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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.

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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.)

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    $\begingroup$ It's a nice result about supersolvable groups (I did not know about it). Just making the remark that we can equally well use the fact that irreducible complex representations have dimensions that are factors of the order of the group. The rest of your argument settles it from that point on just as well. +1 of course. $\endgroup$ Aug 22, 2022 at 5:57
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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.

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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.

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  • $\begingroup$ Ah, this is nice. I think if you apply this argument to a $3$-dimensional irrep you’ll show that it’s induced from a $1$-dimensional irrep of $H$? $\endgroup$ Aug 22, 2022 at 22:38
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    $\begingroup$ @QiaochuYuan Yes. The 3-dimensional irreducible representations of $G$ are just the (sums of) orbits of irreducible representations of $H$. The same argument shows that if $G$ is soluble and $A$ is an abelian normal subgroup then the character degrees divide $|G:A|$. (This is true in general, but the proof is not quite as simple.) $\endgroup$ Aug 23, 2022 at 9:21
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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.

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