What are examples (general features) of the finite groups $G$, such that every irrep (irreducible representation) is contained (as constituent) in the representation induced from trivial representation of some non-trivial subgroup? (I allow subgroups to vary - I mean take all subgroups, induce reps from trivial characters and require that any irrep sits in this set).

Remark: If we induce from the subgroup = {identity}, we will get a regular representation of the whole group, and every irrep lives there. So I should not allow this trivial subgroup.

Example 1: $S_n$ satisfies this property (there is some comment by David Speyer which I cannot find now, which says (as far as I remember) that induction from $S_{k1}\times S_{k2} \times\cdots\times S_{kl}$ will do the job. I have another argument but it is more complicated).

Example 2: Take $\mathbb{Z}/p\mathbb{Z}$ for prime p - it does NOT satisfy my requirement - the only subgroups are $\lbrace 1\rbrace$ and $\mathbb{Z}/p\mathbb{Z}$ itself; first one is forbidden by the rules of the game, and induction from the second will give trivial irrep.

Example 3: If we take PSL(2,7)=GL(3,2) and induce from the 3-Sylow subgroup - it will contain all irreps as constituents (see MO 104939).

In particular: what about $GL(n,\mathbb{F}_p)$ and $A_n$?

  • $\begingroup$ Do you want one subgroup to give all the irreducible representations or is the subgroup allowed to vary? $\endgroup$ – Benjamin Steinberg Aug 26 '12 at 20:16
  • $\begingroup$ Subgroup is allowed to vary. Sorry for being unclear about this. I will edit. $\endgroup$ – Alexander Chervov Aug 26 '12 at 20:18
  • $\begingroup$ @David: The OP is, if I understand correctly, only asking that each irrep appear as a summand in one of the induced representations. It's OK if there are lots of other summands (like the trivial one) along with it. For example, if $G$ is the Klein four-group, then all 4 irreps occur within the 4 induced reps (in fact, within the 3 non-trivial induced reps). $\endgroup$ – Andreas Blass Aug 26 '12 at 21:30
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    $\begingroup$ A slight extension of example 2: If $G$ has a smallest non-trivial subgroup (e.g., the quaternion group), then the requirement isn't satisfied. The smallest non-trivial subgroup is a normal subgroup all of whose elements act trivially in all the induced representations that you consider. So you won't get any of the irreps in which elements of this subgroup act nontrivially. $\endgroup$ – Andreas Blass Aug 26 '12 at 21:33
  • $\begingroup$ A simpler example is $\mathbb Z/4$. An abelian group has this property if and only if it fails to be cyclic. Even $\mathbb Z/2$ fails, for having no proper nontrivial subgroup. $\endgroup$ – Will Sawin Aug 26 '12 at 21:49

EDITED IN RESPONSE TO COMMENTS BY DAVID SPEYER AND F. LADISCH: An example, which is effectively definitive, is the class of Frobenius complements. These are the finite groups which admit a (necessarily faithful) representation in which every non-identity element acts without the eigenvalue $1$. Such groups have cyclic Sylow $p$-subgroups for all odd primes $p,$ and cyclic or generalized quaternion Sylow $2$-subgroups, properties which also occur in Will Sawin's answer . This is a very restricted class of groups. For example, the only perfect Frobenius complement is ${\rm SL}(2,5).$ In any case, if $G$ is a Frobenius complement, and $\chi$ is a faithful complex irreducible character such that $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle =0$ for each non-identity cyclic subgroup $H$ of $G$ (and such a $\chi$ must exist), then $\chi$ is not a constituent of any permutation character induced from the trivial character of a non-identity subgroup of $G.$

Let me justify that Frobenius complements are just those groups with an irreducible complex representation where every non-identity element acts without eigenvalue $1$ . Recall that a Frobenius group $G$ has the form $G = KH,$ where $K \lhd G$ and $H \cap K = 1,$ and, furthermore, $H \cap H^{g} = 1$ for all $g \in G \backslash H.$

