I'm trying to prove the following: Let $G$ be a cyclic group of order $n$. For each divisor $d$ of $n$, denote by $G_d$ the subgroup of $G$ of index $d$.
Show that $G$ has an irreducible representation over $\mathbb{Q}$, unique up to isomorphism, whose kernel is equal to $G_d$.
My Try:
Existence: I know that the number of irreducible representations over $\mathbb{Q}$ of the cyclic group is iqual to the number of subgroups (Indeed is the number of conjugacy classes of cyclic subgroups of $G$.) The kernel of those Reps. should be a subgroup of $G$. How to conclude from here?
Uniqueness: Also, if $\rho$ is the representation whose kernel $G_d$, the characteristic polynomial of $\rho_g$ divides $x^n-1$, since it has no invariant spaces, it must be the $d-th$ cyclotomic polynomial. This proves the uniqueness?
Thank you in advance.
If you now some commutative algebra, here is a solution: representations of $G$ over $\mathbb Q$ is exactly the same as $\mathbb Q[x]/(x^n-1)$-modules. Recall the classical fatcorization (here $\Phi_d$ are cyclotomic polynomials) $$x^n-1=\prod_{d|n}\Phi_d(x),$$ so the Chinese remainder theorem tells us $$\mathbb Q[x]/(x^n-1)\simeq\prod_{d|n}\mathbb Q[x]/(\Phi_d(x)),$$ where $K_d:=\mathbb Q[x]/(\Phi_d(x))$ is a field since $\Phi_d(x)$ is irreducible. Now, it is easy to see that irreducible modules of a product of fields $\prod_dK_d$ is just $K_{d'}$ for divisors $d'|n$.