Suppose that $L/K$ be a normal extension such that $\mathrm{Gal}(L/K)$ is cyclic of order $30$. Prove that for every $d \in \mathbb{Z}$ such that $d$ divides $30$, there is a cyclic extension $M_{d}/K$ with $[M_{d}:K] = d$. Moreover, prove that $M_{d}/K$ is normal.
We have $G = \mathrm{Gal}(L/K) \simeq C_{30}$. We know that there is a unique $2$-Sylow, $3$-Sylow, $5$-Sylow. Let $I = 2$-Sylow $H = 3$-Sylow, and $J = 5$-Sylow, so $HJ$ is a subgroup of order $15$.
It's not difficult to show that $J$ is normal to $G$. Also, $[G:HJ] = 2$ so, $HJ$ is normal to $G$. Since, $H$ is normal to $HJ$ and $HJ$ is normal do $G$ with $H$ a $3$-Sylow subgroup, $H$ is normal to $G$. Thus, $IH$ is normal to $G$.
By Fundamental Theorem of Galois Theory, we have a correspondence between $I,H,J$, $IH$, $HJ$ and $M_{2}, M_{3},M_{5},M_{6},M_{15}$ such that $$[M_{2}:K] = [G:HJ] = 2,$$ $$[M_{3}:K] = [G:H] = 3,$$ $$[M_{5}:K] = [G:IH] = 5,$$ $$[M_{6}:K] = [G:J] = 6,$$ $$[M_{15}:K] = [G:I] = 15.$$ Since $\mathrm{Gal}(L/K)$ is cyclic, every subgroup is cyclic. The Fundamental Theorem of Galois Theory says that $``H$ is normal to $G$ if only if $F$ is Galois over $K$ where $K$ is an extension of $K$ and $F \leftrightarrow H"$. Therefore, $M_{d}/K$ is normal.
Is correct? Or did I make a mistake? Thanks for the help!