Compositum of Galois extension

Question

Suppose $p,q$ are distinct primes. $K_1$ is a Galois extension of $\mathbb{Q}(\mu_{pq})$ of degree $p$, $K_1 \subseteq \mathbb{C}$. $K_2$ is a Galois extension of $\mathbb{Q}(\mu_{pq})$ of degree $q$, $K_2 \subseteq \mathbb{C}$. Let $L$ be the compositum of $K_1$ and $K_2$ in $\mathbb{C}$.

I could not figure out the reasoning behind this argument: ($p$, $q$) = 1, $K_1 \cap K_2 = \mathbb{Q}(\mu_{pq})$, so $Gal(L/\mathbb{Q}(\mu_{pq}))$ $\cong$ $C_p$ × $C_q$ $\cong$ $C_{pq}$. Why is $Gal(L/\mathbb{Q}(\mu_{pq}))$ $\cong$ $C_p$ × $C_q$? Maybe there's a theorem regarding this but I can't remember.

The concept you are missing is that of linearly disjoint extension. A general result is that two Galois extensions with a trivial intersection have a compositum that is Galois, and the Galois group is the direct product. For a thorough introduction to the topic see Pete L. Clarks Lecture notes on field theory. You want Chapter 12 in particular., or his more refined question. For a quick'n-dirty intro, my first-aid seminar notes may help. — Jyrki Lahtonen, May 18 '23 at 19:22

Compositum of Galois extension

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