Notice then that $|K| \equiv 1$ (mod $|H|$), so that ${\rm gcd}(|K|,|H|) = 1.$ Also, we certainly have $C_{G}(h) \leq H$ whenever $h$ is a non-identity element of $H$ since $h \in H \cap H^{c}$ for all $c \in C_{G}(h).$

Let $V$ be a minimal non-identity $H$-invariant subgroup of $K.$ Then using Thompson's theorem that a Frobenius kernel is nilpotent, (which is actually overkill here, since by general properties of coprime automorphism groups to be found in Gorenstein's book "Finite Groups", for example, $H$ normalizes a Sylow $q$-subgroup of $V$ for each prime divisor $q$ of $|V|$), we see that $V$ is an elementary Abelian $p$-group for some prime $p$. Then $V$ is a faithful $FH$-module, where $F = {\rm GF}(p).$ Since $p$ does not divide $|H|$, $V$ "lifts" to a complex representation, and by general (indeed the defining) properties of Brauer characters, it is still the case that each non-identity element of $H$ acts without eigenvalue $1.$ The "lifted" representation need not be irreducible as a complex representation, but its irreducible components all have the property that each non-identity element of $H$ acts without eigenvalue $1$ on them (and each is faithful).

Conversely, if $H$ is a finite group which has a complex irreducible character $\chi$ which does not contain the trivial character on restriction to any non-identity cyclic subgroup of $H,$ then a complex representation of $\chi$ may be reduced (mod $p$) for any prime $p$ which does not divide $|H|$ to afford a $kH$-module $W$ on which each non-identity element of $H$ acts without non-zero fixed points, where $k$ is algebraically closed of characteristic $p.$ Then $W$ may be realised over a finite field, and the sum of its distinct Galois conjugates may be realised over ${\rm GF}(p),$ say by module $V$ over ${\rm GF}(p).$ It is still the case that each non-identity element of $H$acts without non-trivial fixed points on $V,$ so the semidirect product $VH$ is a Frobenius group with kernel $V$ and complement $H.$ Frobenius complements are precisely the groups which have an irreducible character $\chi$ which does not occur as a constituent of ${\rm Ind}_{H}^{G}(1)$ for any non-trivial subgroup $H$ of $G.$ For if $\chi$ is such a character, then ${\rm Res}^{G}_{H}(\chi)$ has no trivial constituent for each non-trivial subgroup $H$ of $G$, in particular for each non-trivial cyclic subgroup of $G.$ Hence each non-identity element of $G$ acts without the eigenvalue $1$ in any complex representation affording $\chi.$ Conversely, if each non-identity element of $G$ acts without eigenvalue $1$ in a representation of $G$, then there is an irreducible representation $\sigma$ with that property, and if $\sigma$ affords character $\chi,$ then $\chi$ does not occur as a constituent of ${\rm Ind}_{H}^{G}(1)$ for any non-trivial subgroup $H$ of $G.$ There are examples of non-Abelian Frobenius complements of odd order: for example, let $G = \langle x,y : x^{9} = y^{7} = 1, x^{-1}yx = y^{2} \rangle.$ Note that $G$ has an irreducible character $\chi$ of degree $3$ such that $x^{3}$ acts, as a non-identity scalar matrix, so that no non-identity $3$-element of $G$ has eigenvalue $1$ in the associated representation, while also each non-identity power of $y$ has three different primitive $7$-th roots of unity as its eigenvalues in the associated representation. However, it is not true that if a finite group of odd order has all its Sylow subgroups cyclic, then it is a Frobenius complement: for example, a non-Abelian group of order $21$ is not a Frobenius complement (though it is a Frobenius GROUP!)

