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Is there a finite group $G$, an element $c$ of order 2 in $G$, and an irreducible 2-dimensional complex representation $\rho$ of $G$ such that all the following are true:

1) $\rho(c)$ has trace zero

2) There is an index 2 subgroup $H$ of $G$ containing $c$, and a 1-dimensional representation $\psi$ of $H$, such that $\rho\cong Ind(\psi)$, the representation of $G$ induced by $\psi$.

3) If $K$ is any index 2 subgroup of $G$ and $\chi$ any 1-dimensional representation of $K$ such that $\rho\cong Ind(\chi)$, then $c\in K$.


Here are some thoughts, but nothing really concrete.

If a 2-dimensional representation is induced from a 1-dimensional representation of an index 2 subgroup, then one checks easily that the trace of the 2-d representation vanishes off the index 2 subgroup. So assumption (2) above implies that the trace of $\rho$ will vanish off $H$, and (1) says "the trace of $\rho$ vanishes on $c$ too", but then (3) says "...but that is not because $\rho$ is induced from an index 2 subgroup not containing $c$".

Note for people wondering about the difference between (2) and (3): it is possible that a 2-dimensional representation of a finite group can be induced from 1-dimensional representations of several distinct subgroups (consider for example the faithful 2-dimensional representation of the quaternion group of size 8).

[ Background: this question came up because the question arose about whether a certain kind of theta series could exist; the theta series, if it existed, would give rise to a certain kind of modular form; if the theta series existed, and my understanding is correct, then $G$ would be the Galois group of a certain extension of number fields, $\rho$ would be the representation attached to the form and $c$ would be some complex conjugation. The details are a bit technical, but basically if one could rule out $\rho$ above using purely representation-theoretic methods then one could rule out the existence of the theta series, and the question above seems to me to be more tractible. If however $\rho$ does exist then the question I'm actually interested in would still be open because the implication only goes one way.]

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Interesting question. I don't have the time now, but if there are no answers tomorrow, I'll check for examples with GAP. –  Tobias Kildetoft Jan 28 '13 at 21:32
    
Thanks! I would imagine that for someone fluent in GAP it would be a triviality to check e.g. all groups of size less than 100 very quickly. And then one would probably know the answer, because surely if there were an example there would be one with $G$ of smallish order. The issue then would be proving something if no example showed up. –  armpit Jan 28 '13 at 21:35
    
Still haven't had time to run it through GAP, but here are a few things one can say about a possible minimal such group: Clearly we can assume that the representation in question is faithful, and we get that the group has an abelian subgroup of index $2$, and the group has a normal abelian $2$-complement. I have a feeling that one can probably rule out the case of the group being nilpotent from this. –  Tobias Kildetoft Jan 29 '13 at 2:16
    
Update: Still haven't run it through GAP, but I am starting to think such groups do not exist, based on how close I am to proving it. I would be done if I could show the following (which I am not sure is true and which is not necessary but sufficient): Let $G$ be a semidirect product of an abelian group of odd order and a non-cyclic $2$-group, such that $G$ has an abelian subgroup $N_1$ of index $2$. Does $G$ have an abelian subgroup $N_2$ of index $2$ with $N_1 \neq N_2$? –  Tobias Kildetoft Jan 29 '13 at 14:14
    
@Tobias: re: your question. If $G$ is $C_2\times C_2$ I think the answer's "yes" and if $G$ is $S_3\times C_2$ I think the answer's "no", so in general I think the answer to your question is "sometimes". Did I make a slip? –  armpit Jan 29 '13 at 14:48

1 Answer 1

I have now had time to run this through GAP, and such groups do indeed exist. The smallest example has order $24$, and if you run the code at https://www.dropbox.com/s/v448jvr6ip2v4k6/armpit.g in GAP, it will return a list consisting of that group, the element on which the character vanishes, the character and the subgroup from which it is induced.

The code will take a few seconds to run, as it is not written very optimally, and it searches all groups from order $1$ and up. If you wish to know more (such as how many there are of such groups and similar things), let me know, and I will try to modify the code to find more of them.

Edit: I forgot to mention that to actually get it to return the results, one needs to type test(6); after reading the code into GAP. If you downloaded the code prior to this edit, you need to type galois(6); instead.

One can get a description of the structure of the group from GAP by (after getting the results) typing StructureDescription(last[1]); (last simply contains the last thing returned by GAP). This gives a description as a semidirect product, which however is not exactly the same as that on the list on groupprops. To compare to that list, one can use the command IdSmallGroup which shows that it is number 8 on that list (as it has Id 8)

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Thanks for this. It would be nicer to have a solution which was self-contained (i.e. didn't rely on "download this file and download a computer program to understand my answer"). Do you happen to know a name for this group of order 24? For example, which one of groupprops.subwiki.org/wiki/Groups_of_order_24 those is it? Maybe one can do the calculations by hand after that and get an independent check. –  armpit Feb 1 '13 at 7:33
    
An explicit permutation representation is given by using $a=(1,2,3,4,5,6)(7,8,9,10,11,12)$; $b=(1,4)(2,5)(3,6)$; and $c=(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)$. Then your group is $\langle a,b,c\rangle$. –  user641 Feb 1 '13 at 21:19

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