Arguing that $\mathrm{GL}_2(\mathbb{C})$ cannot contain a copy of the Heisenberg group The finite subgroups of $\mathrm{GL}_2(\mathbb{C})$ are pretty much classified in several expository papers in the web as well as posts on this site. However, I would like to know that how does one show that the Heisenberg group $H_3$ cannot be one such subgroup?
Recall that $H_3$ is a non-abelian group of order $27$ where every element can be given in the form of an upper triangular $3$-by-$3$ matrix with $1$'s in the diagonal, and variables $a,b,c$ filling up the remaining three slots with $a,b,c \in \mathbb{Z}/3\mathbb{Z}$. So every non-identity element has order $3$.
This presentation of elements of $H_3$ isn't helpful, since the closest thing to $\mathbb{Z}/3\mathbb{Z}$ in $\mathbb{C}$ is $\{1,\omega,\omega^2\}$ where $\omega$ is a primitive cube root of unity, but their operations differ (sum for the former, product for the latter), and so I cannot explicitly construct a map $\varphi: H_3 \rightarrow \mathrm{GL}_2(\mathbb{C})$ and argue. I also thought of not considering the matrix presentation structure of $H_3$, rather just as a non-abelian group of order $27$, then $\varphi$ would be a $2$-dimensional representation of $H_3$ which I think we can show does not exist.
Any help would be appreciated.
 A: The Heisenberg group $H_3$ is a $3$-group, hence supersolvable. By Theorem 16 of Serre Linear Representations of Finite Groups, any irreducible representation of $H_3$ is induced from a character of a subgroup, hence in particular has dimension $1,3,9,$ or $27$.
Thus, your representation $\varphi$ must be reducible, i.e., a direct sum of characters. Thus it factors through the abelianization of $H_3$ (and, in particular, is not faithful.)
A: Here is another way:
If $\varphi: H_3 \rightarrow GL_2(\mathbb{C})$ is faithful, it must be irreducible. (Otherwise the representation would decomposable into a direct sum of two $1$-dimensional representations, in which case $\operatorname{Im} \varphi$ is an abelian group.)
So suppose that $\varphi$ is irreducible. By Schur's lemma, the center $Z(H_3) = \langle z \rangle$ acts by scalar matrices, say $\varphi(z) = \lambda I_2$ with $\lambda \in \mathbb{C}$.
On the other hand $Z(H_3) = [H_3,H_3]$ and commutators have determinant  $1$, so $\det(\lambda I_2) = \lambda^2 = 1$. But $z^3 = 1$, so we have $\lambda = 1$. Hence $\varphi(z) = 1$, so $\varphi$ is not faithful, a contradiction.
A: Here is a completely different argument, because we seem to be collecting proofs here.
Let $G$ be the Heisenberg group, and suppose that $G$ has a faithful $2$-dimensional representation $\rho$. This must be irreducible as $1$-dimensional representations all have $G'\neq 1$ in the kernel. Let $H$ be a (normal) non-cyclic subgroup of order $9$, which is abelian. Then $\rho$ restricts to $H$ as the sum of two irreducible representations, $\tau_1$ and $\tau_2$. Now $\rho$ is irreducible, so there must be $g\in G$ such that $g$ does not stabilize $\tau_1$, and since $H$ is a normal subgroup, $\tau_1\cdot g$ is another $H$-subrepresentation of $\rho$.
If $\tau_1\neq \tau_2$ then these are the only two $H$-subrepresentations, so $\tau_1 \cdot g=\tau_2$ and vice versa. But then $g^2$ stabilizes each $\tau_i$, and thus $\langle g^2\rangle=\langle g\rangle$ stabilizes each $\tau_i$, a contradiction.
So $\tau_1=\tau_2$, and $H$ acts on $\rho$ as scalars. But $H$ is non-cyclic and all finite subgroups of scalars are cyclic, so $\rho$ cannot be faithful.
A: Here is yet another argument. You can compute that the abelianization of $H_3$ has order $9$, so $H_3$ has nine $1$-dimensional irreducible representations. If $d_i$ are the dimensions of the irreducibles then we have $\sum d_i^2 = 27$ so the squares of the dimensions of the remaining irreducibles, which must have dimension $\ge 2$, sum to $27 - 9 = 18$.
$4^2 = 16$ can't appear in this sum because $18 - 16 = 2$ is less than $4$, so the remaining irreducibles must have dimension either $2$ or $3$. Since $18$ is even, there must be an even number of $3$-dimensional irreducibles; there must also be at least one, because $18$ is not divisible by $4$. So there are at least two, and since $18 = 3^2 + 3^2$, there are exactly two.
So there are nine $1$-dimensional irreducibles and two $3$-dimensional irreducibles; this means, as others have already said, that any $2$-dimensional representation must be a direct sum of two $1$-dimensional irreducibles and hence can't be faithful.
