Fixed fields,cyclic groups and Galois theory Let $G=S \times T$ where $S$,$T$ are both finite cyclic groups. 
Question 1: Is it true that there exists a Galois finite extension $L/F$ such that $Gal(L/F) \cong S \times T$? (I can't recall if such extension can be taken to be finite).
Set $H_{1}=\{e\} \times T$. Then the corresponding fixed field of $H_{1}$ has the property that $Gal(H_{1}/F) \cong S$.
Question2: what is the degree of the extension $H_{1}/F$? Is this always equal to $|S|$?
Question 3: Suppose we drop the finiteness condition and at least one of $S$ or $T$ is equal to $\mathbb{Z}$ (the integers). Does the Galois extension (not sure if it even exists in this case) $L/F$ in question $1$ can be taken to be finite?
 A: Question 1: Yes. More generally, if $G$ is any finite group, then there exists a finite Galois extension $L/K$ with Galois group $G$. To see this it suffices to construct, for every $n$, a finite Galois extension $L/K$ with Galois group $S_n$; once you have this, you can pick an embedding $G \to S_n$ and take the fixed field of $G$. And Galois extensions with Galois group $S_n$ are not hard to construct: a nice example is given by $\mathbb{Q}(x_1, x_2, \dots x_n)$ as a Galois extension of $\mathbb{Q}(e_1, e_2, \dots e_n)$, where the $e_i$ are the elementary symmetric polynomials in the $x_i$.
Question 2: Yes. More generally, a finite Galois extension with Galois group $G$ has degree $|G|$.
Question 3: No, by Question 2. In fact a countably infinite group can never occur as a Galois group; as Lubin says in the comments, infinite Galois groups are naturally equipped with topologies with respect to which they are profinite, and in particular compact Hausdorff, and an infinite compact Hausdorff group is necessarily uncountable. 
