Proving the Galois Group of an extension is abelian Let $E_{1}, E_{2}$ be subfields of $\mathbb{C}$. Suppose $E_{1}|\mathbb{Q}$ and $E_{2}|\mathbb{Q}$ are finite Galois extensions and $G(E_{1}:\mathbb{Q})\cong$ $\mathbb{Z}_{6}\cong$ $G(E_{2}:\mathbb{Q})$. Suppose $[E_{1}\cap E_{2}:\mathbb{Q}]=2$. If $F$ is the smallest field that contains $E_{1}$ and $E_{2}$, show that $G(F:\mathbb{Q})$ is abelian.
I have already proved that $G(F:\mathbb{Q})$ is a Galois extension and that $[F:\mathbb{Q}]=18$.
 A: Some general Galois theory:


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*If $L\mid M\mid K$ is a tower of fields with $L,M\mid K$ Galois then there is a canonical projection map $G(L\mid K)\to G(M\mid K)$ given by restriction of the actions on $L$ to $M$. As $L\mid K$ Galois implies $L\mid M$ Galois, we have an inclusion $G(L\mid M)\hookrightarrow G(L\mid K)$.

*If $E_1,E_2\mid F$ are Galois extensions in a field containing $F$ then the compositum $E_1E_2\mid F$ is also a Galois extension, and using the projections $G(E_1E_2\mid F)\to G(E_1\mid F), G(E_2|F)$ we can form the diagonal map $G(E_1E_2\mid F)\to G(E_1\mid F)\times G(E_2\mid F)$. If a $\sigma$ is in the kernel then it acts trivially $E_1,E_2$ hence trivially on $E_1E_2$, so the kernel is trivial and the map is injective. (One may check that for $(\sigma,\tau)$ to be in the image it suffices for $\sigma,\tau$ to agree as actions on $E_1\cap E_2$ which is also Galois $\mid F$.)
Therefore any compositum of abelian extensions is abelian: if $G(E_1\mid F),G(E_2,F)$ are abelian then $G(E_1E_2\mid F)$ is a subgroup of the abelian group $G(E_1\mid F)\times G(E_2\mid F)$.
