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Let $G$ be a finite subgroup of the group $U_d(\mathbb{C})$ of unitary transformations of $\mathbb{C}^d$. Suppose that $G$ acts irreducibly but is imprimitive, meaning that there is a nontrivial direct sum decomposition $\mathbb{C}^d = \bigoplus_{i = 1}^r V_i$ such that each $g \in G$ permutes the $V_i$.

Then it seems that the $V_i$ are necessarily orthogonal: I wrote up a proof here. However I'm happy to admit that it took me quite some time to find this proof, and I still don't know of a reference. This must surely be well-known, and was probably known to Frobenius. Can anyone supply me with a reference?

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    $\begingroup$ Interestingly this is not necessarily true over finite fields, where it is possible for the $V_i$ to be totally singular. For example, ${\rm SU}(2n,q)$ has subgroups that are extensions of ${\rm SL}(n,q^2)$, which interchange two totally singular subspaces of dimension $n$. $\endgroup$
    – Derek Holt
    Commented Jan 10, 2018 at 13:26

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This is not an answer, but here is another proof in the same spirit as yours. Write $(-\vert -)$ for the canonical scalar product of ${\mathbb C}^d$. Since the $G$-representation ${\mathbb C}^d$ is irreducible, we have $$ {\mathbb C}^d = \bigoplus_{g\in G/G_1} g.V_1 $$ where $G_1$ is the stabilizer of $V_1$ in $G$. Write $(-\vert -)_1$ for the restriction of $(-\vert -)$ to $V_1\times V_1$; it is $G_1$-invariant. For all $g\in G$ define a scalar product on $g.V_1$ by $$ (gv_1 \vert gw_1 )_{gV_1} = (v_1 \vert w_1 ) $$ It does not depend on the choice of $g$. Let $<-\vert ->$ be the scalar product on ${\mathbb C}^d$ given by the orthogonal sum of the $(-\vert -)_{gV_1}$, $g\in G/G_1$. It is clearly $G$-invariant. Now since the representation ${\mathbb C}^d$ is irreducible the scalar products $(-\vert -)$ and $<-\vert ->$ are proportional and we are done. N.B. Of course Schur's lemma is hidden in my proof.

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  • $\begingroup$ Paul, thanks for this. In fact, I think it's exactly the same proof, and in fact when I first came up with the proof I formulated it this way. Maybe it's more natural. Schur's lemma is hidden in the last step and I think < x, y > = (phi(x), y) with my notation. I don't know how you got to this argument, but for me it was via the fact that imprimitive => induced from G_1, and then look at how induced representations are given a unitary structure. $\endgroup$
    – Ben Green
    Commented Jan 10, 2018 at 14:51

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