9 Changed $|G|$ to $|K|$.

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!)

8 added 751 characters in body

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!) 7 simplified argument at end  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$to G.$

 
 
 
6 fixed paragraphs
 
5 Included proof of characterization of Frobenius complements.
4 Showed that everything works if trivial is replaced by linear
3 gave other examples
2 typos corrected
1