# Groups with one simple projective representation

I'm still trying to learn projective representation theory,  so please bear with me. Thanks to YCor and David Speyer for pointing out the issues with my previous formulation of this question. Hopefully this is better.

Below my field will always be $\mathbb{C}$.

Is there a classification of finite groups G with the property that there exist $c\in H_{2}(G,\mathbb{Z})$, such that there is a unique irreducible linear representation of the central extension of G associated to c that descends to an irreducible projective representation of G?

The trivial group trivially has this property.

The first nontrivial example if this behavior is the klein four group. Specifically, $D_8$ is the central extension of klein four associated to the unique contrivial homology class. Moreover, there is exactly one 2-dimension irreducible representation that descends to an irreducible projective representation of klein four.

As a non-example take a finite nontrivial cyclic group G of order n. Then the Schur multiplier is trivial so there are exactly n inequivalent linear representations G that descend the (trivial) Central extension of G.

• From your examples, it appears that the one-dimensional trivial representation counts as an irreducible projective representation for you. If every other irreducible representation is required to be projectively equivalent to this one, then in particular all irreducible representations are one dimensional, and the group is abelian. Conversely, if the group is abelian than of course all irreducible representations are projectively equivalent. I suspect this is not what you meant to ask, but your including the Klein 4-group as an example leaves me confused as to what you did mean. – David E Speyer Jan 10 '17 at 19:23
• One interesting question I could imagine is groups for which all irreps of dimension $>1$ are projectively equivalent to each other. (The quaternion $8$-group, for example.) But I don't know if this is your question. – David E Speyer Jan 10 '17 at 19:24
• @DavidSpeyer in the last interpretation you seem to restrict to projective reps lifting to a representation. – YCor Jan 10 '17 at 19:28
• I'm not well versed in projective reps (clearly), so I'm struggling to properly phrase the question. My limited understanding is that if the Schur multiplier is trivial, then projective reps and linear reps are basically the same thing (you can lift them to linear reps on the same group). So finite cyclic groups won't have the property I want. However, the Klein-4 group has one 2-dimensional projective corresponding to the nontrivial Schur multiplier – Paul Jan 10 '17 at 19:29
• possibly you just want to add the adjective "positive-dimensional" to your question (to rule out the 0-dimensional projective representation on the point). Btw this rules out the trivial group from examples... – YCor Jan 10 '17 at 19:31

Previous answer was broken, here is what I can salvage.

To be clear, I am taking a projective representation to be a map $G \to PGL(V)$ with $\dim V \geq 2$. I call it irreducible if the corresponding representation $\tilde{G} \to GL(V)$, where $\tilde{G}$ is a central extension of $G$, is irreducible. I call two projective representations $G \to PGL(V)$ and $G \to PGL(W)$ isomorphic if I can choose an isomorphism $V \cong W$ to make them coincide. The goal is to classify $G$ such that $G$ only has one irreducible projective representation up to isomorphism.

Let $A=[G,G]$ and $B = G^{ab}$, so we have $1 \to A \to G \to B \to 1$. I can show

• $A$ is elementary abelian (possibly trivial).

• The conjugacy action of $B$ on $A \setminus \{ e \}$ is transitive.

• $B$ is either of the form $C_m$ or $C_{2m} \times C_2$. (Here $C_k$ is the cyclic group of order $k$.)

This is a pretty limited list; I am not sure which groups on this list have the stated property.

So, let's prove this. Define $A=[G,G]$ and $B = G^{ab}$ and suppose $G$ has the claimed property.

Lemma All non-identity elements of $A$ are conjugate in $G$. (The old version of this answer said conjugate in $A$; that's false.)

Proof Suppose for the sake of contradiction that $A$ has $\geq 3$ conjugacy classes in $G$. Since the character table is invertible, there are two characters, $\chi_1$ and $\chi_2$ with $\chi_i$ coming from $\rho_i : G \to GL(V_i)$ which are linearly independent of each other and of the trivial character. Any one dimensional rep of $G$ is trivial on $A$, so $\dim V_1$ and $\dim V_2 \geq 2$.

By assumption, the projective representations $\bar{\rho}_i : G \to PGL(V_i)$ are isomorphic, choose an isomorphism $V_1 \cong V_2$ such that $\bar{\rho}_1 = \bar{\rho}_2$. Then $\rho_1(g) = \chi(g) \rho_2(g)$ for some $\chi: G \to \mathbb{C}^{\ast}$. It is easy to see that $\chi$ is a (one-dimensional) character of $G$. But then $\chi$ must be trivial on $A$, so $\chi_1 = \chi_2$ on $A$, contradicting our linear independence. $\square$

So the conjugacy action of $G$ on $A$ is transitive on $A \setminus \{ e \}$. By a standard lemma, this implies $A$ is elementary abelian. We've proved the first two bullet points.