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    $\begingroup$ Due to the classification of Hamiltonian groups, only the quaternions times an odd cyclic subgroup have a faithful irreducible representation. $\endgroup$ – Will Sawin Aug 26 '12 at 23:24
  • $\begingroup$ @Will: Yes, that is true. $\endgroup$ – Geoff Robinson Aug 26 '12 at 23:28
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    $\begingroup$ Isn't the answer precisely "The Frobenius complements are the only groups which fail to have this property?" Suppose $G$ is a group and $V$ is a representation which does not occur in $Ind_H^G 1$ for any $H$. Then, for any $g \in G$, the restriction $V|_{\langle g \rangle}$ must have no trivial component, so $g$ acts on $V$ without eigenvalue $1$ and the group is a Frob. complement. Conversely, suppose that there is some $H \subset G$ such that $V|_H$ has a trivial component. Then choose $h \in H$, and $V|_{\langle h \rangle}$ has a trivial component, hence $h$ has eigenvalue $1$. $\endgroup$ – David E Speyer Aug 27 '12 at 0:43
  • $\begingroup$ I didn't know this characterization of Frob. complements before reading Geoff's answer, but it seems to be exactly what we need. $\endgroup$ – David E Speyer Aug 27 '12 at 0:44
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    $\begingroup$ Perhaps it is a good idea to edit this answer so that it contains the result of David's comment in the first paragraph? Your answer contains the proof anyway, and this characterization is, imho, the definitive answer to this question. For example, from the known properties of Frobenius complements it follows at once that lots of groups have the property of the question (all simple groups, all perfect groups but one, all groups containing a non-cyclic group of order $pq$...). $\endgroup$ – Frieder Ladisch Sep 4 '12 at 20:56

By Frobenius reciprocity, a representation is induced from the trivial representation of the subgroup $H$ if and only its restriction to $H$ includes the trivial representation. So every non-cyclic abelian group has this property, because an irreducible representation is one-dimensional, so factors through a map to a cyclic group, so has a kernel.

Moreover, this implies that if $H\subset G$ and $H$ has this property, then $G$ does as well, so a sufficient condition is that a group have an abelian subgroup that's not cyclic.

This resolves the case for $GL_n(\mathbb F_p)$, $n>1$ and $p>2$ (the diagonal subgroup), and $n>2$ and $p=2$ (an abelian subgroup of the $2$-Sylow subgroup). n=2, p=2 is just $S_3$, which manifestly has this property. (Showing that this sufficient condition is not necessary) So $GL_n$ for $n>1$ has this property. $GL_1$ does not, since it is always cyclic. (For cyclic groups, every induced representation fails to be faithful, so take a faithful 1-dimensional irrep.)

$A_n$ for $n\geq 4$ contains the Klein four subgroup of $A_4$ and so has this property. $A_3$ is cyclic and so does not.

I do not know an easy-to-check necessary condition.

  • $\begingroup$ @Will Sawin Thank you very much it is very cool and unexpected answer. Let me take some time to wrap my mind over it. Is my understanding correct that in the first sentence you mean that the representation is such that we already now that it is induced from some character of H, but we do not known from trivial or not. Otherwise representation may be trivially restricted to H, but not induced from H at all. As far as I understand... $\endgroup$ – Alexander Chervov Aug 27 '12 at 15:10
  • $\begingroup$ Well every irrep of $G$ is a subrepresentation of some induced representation of $H$, because the regular repreaentation of $G$ is induced from the regular representation of $H$. $\endgroup$ – Will Sawin Aug 27 '12 at 15:16
  • $\begingroup$ @Will Sawin May I ask you to comment on "and n>2 and p>3 (an abelian subgroup of the 2-Sylow subgroup)." 1) may be misprint here: p>1 ? Otherwise why you mention it since diagonal subgroups is Okay for this case. 2) "an abelian subgroup of the 2-Sylow subgroup" - it seems you know something about 2-Sylow subgroup... I'am not expert in finite group... Can you explain what you mean ? $\endgroup$ – Alexander Chervov Aug 28 '12 at 12:11
  • $\begingroup$ "So GLn for n>1 has this property." How do you treat GL_n(F_2) ? There is no diagonal subgroup there ... May be it is related to previous question... $\endgroup$ – Alexander Chervov Aug 28 '12 at 12:13
  • $\begingroup$ Sorry that should have been clearer in the question but there was a typo. For $n>2$, consider matrices with $1$s on the diagonal and $0$s everywhere else but the top row. This is a $\mathbb (Z/2)^{n-1}$ subgroup. For $n=2$ this is $S_3$ so we just check explicitly. I guess it would be simpler to check explicitly for $GL_2$ and then use the fact that $GL_2\subset GL_n$. $\endgroup$ – Will Sawin Aug 28 '12 at 13:37

