TITLE: Condition for normalizer to be finite
QUESTION [2 upvotes]: Let $ \Gamma $ be a finite subgroup of $ SU_d $. Is the normalizer $ N_{SU_d}(\Gamma) $ finite if and only if the action of $ \Gamma $ on $ \mathbb{C}^d $ is irreducible?

REPLY [2 votes]: Yes. First, $N(\Gamma)/C(\Gamma)$ is finite since it's isomorphic to a subgroup of $\mathrm{Aut}(\Gamma)$, which is finite because $\Gamma$ is. Second, $C(\Gamma)$ commutes with all of $\Gamma$, so it commutes with the span of $\Gamma$ in $M_d(\mathbb{C})$, which is all of $M_d(\mathbb{C})$ because $\Gamma$ is irreducible (Burnside's irreducibility criterion), so $C(\Gamma)$ is in the center $Z(M_d(\mathbb{C}))$ which consists of scalar matrices, and the scalar matrices in $SU(d)$ are all $d$th roots of unity, so $|C(\Gamma)|\le d$ is finite too. In particular, this shows $|N(\Gamma)|\le d\,|\mathrm{Aut}(\Gamma)|$.
(This shows $\Gamma$ irreducible $\implies N(\Gamma)$ finite.)
Or, suppose $N(\Gamma)$ is infinite. As $N(\Gamma)/C(\Gamma)$ is finite, $C(\Gamma)$ must be infinite. As the only scalar matrices in $SU(d)$ are the $d$th roots of unity, $C(\Gamma)$ must have non-scalar matrices. But every element of $C(\Gamma)$ is an intertwiner, and Schur's lemma says if $\Gamma$ were irreducible then all intertwiners would be scalar transformations, so $\Gamma$ cannot be irreducible.
(This shows $N(\Gamma)$ infinite $\implies\Gamma$ reducible, the contrapositive.)
OTOH, suppose $\Gamma$ were reducible. Any scalar transformation of one (proper, nonzero) subrep can be extended by a corresponding scalar transformation of the complementary subrep to yield an element of $\mathrm{SU}(d)$. There are infinitely many matrices constructed this way, and all of them are in the centralizer $C(\Gamma)$. (This is how your example was constructed.)
(This shows $\Gamma$ reducible $\implies N(\Gamma)$ infinite, the contrapositive of the converse.)
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