Now, we study the structure of the abelian group $B$. Any irreducible projective representation of $B$ is an irreducible projective representation of $G$, so we deduce that $B$ has at most one irreducible projective representation (none if $B$ is cyclic.)

Now, if $m>2$, then $C_m^2$ has multiple irreducible projective representations corresponding to different representations of the Heisenberg group of order $m^3$. (We can see this from the above argument: the commutator subgroup of the Heisenberg group is central, so the conjugacy action on the commutator subgroup is trivial and can't be transitive.) So $B$ must not have $C_m^2$ as a quotient for $m>2$. Also, $C_2^3$ has multiple irreducible representations, by projecting onto $C_2^2$ in different ways. Looking at the classification of finite abelian groups, this limits us to $C_m$ or $C_2 \times C_{2m}$.

I have not figured out which groups on the above list actually have the specified property.

I now have the list of candidates down to the following:

$C_m$ -- this has no irreducible projective representations, so you can decide whether to count it or not.

$C_{2m} \times C_2$. The only irreducible projective representation of this is through the quotient $C_2 \times C_2$, so it works.

$C_p^k \rtimes C_{(p^k-1)t}$, where the action is by a cyclic subgroup of $GL(\mathbb{F}_p)$ of order $p^k-1$. I have not yet figured out whether or not this works.

$G \times C_s$ where $s$ is odd and $G$ is a non-abelian central extension of the form $1 \to C_2 \to G \to C_{2^k} \times C_{2} \to 1$. I can show that all projective representations of this factor through the quotient $G$, but I haven't figured out for which $G$ there is a unique irreducible projective representation.

I'll try to add proofs of what I have later today.

I interprete the question as asking for projective reps associated to a fixed cohomology class, as follows: $\DeclareMathOperator{GL}{GL} \DeclareMathOperator{PGL}{PGL} \newcommand{\CC}{\mathbb{C}}$
By a projective representation, I understand a map $R\colon G\to \GL(V)$ such that the composed map $G \to \PGL(V)$ is a group homomorphism. Thus there is a map $c\colon G\times G\to \CC^*$ such that $$R(g)R(h) = c(g,h) R(gh) \quad \text{for all } g, h \in G ,$$ and $c\in Z^2(G,\CC^*)$ is a cocycle.
(Recall that the Schur multiplier $H^2(G,\CC^*) \cong H_2(G,\mathbb{Z})$. In fact, there is a natural isomorphism from $H_2(G,\mathbb{Z})$ onto the the dual group of $H^2(G,\CC^*)$.)

For which groups is there a cocycle $c$ such that there is only one irreducible projective representation with that given cocycle $c$, up to "equivalence"?
There are, however, two possible notions of equivalence here: We may say that the projective representations $R_i\colon G\to \GL(V_i)$ are equivalent when there is an isomorphism $T\colon V_1\cong V_2$ such that $T \circ R_1(g) = R_2(g) \circ T$ for all $g\in G$. Or we may require only that there are scalars $\mu(g)\in \CC^*$ such that $T\circ R_1(g) = \mu(g) R_2(g)\circ T$ for all $g\in G$ (as in David's answer).
In the first case, the question is equivalent to having a cocycle $c$ such that the twisted group algebra $\CC^{c}[G]$ is simple as an algebra, i.e. isomorphic to a matrix ring over $\CC$. Equivalently, there is a central extension $1 \to Z \to \tilde{G} \to G \to 1$, such that $Z$ has a linear character $\lambda$ such that there is exactly one irreducible representation of $\tilde{G}$ lying over $\lambda$. (Then $\tilde{G}$ is called a group of central type and $G$ a central factor group. Necessarily $Z=Z(\tilde{G})$.)
A celebrated result of Howlett and Isaacs says that then $G$ is solvable (the proof depends on the classification of finite simple groups). To the best of my knowledge, central factor groups are not classified, but a number of necessary conditions are known. An abelian group $G$ is a central factor group iff it is of the form $G=A\times A$ for some abelian group $A$. A group $\tilde{G}$ is of central type iff for each Sylow $P$ of $\tilde{G}$ we have $Z(P) = P\cap Z(\tilde{G})$ and $P$ is of central type as well.
If we take the coarser equivalence relation, we get an even larger class of groups. It's equivalent to having a central extension $1 \to Z \to \tilde{G}\to G \to 1$ and a linear character $\lambda$ of $Z$, such that all irreps of $\tilde{G}$ over $\lambda$ come from one irrep by tensoring with linear reps of $\tilde{G}$. I don't know if this can be somehow reduced to the case of central factor groups.