho(r) = \begin{bmatrix} – \tfrac 1 2 & – \tfrac {\sqrt{3}} 2 \\ \tfrac{\sqrt{3}} 2 & – \tfrac 1 2 \end{bmatrix}, \qquad \qquad \rho(s) = \begin{bmatrix} – 1 & 0 \\ 0 & 1 \end{bmatrix},$$
which gives $V$ the structure of a $G$-module.
If we take $K$ to be the order-$2$ subgroup $\{ e, s \}$, then clearly, the one-dimensional subspaces $\langle (1, 0) \rangle $ and $\langle (0, 1) \rangle$ of $V$ are both invariant under the action of $K$. As $K$-modules, these one-dimensional subspaces are inequivalent, since $\left(\rho|_K\right)|_{\langle (1, 0) \rangle}$ is the trivial representation of $K$ whereas $\left(\rho|_K\right)|_{\langle ( 0, 1) \rangle}$ is the alternating representation of $K$.
Further evidence: I verified that my statement holds for all irreducible representations of all groups $G$ of order less than or equal to $15$. I found the relevant character tables on the internet and turned the handle. To be fair, this isn't terribly strong evidence, since none of these groups are very complicated!
Idea for generalising the example: In the general case, if $K$ is an abelian subgroup of $G$, then certainly, all irreducible $K$-modules are one-dimensional. It remains to show that it is possible to find an abelian subgroup $K$ of $G$ that is "large enough" such that when we decompose $V$ as a direct sum of irreducible $K$-submodules, these $K$-submodules are all inequivalent. I don't know how to prove this, and I'm not even sure if this is true.
Motivation: In chemistry, the quantum states of an electron in a molecule are elements of a complex Hilbert space $V$, which, under certain approximations, is finite dimensional. The energy states of the electron are eigenvectors of a hermitian operator $\hat H : V \to V$ known as the Hamiltonian operator. Geometric symmetries of the molecule are described by a group representation $\rho : G \to V$, such that each $\rho(g)$ is unitary and commutes with $\hat H$.
The existence of a symmetry group drastically simplifies the task of determining the energy states, since it makes it possible to find subspaces of $V$ that are invariant under the action of $\hat H$. To illustrate this, consider a hypothetical example where $V$ decomposes as a direct sum $V_1 \oplus V_2 \oplus V_3$, where $V_1, V_2, V_3$ are irreducible $G$-submodules of $V$, with $V_1$ and $V_2$ being equivalent to one another and $V_3$ being inequivalent from $V_1$ and $V_2$. There exist elements of the group algebra $\mathbb C[G]$ that are projections onto $V_1 \oplus V_2$ and $V_3$ respectively (namely, the primitive central idempotents), and these elements commute with the Hamiltonian operator. Therefore, the subspaces $V_1 \oplus V_2$ and $V_3$ are invariant under the action of the Hamiltonian operator $\hat H$. So the chemist's calculation reduces to the easier calculation of diagonalising $\hat H|_{V_1 \oplus V_2}$ and $\hat H|_{V_3}$.
Now, suppose that the question I'm asking in this post has an affirmative answer. Then there exists a subgroup $K$ of $G$ such that $V_1$ decomposes as a direct sum $\mathbb C v_1 \oplus \mathbb C v_2 \oplus \cdots \oplus \mathbb C v_n$, where the $\mathbb C v_i$'s are inequivalent one-dimensional $K$-submodules of $V_1$. Let $\tau : V_1 \to V_2$ be the (unique up to rescaling) $G$-module isomorphism between $V_1$ and $V_2$. By a similar argument as earlier, $\mathbb C v_i \oplus \mathbb C \tau(v_i)$ is an invariant subspace under the action of $\hat H$, for any $i$. This simplifies the chemist's calculation even further, since the task of diagonalising $\hat H|_{V_1 \oplus V_2}$ reduces to the task of diagonalising $\hat H|_{\mathbb C v_i \oplus \mathbb C \tau (v_i)}$, for each $i$.
Miessler's classic chemistry textbook uses this technique over and over again to characterise the electron energy levels for certain molecules with symmetry – for example, ammonia, boron trifluoride and various octahedral transition metal complexes. Of course, the technique is not formulated in a mathematically precise manner, and there is no proof that the technique works for arbitrary molecules with symmetry. After all, the book is for chemists, not mathematicians. This prompted me to have a go at filling in the details myself.
Best Answer
The group $\mathrm{SL}(2,5)$ with a character of degree $6$ is a counterexample according to GAP.