Famously, the finite subgroups of $\operatorname{SU}(2)$ admit an ADE classification.

Question. Is there a similar result for finite subgroups of $\operatorname{U}(2)$? Are they classified? If this is the case, what is the complete list of them?

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    $\begingroup$ There is a short exact sequence $$1 \to SU(2) \to U(2) \xrightarrow{\det} U(1) \to 1.$$ Although it does not split naturally, anyway it admits a splitting exhibiting $U(2)$ as a semidirect product $SU(2) \rtimes U(1)$, see math.stackexchange.com/questions/1180656/is-u2-su2-times-u1 $\endgroup$ – Francesco Polizzi Apr 9 at 18:31
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    $\begingroup$ Probably, using this information and the knowledge of the finite subgroups of $SU(2)$, one should be able to classify those of $U(2)$. $\endgroup$ – Francesco Polizzi Apr 9 at 18:33
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    $\begingroup$ I agree that $U(2)$ is isomorphic (as a Lie group) to $(SU(2)\times U(1)) / \mathbb{Z}_2$ (despite the proof I know doesn't involve short exact sequences). At the moment I fail to see how the short exact sequence you wrote could be useful in solving the problem, but I must admit I am very ignorant in homological algebra, in general. $\endgroup$ – Federico Carta Apr 9 at 18:35
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    $\begingroup$ @FedericoCarta Well, that means that finite subgroups of $U(2)$ correspond to $\mathbb{Z}_2$-invariant finite subgroups of $SU(2)\times U(1)$. Every finite subgroup of $SU(2)\times U(1)$ is contained in one of the form $A\times B$, for $A,B$ finite subgroups of $SU(2),U(1)$, respectively. It seems very likely to me that you can use this to give a complete classification. $\endgroup$ – dhy Apr 9 at 18:40

Every finite subgroup of $\operatorname{GL}(2,\mathbb{C})$ is conjugate to a subgroup of $U(2)$, so you are asking first for the isomorphism types of finite subgroups of $\operatorname{GL}(2, \mathbb{C}).$

These were already known to C. Jordan. They are easy to recover.

There are three types:

I: Reducible subgroups: these are conjugate to groups of diagonal matrices, so (since finite), they are finite Abelian groups with at most two generators, and all of these can occur. If we restrict to groups in which all matrices are unimodular, we get all finite cyclic groups as possibilities.

II: Irreducible but imprimitive subgroups: These are equivalent to monomial groups, and have a finite Abelian normal subgroup of index $2$ which (after replacing the representation by an equivalent one if necessary) is in fact cyclic (in modern terminology (for group theorists) this is a consequence of Clifford's Theorem). Each finite non-Abelian group $G$ which has an Abelian (necessarily normal) subgroup $A$ of index $2$ does occur as a finite subgroup of $\operatorname{GL}(2,\mathbb{C})$.

For such a group $G$ has all irreducible characters of degree at most two, but (since $G$ is non-Abelian) there is a non-linear irreducible character of $G$. Indeed, since $A$ is cyclic, we may choose a faithful linear character of $A$ and induce it to $G$. The representing matrices for elements of $G-A$ are not even diagonal, and no non-identity element of $A$ is represented by the identity matrix in this induced representation.

Hence type II gives rise to every finite non-Abelian group which has a cyclic normal subgroup of index $2$. However, if we restrict to type II groups consisting of unimodular matrices, then it is easy to check (using similar arguments) that the groups we get are just those finite non-Abelian groups $G = \langle A, u \rangle$ where $A$ is cyclic of even order and $u$ is an element of order $4$ with $u^{-1}au = a^{-1}$ for all $a \in A$.

III: Irreducible primitive groups: these are the irreducible subgroups of $\operatorname{GL}(2,\mathbb{C})$ which are not conjugate to monomial groups (i.e., the $2$-dimensional representation is not equivalent to one induced from a proper subgroup).

It was known to Jordan that if $G$ is such a group, then $G/Z(G)$ is isomorphic to $A_{4}$, $S_{4}$ or $A_{5}$. But we need to think explicitly how we make this identification. We take our group $G$ (in such a representation), and we multiply each element of a chosen set of generators by a (root of unity) scalar multiple so that the result has determinant one (this can be done because the original determinant is some root of unity). We get a resulting group $H \subseteq \operatorname{SL}(2,\mathbb{C})$ with $H/Z(H) \cong G/Z(G)$.

There are (up to equivalence) three possible choices for $H$: we can have $H \cong \operatorname{SL}(2,3)$, $H \cong \operatorname{BO}$ or $H \cong \operatorname{SL}(2,5)$. The group I call $\operatorname{BO}$ has a generalized quaternion Sylow $2$-subgroup of order $16$ and has order $48$. It has $\operatorname{SL}(2,3)$ as a normal subgroup of index $2$.

