Finite groups such that every irrep can be induced from trivial irrep of a subgroup ?

By Frobenius reciprocity, a representation is induced from the trivial representation of the subgroup $H$ if and only its restriction to $H$ includes the trivial representation. So every non-cyclic abelian group has this property, because an irreducible representation is one-dimensional, so factors through a map to a cyclic group, so has a kernel.

Moreover, this implies that if $H\subset G$ and $H$ has this property, then $G$ does as well, so a sufficient condition is that a group have an abelian subgroup that's not cyclic.

This resolves the case for $GL_n(\mathbb F_p)$, $n>1$ and $p>2$ (the diagonal subgroup), and $n>2$ and $p=2$ (an abelian subgroup of the $2$-Sylow subgroup). n=2, p=2 is just $S_3$, which manifestly has this property. (Showing that this sufficient condition is not necessary) So $GL_n$ for $n>1$ has this property. $GL_1$ does not, since it is always cyclic. (For cyclic groups, every induced representation fails to be faithful, so take a faithful 1-dimensional irrep.)

$A_n$ for $n\geq 4$ contains the Klein four subgroup of $A_4$ and so has this property. $A_3$ is cyclic and so does not.

I do not know an easy-to-check necessary condition.

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 element 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 $|K| \equiv 1$ (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 $F = {\rm GF}(p).$ Since $p$ 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" representation need not be irreducible as a complex representation, but its irreducible components all have the property that each non-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 the 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 the 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 cyclic, then it is a Frobenius complement: for example, a non-Abelian group of order $21$ is not a Frobenius complement (though it is a Frobenius GROUP!)

Concerning the finite general linear groups (or other finite groups of Lie type), it's been known for a long time that you can't get every irreducible character as a constituent of one induced from the trivial character of a proper parabolic subgroup. This is what makes the whole subject so challenging, going back as far as the work of Frobenius for the groups $\mathrm{SL}_2(\mathbb{F}_p)$ and extending through the work of J.A. Green on characters of finite general linears to the much more complicated work coming from Deligne-Lusztig theory. Caveat: In this situation I'm not considering all possible proper subgroups, just those relevant to the BN-pair structure, so it's of course possible to find exceptions. But in Lie theory, including finite general linear groups, one is really looking for uniform methods to produce character tables.

On the other hand, for groups of Lie type there is a rich theory of what can be done if you induce up from the trivial character of a parabolic subgroup and then decompose the induced character using Hecke algebra methods. The problem is that it doesn't get everything you want.

By the way, I'd be curious to know whether there is a reasonable necessary and sufficient condition on a finite group to make the answer to your question affirmative. (Probably not.) In any case, your header does suggest that you want the induced representations involved to be irreducible, which misleads people at first.