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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 elements 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 $|G| |K| \equiv 1$ (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 $F = {\rm GF}(p).$ Sin $p$ 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" repesentation need not be irreducible as a complex representation, but its irreducible components all have the property that each on-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 te 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 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 Abelian, then it is a Frobenius complement: for example, anon-Abelian group of order $21$ is not a Frobenius complement (though it is a Frobenius GROUP!)
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Let $G$ be a finite Hamiltonian group- that is- every subgroup of $G$ is normal EDITED IN RESPONSE TO COMMENTS BY DAVID SPEYER AND F. Then no faithful irreducible complex character of $G$ can occur as a constituent of a permutation character induced from a non-trivial subgroup $H$ of $G.$ For if $\chi$ is faithful and irreducible, and $\langle {\rm Res}^{G}_{H}(\chi)LADISCH: An example, 1 \rangle \neq 0,$ then we have $H \leq {\rm ker} \chi$ by Clifford's theoremwhich is effectively definitive, as $H \lhd G.$ Hence if $H \neq 1,$ then $\chi$ can't be faithful. Another example is the class of Frobenius complements. These are the finite groups which admit a (necessarily faithful) representation in which every non-identity elements 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. Again, this 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 $G.$ Since it came up in comments, let 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 $g \in G \backslash H.$ 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 $p$)$ fo 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 te 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 $H.$ Every Frobenius complements are precisely the groups which have an irreducible character $\chi$ of a non-trivial finite group $G$ which does not occur as a constituent of a character induced from a degree $1$ character {\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 $G$ (though acts without eigenvalue $1$ in any complex representation affording $\chi.$ Conversely, if each non-identity element of $\chi$ G$ acts without eigenvalue $1$ in a representation of $G$, then there is itself linear, we need to allow an irreducible representation $H = G$). For \sigma$ with that property, and if we restrict $\sigma$ affords character $\chi,$ then $\chi$ to does not occur as a cyclic constituent of ${\rm Ind}_{H}^{G}(1)$for any non-trivial subgroup $\langle g \rangle$ H$ of $G,$ it 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 linear constituentnon-identity scalar matrix, say so that no non-identity $\lambda,$ and by Frobenius reciprocity3$-element of $G$ has eigenvalue $1$ in the associated representation, while also each non-identity power of $\chi$ occurs with positive multiplicity when we induce y$ has three different primitive $\lambda$ from 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 Abelian, then it is a Frobenius complement: for example, anon-Abelian group of order $\langle g \rangle$ to G.$21$ is not a Frobenius complement (though it is a Frobenius GROUP!)
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I also remark that Brauer's characterization of characters shows that every non-linear Every irreducible character $\chi$ of a non-trivial finite group $G$ occurs does occur as a constituent of a character induced from a linear degree $1$ character of a non-trivial subgroup of $G.$ For choose a prime H$ of $p$ which divides G$ (though if $\chi(1).$ There must be a \chi$ is itself linearcharacter , we need to allow $\lambda$ of H = G$). For if we restrict $\chi$ to a Brauer elementary cyclic subgroup $E$ \langle g \rangle$ of $G$ such that $p$ does not divide `$\langle {\rm Ind}_{E}^{G}(\lambda), \chi \rangle.$` Then `$\langle {\rm Res}^{G}_{E}(\chi)G,$ it has a linear constituent, \lambda \rangle$` is not divisible by say $p.$ In particular, `$\langle {\rm Res}^{G}_{E}(\chi)\lambda,$ and by Frobenius reciprocity, \lambda \rangle \neq \chi(1),$` so that $E$ is not the trivial subgroup.\chi$ occurs with positive multiplicity when we induce $\lambda$ from $\langle g \rangle$ to G.$
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Let $G$ be a finite Hamiltonian group- that is- every subgroup of $G$ is normal. Then no faithful irreducible complex character of $G$ can occur as a constituent of a permutation character induced from a non-trivial subgroup $H$ of $G.$ For if $\chi$ is faithful and irreducible, and $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle \neq 0,$ then we have $H \leq {\rm ker} \chi$ by Clifford's theorem, as $H \lhd G.$ Hence if $H \neq 1,$ then $\chi$ can't be faithful.
