I became interested in the following question and realized that it was asked by Niko Bellic in the comment to previous question of mine. For which finite groups a complex representation which is free on the complement of the origin does exist? Of course it may be a priori irreducible. In other words, how to describe finite matrix groups for which $A-I$ is invertible for all $A\ne I$ in the group? For Abelian groups only cyclic groups satisfy this property.
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asked Oct 12, 2016 at 9:21
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1$\begingroup$ It looks to me like finite Heisenberg groups also work. For an integer $\ell > 1$, consider the group $G$ with cyclic center $Z(G) = \langle z | z^\ell \rangle$ given by $G=\langle x,y,z | z^\ell, x^\ell = z = y^\ell, yx = z xy \rangle$. Consider the $\ell$-dimensional representation that sends $x$ to a "cyclic" shift matrix $\mathbf{e}_i = \zeta \mathbf{e}_{i+1}$ for $\zeta$ a primitive $\ell^2$ root of unity, and that sends $y$ to the diagonal matrix $\mathbf{e}_i \mapsto \zeta^{1+i\ell} \mathbf{e}_i$. $\endgroup$– Jason StarrCommented Oct 12, 2016 at 9:38
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4$\begingroup$ These are precisely the Frobenius complements, well-studied in the literature, and they appear in some previous questions on here ( some of whose answers I have been involved with). They were already studied by Burnside though he made a famous error in one of his statements about them. It is true that the Sylow $q$-subgroups of a Frobenius complements are cyclic for odd $q$, and cyclic or generalized quaternion for $q = 2$. The only perfect Frobenius complement is ${\rm SL}(2,5)$ in either of its faithful $2$-dimensioal complex representation. A Frobenius complement of odd order is metacyclic, $\endgroup$– Geoff RobinsonCommented Oct 12, 2016 at 10:01
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$\begingroup$ Since finite Heisenberg groups are not metacyclic, that means my example does not work: even though $x$ and $y$ act freely for that representation, already $xy^{-1}$ has a nonzero fixed vector. $\endgroup$– Jason StarrCommented Oct 12, 2016 at 10:08
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$\begingroup$ @GeoffRobinson. Our comments were posted at just about the same moment. I also realized that my examples don't work. $\endgroup$– Jason StarrCommented Oct 12, 2016 at 10:12
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1$\begingroup$ @Jason Starr: OK, I deleted my second comment. $\endgroup$– Geoff RobinsonCommented Oct 12, 2016 at 10:48
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