What is the smallest group which is not a complex reflection group?
Many well known families of finite groups are complex reflection groups https://en.wikipedia.org/wiki/Complex_reflection_group For example: finite abelian groups, dihedral groups, symmetric groups.
I think $ A_5 $ is not a complex reflection group. But I imagine there are smaller examples.
@MoisheKohan is right
All groups of order 1,2,3,5,7 are cyclic
And all groups of order 4 are abelian.
The only nonabelian group of order 6 is the dihedral group $ D_3=S_3 $.
Of the $ 5 $ groups of order $ 8 $, $ 3 $ abelian $ C_8,C_4 \times C_2, C_2 \times C_2 \times C_2 $ the fourth is dihedral group $ D_4 $.
All the groups listed so far are complex reflection groups (groups generated by complex reflections). That is because al finite abelian groups are complex reflection groups. And dihedral groups are not only complex reflection groups they are actually real reflection groups (aka Coxeter groups).
However the fifth group of order $ 8 $, the quaternion group, is not a complex reflection group.
To see why, recall that a unitary matrix is a complex reflection if all its eigenvalues are $ 1 $ except for a single eigenvalue which is an $ n $th root of unity. The quaternion group has no representation of this form. Indeed the only faithful irreducible representation of $ Q_8 $ is the standard one of degree $ 2 $. In this representation $ 6 $ of the matrices have spectrum $ (i,-i) $ while one matrix has spectrum $ (-1,-1) $ and one matrix has spectrum $ (1,1) $. So the only complex reflection is the identity and that certainly does not generate the whole group.
Suppose that $\rho: Q_8\to U(n)$ is a faithful representation whose image is a reflection subgroup. Let $g_1,...,g_k$ denote the generators of $Q_8$ which map to complex reflections. Then at least two of these, say, $g_1, g_2$ have to be noncentral and not $\pm 1$ of each other. But then $g_1, g_2$ already generate $Q_8$. Let $H_1, H_2\subset {\mathbb C}^n$ be hyperplanes fixed by $\rho(g_1), \rho(g_2)$. Their intersection is fixed by the entire $\rho(Q_8)$ elementwise. Set $H:=H_1\cap H_2$. Then $\rho(Q_8)$ preserves the orthogonal complement $V=H^\perp$ and acts on it faithfully; it follows that $V$ is necessarily 2-dimensional. It has to be irreducible since $Q_8$ is nonabelian. Now, we are in business since, as you observed, there is, up to an isomorphism, unique faithful 2-dimensional representation of $Q_8$ and it is not a reflection representation.
A less conceptual answer: complex reflection groups are completely classified as direct products of an explicit list of irreducibles. All of the exceptional groups have order at least $24$, so the only possibilities are that $Q_8$ could be a direct product of groups $G(m, p, n)$ (where $m, p, n$ are positive integers with $p \mid m$). Since $Q_8$ is not a nontrivial direct product, it would have to be a single $G(m, p, n)$. The order of $G(m, p, n)$ is $m^n \cdot n! / p$, so we would have to have $n! \mid 8$. The case $n = 1$ is out (we get only cyclic groups), so it must be that $n = 2$ and $m^2 / p = 4$. This equation has two solutions, $G(2, 1, 2)$ and $G(4, 4, 2)$. However, both of these are actually the dihedral group $D_{2 \times 4}$.