Another example is the class of Frobenius complements. These are the finite groups which admit a (necessarily faithful) representation in which every non-identity elements 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. Again, 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 $G.$
Since it came up in comments, let me justify that Frobenius compldements 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 $|G| \equiv 1$ (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 $F = {\rm GF}(p).$ Sin $p$ 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" repesentation need not be irreducible as a complex representation, but its irreducible components all have the property that each on-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 $p$)$ fo 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 te 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 $H.$
I also remark that Brauer's characterization of characters shows that every non-linear irreducible character $\chi$ of a finite group $G$ occurs as a constituent of a character induced from a linear character of a non-trivial subgroup of $G.$ For choose a prime $p$ which divides $\chi(1).$ There must be a linear character $\lambda$ of a Brauer elementary subgroup $E$ of $G$ such that $p$ does not divide $\langle `$\langle {\rm Ind}_{E}^{G}(\lambda), \chi \rangle.$rangle.$` Then $\langle `$\langle {\rm Res}^{G}_{E}(\chi), \lambda \rangle$rangle$` is not divisible by $p.$ In particular, $\langle `$\langle {\rm Res}^{G}_{E}(\chi), \lambda \rangle \neq \chi(1),$chi(1),$` so that $E$ is not the trivial subgroup.
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Since it came up in comments, let me justify that Frobenius compldements 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 $|G| \equiv 1$ (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 $F = {\rm GF}(p).$ Sin $p$ 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" repesentation need not be irreducible as a complex representation, but its irreducible components all have the property that each on-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 $p$)$ fo 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 te 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 $H.$
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Let $G$ be a finite Hamiltonian group- that is- every subgroup of $G$ is normal. Then no faithful irreducible complex character of $G$ can occur as a constituent of a permutation character induced from a non-trivial subgroup $H$ of $G.$ For if $\chi$ is faithful and irreducible, and $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle \neq 0,$ then we have $H \leq {\rm ker} \chi$ by Clifford's theorem, as $H \lhd G.$ Hence if $H \neq 1,$ then $\chi$ can't be faithful.
Another example is the class of Frobenius complements. These are the finite groups which admit a (necessarily faithful) representation in which every non-identity elements 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. Again, 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 $G.$
I also remark that Brauer's characterization of characters shows that every non-linear irreducible character $\chi$ of a finite group $G$ occurs as a constituent of a character induced from a linear character of a non-trivial subgroup of $G.$ For choose a prime $p$ which divides $\chi(1).$ There must be a linear character $\lambda$ of a Brauer elementary subgroup $E$ of $G$ such that $p$ does not divide $\langle {\rm Ind}_{E}^{G}(\lambda), \chi \rangle.$ Then $\langle {\rm Res}^{G}_{E}(\chi), \lambda \rangle$ is not divisible by $p.$ In particular, $\langle {\rm Res}^{G}_{E}(\chi), \lambda \rangle \neq \chi(1),$ so that $E$ is not the trivial subgroup.
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Let $G$ be a finite Hamiltonian group- that is- every subgroup of $G$ is normal. Then no faithful irreducible complex character of $G$ can occur as a constituent of a permutation character induced from a non-trivial subgroup $H$ of $G.$ For if $\chi$ is faithful and irreducible, and $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle \neq 0,$ then we have $H \leq {\rm ker} \chi$ by Clifford's theorem, as $H \lhd G.$ Hence if $H \neq 1,$ then $\chi$ can't be faithful.
Another example is the class of Frobenius complements. These are the finite groups which admit a (necessarily faithful) representation in which every non-identity elements 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. Again, 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 $G.$
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Let $G$ be a finite Hamiltonian group- that is- every subgroup of $G$ is normal. Then no faithful irreducible complex character of $G$ can occur as a constituent of a permutation character induced from a non-trivial subgroup $H$ of $G.$ Fr For if $\chi$ i fainthful is faithful and irreducible, and $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle \neq 0,$ then we have $H \leq {\rm ker} \chi$ by Clifford's theorem, as $H \lhd G.$ Hence if $H \neq 1,$ then $\chi$ can't be faithful.
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Let $G$ be a finite Hamiltonian group- that is- every subgroup of $G$ is normal. Then no faithful irreducible complex character of $G$ can occur as a constituent of a permutation character induced from a non-trivial subgroup $H$ of $G.$ Fr if $\chi$ i fainthful and irreducible, and $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle \neq 0,$ then we have $H \leq {\rm ker} \chi$ by Clifford's theorem, as $H \lhd G.$ Hence if $H \neq 1,$ then $\chi$ can't be faithful.
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