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Obvious necessary condition is that the center must be a cyclic group. Is it sufficient (doubt here)? If not, is there any nice characterization in terms of group structure, without appealing to representations?

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    $\begingroup$ I'm assuming that "exact" here means "faithful" (representation has trivial kernel). For finite $p$-groups, it's a standard fact that having a faithful irreducible representation is equivalent to having a cyclic center. I'm not sure about the general case, but it's been discussed in many books and papers. My impression is that there is no known definitive structural condition for sufficiency. $\endgroup$ – Jim Humphreys Mar 2 '11 at 16:52
  • $\begingroup$ P.S. It's also important here to specify the field or its characteristic, since that affects such existence questions. $\endgroup$ – Jim Humphreys Mar 2 '11 at 17:01
  • $\begingroup$ I thought about complex numbers, but other fields are of interest too. $\endgroup$ – Fedor Petrov Mar 2 '11 at 18:25
  • $\begingroup$ Are the distributive permutation groups linearly primitive? $\endgroup$ – Sebastien Palcoux Jan 3 '15 at 12:14
  • $\begingroup$ I would like to ask a variation on this question: Characterize finite groups G which admit an irreducible representation in which the group action is free on the compliment of the origin. $\endgroup$ – Nico Bellic Jul 17 '15 at 2:38
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A finite abelian group has a faithful irreducible representation if and only if it is cyclic. The case of finite groups was solved by Gaschütz in

W. Gaschütz, Endliche Gruppen mit treuen absolut-irreduziblen Darstellungen. Math. Nach. 12 (1954)

From Mathematical Reviews:

"In the present formulation the author calls the direct product $$ S= M_1\times M_2\times\cdots\times M_t $$ of the minimal normal subgroups $M_i$ of $G$ the `base' of $ G$, and writes $S=A\times H$, where $A$ is abelian and $H$ contains no normal abelian subgroup. The condition is as follows: a finite group $G$ has a faithful irreducible representation in an algebraically closed field of characteristic zero if and only if the base $S$ (or alternatively, $A$) of $G$ is generated by a single class of conjugates in $G$. The proof is based on an elegant application of the exclusion principle."

Maybe you also want to look at

Bekka, Bachir; de la Harpe, Pierre, Irreducibly represented groups. Comment. Math. Helv. 83 (2008), no. 4, 847–868

where the case of infinite groups and unitary representations on Hilbert spaces was studied. One of the main results of the paper is the following:

Theorem: A countable group $G$ is irreducibly represented, if one of the following conditions hold:

  1. $G$ is torsion-free
  2. $G$ is icc; this means the all non-trivial conjugacy classes are infinite,
  3. $G$ has a faithful primitive action on an infinite set.
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    $\begingroup$ The result of Gaschütz can be found also in Huppert's character theory book, Theorem 42.7. $\endgroup$ – Steve D Mar 2 '11 at 19:41
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    $\begingroup$ The subgroup generated by all the minimal normal subgroups is not the direct product of all of them, but of some of them (see Huppert's book, theorem 42.9). $\endgroup$ – Sebastien Palcoux Jan 29 '16 at 14:19
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I thought I'd add a specific example of a finite group with cyclic centre (trivial, in fact), yet no faithful irreducible complex representation (the example is from problem 2.19 of Isaacs' Character theory of finite groups, MR460423).

The group $(C_2)^4\rtimes C_3$, where $C_n$ denotes the cyclic group of order $n$ and $C_3=\langle \sigma\rangle$ acts on $(C_2)^4=\langle \tau_1,\tau_2,\tau_3,\tau_4\rangle$ via

$$\begin{align*} \sigma\cdot\tau_1=\tau_2 \hspace{0.5in}&\sigma\cdot\tau_2=\tau_1\tau_2 \newline \sigma\cdot\tau_3=\tau_4 \hspace{0.5in}&\sigma\cdot\tau_4=\tau_3\tau_4. \end{align*}$$

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    $\begingroup$ A handy sufficient condition is this: $G$ has a faithful irreducible complex representation if each Sylow subgroup of $G$ has a cyclic center. $\endgroup$ – Marty Isaacs Jan 29 '12 at 20:35

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