Timeline for A question on groups having a subgroup which fixes a vector in every irreducible representations
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 2 at 13:37 | comment | added | Aurel | @SoumyadipSarkar I've added details in my answer below. Let me know if it is clear. | |
| Oct 2 at 8:13 | comment | added | Soumyadip Sarkar | @Aurel Sorry, I did not get you. Which $H$ works in every non-abelian simple group? | |
| Oct 1 at 22:48 | answer | added | Aurel | timeline score: 11 | |
| Oct 1 at 21:58 | comment | added | Aurel | A trivial remark is that if $G$ admits such an $H$, then for every $G'$, you can take $H\times 1$ in $G\times G'$. | |
| Oct 1 at 21:56 | comment | added | Aurel | For the same reason, a double transposition works in $S_n$ for $n\ge 4$. | |
| Oct 1 at 21:50 | comment | added | Aurel | If $\sigma\in G$ has order $2$ and $H=\langle\sigma\rangle$ does not work, then there exists an irrep $\rho$ such that $\rho(\sigma)=-{\rm Id}$, so that $\sigma$ is central of order $2$ in $G/\ker\rho$. In particular, in every nonabelian simple group, such an $H$ always works. | |
| Sep 27 at 15:24 | comment | added | Mikael de la Salle | @LSpice in a corner, say the upper-left. | |
| Sep 27 at 15:03 | comment | added | LSpice | @MikaeldelaSalle, re, how is $\operatorname{SL}_2(R)$ sitting inside $\operatorname{SL}_4(R)$? Blocks, upper left-hand corner, or something else? | |
| Sep 27 at 14:15 | comment | added | Mikael de la Salle | (but for us, the mere existence of some $H$ was not by itself interesting, what it meaningful in these examples is that $H$ is large, eg $\mathrm{SL}_2(\mathbf{Z})$ contains a nonabelian free group). | |
| Sep 27 at 14:06 | comment | added | Mikael de la Salle | I do not have a good answer, but with Michael Magee, we have an example that we have found meaningful, for operator algebraic considerations: the pairs $(G,H)$ equal to $(\mathrm{SL}_4(\mathbf Z/N\mathbf Z),\mathrm{SL}_2(\mathbf Z/N\mathbf Z))$ for integers $N$ (or equivalently $\mathrm{SL}_4(\mathbf Z_p),\mathrm{SL}_2(\mathbf Z_p))$ for primes $p$) have the property that all nonzero representation over $\mathbf C$ of $G$ have a nonzero $H$-invariant vector. This is false for $4$ replaced by $3$, see arxiv.org/abs/2312.03220 | |
| Sep 27 at 13:22 | history | edited | Soumyadip Sarkar |
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| Sep 27 at 12:09 | comment | added | Peter Mueller | Another necessary condition is that $H$ need to be a subgroup of the derived subgroup $G'$ of $G$. But these conditions are not necessary, the smallest examples have order $32$ where nevertheless the character theoretic condition does not hold. | |
| Sep 27 at 10:15 | comment | added | Geoff Robinson | If $G$ is nilpotent, then every irreducible character $\chi$ of $G$ has the form ${\rm Ind}_{X}(G)(\lambda),$ where $X$ is a subgroup of $G$ necessarily containing $Z(G)$, and $\lambda$ is a linear character of $X$. Mackey's formula then gives a criterion to check whether $H$ has fixed points on the induced representation: something like, there are fixed points if and only if $X \cap gHg^{-1} \leq {\rm ker} \lambda$ for some $g \in G.$ Not sure how helpful that is. | |
| Sep 27 at 8:18 | comment | added | Peter Mueller | If such an $H$ exists, we may assume that $H$ has prime order. Also, $H$ is not normal. So for the quaternion group of order $8$, no such $H$ exists. I haven't checked whether the necessary condition, that $G$ has a non-normal subgroup of prime order, is sufficient. | |
| Sep 27 at 6:07 | history | asked | Soumyadip Sarkar | CC BY-SA 4.0 |