Concerning the finite general linear groups (or other finite groups of Lie type), it's been known for a long time that you can't get every irreducible character as a constituent of one induced from the trivial character of a proper parabolic subgroup. This is what makes the whole subject so challenging, going back as far as the work of Frobenius for the groups $\mathrm{SL}_2(\mathbb{F}_p)$ and extending through the work of J.A. Green on characters of finite general linears to the much more complicated work coming from Deligne-Lusztig theory. Caveat: In this situation I'm not considering all possible proper subgroups, just those relevant to the BN-pair structure, so it's of course possible to find exceptions. But in Lie theory, including finite general linear groups, one is really looking for uniform methods to produce character tables.

On the other hand, for groups of Lie type there is a rich theory of what can be done if you induce up from the trivial character of a parabolic subgroup and then decompose the induced character using Hecke algebra methods. The problem is that it doesn't get everything you want.

By the way, I'd be curious to know whether there is a reasonable necessary and sufficient condition on a finite group to make the answer to your question affirmative. (Probably not.) In any case, your header does suggest that you want the induced representations involved to be irreducible, which misleads people at first.

  • $\begingroup$ Induced representations of trivial representations are irreducible only when they're trivial, so that version would be a very silly questions. $\endgroup$ – Will Sawin Aug 26 '12 at 23:06
  • $\begingroup$ @Will: In my first line, I meant to write "parabolic" but didn't, so I've corrected that. I didn't really intend to exhibit a specific counterexample; my caveat meant to indicate that it doesn't really help in constructing character tables (especially for Lie families) just to know in principle that all irreducible characters might show up somehow by inducing trivial characters of subgroups (i.e., permutation characters on cosets). Aside from that, my problem with the header is that it does make the question look silly at first sight as you point out. $\endgroup$ – Jim Humphreys Aug 27 '12 at 0:42
  • $\begingroup$ @Jim thank you for yours answer. Indeed part of motivation to ask is - parabolic induction - why should we restrict to parabolic subgroups ? I guess it is because it is class of the subgroup which is present on the regular basis for all Lie groups ? However, is the end of the story or someone can someday to come with another class of subgroups ? $\endgroup$ – Alexander Chervov Aug 28 '12 at 12:15
  • $\begingroup$ "In any case, your header does suggest that you want the induced representations involved to be irreducible, which misleads people at first." May I ask yours advise - yes, I understand title may mislead, but on the other hand it is already too long ... So it is always bothers me - should I the title be precise or I can abbreviate it for shortness scarifying the clarity ? $\endgroup$ – Alexander Chervov Aug 28 '12 at 12:17
  • $\begingroup$ @Alexander: Constructing headers is an art form, I fear ;-) Concerning groups of Lie type like general linear groups, it's not rewarding to study each one in isolation. For a Lie family the BN-structure gives the best hope of using induction methods efficiently from large proper subgroups. Only subgroups related to Lie structure seem useful for study of an entire family. (But my "answer" isn't a direct answer to your question, only an indirect way to ask how useful the question is.) $\endgroup$ – Jim Humphreys Aug 28 '12 at 15:40

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