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What are the finite groups which admit a non-zero representation in char 0 where every non-zero vector has stabilizer equal to $\left<1\right>$? Cyclic groups of prime order is one obvious class. Is there anything else?

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    $\begingroup$ All nontrivial cyclic groups, ${\rm SL}(2,5)$ and its subgroups. $\endgroup$
    – Derek Holt
    Jul 17, 2015 at 8:27
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    $\begingroup$ Also generalized quaternion groups. $\endgroup$ Jul 17, 2015 at 9:16
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    $\begingroup$ A sylow $2$-subgroup would have to be cyclic or generalized quaternion, which restricts the structure. I wonder if there are any interesting odd order examples. $\endgroup$
    – Derek Holt
    Jul 17, 2015 at 9:20
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    $\begingroup$ After reading Geoff Robinson's answer, I found the earlier post, which is mathoverflow.net/questions/191769 So perhaps this question should be regarded as a duplicate? $\endgroup$
    – Derek Holt
    Jul 17, 2015 at 10:27
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    $\begingroup$ @Derek Holt: I agree that there is not much to be learned from this question which is not covered by the earlier question and its answers. $\endgroup$ Jul 17, 2015 at 10:41

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I think these groups have appeared on MO before. A finite group $H$ has such a representation in characteristic zero if and only if $H$ occurs as a Frobenius complement. This is "well-known folklore" and appears in a book by D. Passman for example.

First, if $H$ is Frobenius complement in a Frobenius group $G$ with Frobenius kernel $K$, any non-trivial minimal $H$-invariant subgroup $V$ of $K$ is an elementary Abelian $q$-group for some prime $q$ not dividing $|H|$, since $|H|$ and $|K|$ are comprime. Also $H$ acts faithfully on $V$. We can pass to an algebraic closure of ${\rm GF}(q)$ and lift the representation of $H$ to characteristic zero, and we obtain a complex representation of $H$ such that all non-identity elements act without the eigenvalue $1$.

Conversely, if $H$ has a complex representation with this last property, then we can reduce the associated module (mod $q$) for some prime $q$ not dividing $|H|$. Then we easily obtain a semidirect product $VH$ with $V$ an elementary Abelian normal $q$-subgroup such that $H$ is a Frobenius complement.

The structure of a (finite) Frobenius complement $H$ is reasonably well understood. For example, Burnside knew that if $p$ and $q$ are different prime divisors of $H$, every subgroup of $H$ of order $pq$ is cyclic. As mentioned in comments, all Sylow subgroups of $H$ are cyclic or (generalized) quaternion.

The only perfect Frobenius complement is ${\rm SL}(2,5).$ There is a Frobenius complement of order $63$ with center of order $3$.

Later edit: This last group has a complex irreducible representation of degree $3$ in which no non-identity element has an eigenvalue $1$. In general, a Frobenius complement of odd order is metacyclic.

Even later edit: Here is a quick direct proof that any finite non-trivial group $G$ which has all subgroups normal necessarily has an irreducible complex representation (which may not be faithful) of the type in the question: take $\chi$ to be an irreducible non-trivial complex character of $G$. Let $N = {\rm ker} \chi$. Then by induction, we may suppose that $N = 1$, as $G/N$ has all its subgroups normal. Hence we may suppose that $\chi$ is faithful. For any $g \in G^{\#} $, Clifford's theorem tells us that ${\rm Res}^{G}_{\langle g \rangle}(\chi)$ is a sum of $G$-conjugate irreducible characters of $\langle g \rangle$, since $\langle g \rangle \lhd G.$ None of these can be the trivial character (or they would all be trivial, since they are all $G$-conjugate, and that would put $g$ in ${\rm ker}\chi$). Hence $g$ acts without the eigenvalue $1$ in the representation affording $\chi$.

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  • $\begingroup$ Is there a reference discussing that Frobenius complement of order 63, and its 3D representation you mention? $\endgroup$
    – rj7k8
    May 9, 2021 at 16:11
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    $\begingroup$ I am not sure about an explicit reference: the group is $\langle a, b : a^{7} = b^{9} = 1, b^{-1}ab = a^{2} \rangle..$ There is a $3$-dimensional representation, where $a$ may be represented by $\left(\begin{array}{clcr} \omega & 0 & 0\\0&\omega^{2} & 0\\0&0&\omega^{4}\end{array} \right)$, ($\omega$ a primitive complex $7$-th root of unity) and $b$ may be represented by $\left( \begin{array}{clcr} 0&\alpha &0\\0&0&\alpha\\ \alpha & 0 & 0 \end{array} \right)$, where $\alpha$ is a primitive $9$-the root of unity. $\endgroup$ May 9, 2021 at 16:40

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