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 $p>3$ 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. 3 deleted 123 characters in body [This answer contradicts Jim Humphreys's, and I'm guessing his is right and mine is wrong, but so far I can't see why.] 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$p>3$(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. 2 added 111 characters in body [This answer contradicts Jim Humphreys's, and I'm guessing his is right and mine is wrong, but so far I can't see why.] 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 every group with a non-cyclic group have an abelian subgroup has this propertythat'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$p>3$(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.