It is easy to show that each possiblity of $H$ is generated by two elements: in fact $\operatorname{SL}(2,3) = \langle a,b \rangle$ with $a$ of order $3$ and $b$ of order $4$ (with some relations), $\operatorname{BO} = \langle a,b \rangle$ with $a$ of order $3$ and $b$ of order $8$ (with relations) and $\operatorname{SL}(2,5) = \langle a,b \rangle$ with $a$ of order $3$ and $b$ of order $5$ (with relations).

To produce ALL finite possibilities for $G$, we must take each choice of $H = \langle a,b \rangle$ and take all choices $G = \langle za,wb \rangle$ where we identify $a$ and $b$ with their representing matrices and where $z$, $w$ are arbitrary roots of unity. There will be repetitions of isomorphism type when we do this. The resulting $G$ has the form $C \times \langle z^{\prime}a,w^{\prime}b \rangle$ where $C$ is a finite cyclic group of scalars whose order is coprime to $30$ and $z^{\prime}$ and $w^{\prime}$ are roots of unity whose order has the form $2^{x}3^{y}5^{z}$ for non-negative integers $x$, $y$, $z$.

Later edit: In fact, I think we can go further on consideration of determinant:

We can always write the right-hand factor (up to isomorphism) as $D \times E{\ast}H$ where $D$ is a cyclic $\{3,5\}$-group, $E$ is a cyclic $2$-group, and $H$ is one of the three possibilities above (and ${\ast}$ denotes the central product, where $Z(H)$ (which has order $2$) is identified with the unique subgroup of order $2$ of $E$ (if $E$ > 1).

It follows that the distinct possibilities for isomorphism types of $G$ of type III are of the form $ C_{1}{\ast} H$ where $H$ runs through the three possibilities above, $C_{1}$ runs over all finite cyclic groups not of twice odd order , and $\ast$ denotes the central product (which is direct when $C_{1}$ has odd order, and identifies the unique subgroup of order $2$ of each factor when $C_{1}$ has order divisible by $4$, and where $C_{1}$ and $H$ are mutually centralizing, so $C_{1}$ is represented by scalar matrices). We may omit the case that the cyclic group $C_{1}$ has twice odd order from the list because in that case, if we write $C_{1} = D \times E$ with $D$ of odd order and $E$ of order $2$, then $C_{1} \ast H \cong D \times H.$

We may note that the order of the finite cyclic group $C_{1}$ is the order of the image ${\rm det}(G)$ when $\{ {\rm det}(g): g \in G \} $ has odd cardinality, and twice the order of the image ${\rm det}(G)$ when $\{ {\rm det}(g): g \in G \} $ has even cardinality

  • $\begingroup$ Why is the number of generators of a finite group of diagonal matrices bounded by the rank? It must be obvious, but I can't seem to see it. $\endgroup$ – LSpice Apr 10 at 16:12
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    $\begingroup$ If the dimension is $n$, the character afforded by the representation decomposes as a sum of $n$ (not necessarily distinct) linear characters $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}.$ If the diagonal group is $A$ ( which is finite), then $A/{\rm ker} \lambda_{i}$ is a finite cyclic group $C_{i}$ for each $i$. Then $ a \to (\lambda_{1}(a), \ldots \lambda_{n}(a))$ gives a faithful embedding of $A$ in the direct product $C_{1} \times C_{2} \times \ldots \times C_{n}$, and we use the theory of finitely generated $\mathbb{Z}$-modules to conclude. $\endgroup$ – Geoff Robinson Apr 10 at 18:20

It seems that this is classically known and that the subgroups of $\operatorname{U}(2)$ have been in fact classified.

For a modern treatment, you can look at the paper

E. Falbel, J. Paupert: Fundamental domains for finite subgroups in $\operatorname{U}(2)$ and configurations of Lagrangians, Geom. Dedicata 109, 221-238 (2004). ZBL1108.51018.

The complete list you are interested in is the content of Theorem 2.2.

In the Introduction, the authors state that the classification was originally worked out in references [1-3], which I list below:

[1] J. H. Conway, D. A. Smith: On Quaternions and Octonions, K. Peters 2003.

[2] H.S.M. Coxeter: Regular Complex Polytopes, Cambridge University Press 1991.

[3] P. Du Val: Homographies, Quaternions and Rotations, Oxford Mathematical Monographs Science Publishers 1964.

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    $\begingroup$ The way I would phrase is it is that the projection to $U(2)$ gives a finite subgroup of $PU(2) = SO(3)$, which have the ADE classification, and then the game is to figure out what can be the inverse image in $U(2)$ of a fixed finite subgroup of $SO(3)$. $\endgroup$ – Will Sawin Apr 9 at 18:51

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