# Prove an isomorphism between two galois groups

Let $E/K$ and $L/K$ be finite extension. E/K is galois. prove that $$\sigma\in Gal(EL/L) \rightarrow \sigma|_{E}\in Gal(E/E\cap L)$$ is a group isomorphism.

First we have show this is well-defined, for $\sigma\in Gal(EL/L)$, we must prove that $\sigma(x)\in E$ $,\forall x\in E$. This follows from that $E/K$ is normal.

For injectivity, if $\sigma|_{E}$ fixes $E$, then $\sigma$ fixes $EL$

For surjectivity, we have to show that any automorphism in $Gal(E/E\cap L)$ can be extended to an automorphism in $Gal(EL/L)$. I'm stuck in this part.

## 1 Answer

For surjectivity, I would be inclined to use the Fundamental Theorem of Galois Theory.

The main idea is as follows. We have a group homomorphism $$\rho : Gal(EL/L) \to Gal(E/K), \ \ \ \ \ \sigma \mapsto \sigma |_E.$$ The image $G = \rho(Gal(EL/L))$ is a subgroup of $Gal(E/K)$. Our task is to show that $G = Gal(E/E \cap L)$.

Let $F$ be the subfield of $E$ containing all elements of $E$ that are fixed by $G$.

Since $E/K$ is a Galois extension, the Fundamental Theorem of Galois Theory tells us that $G = Gal(E/F)$.

So it remains to show that $F = E \cap L$. I'll leave that to you...

[Don't forget that $EL/L$ is also a Galois extension - we showed this in a previous question on math.SE. You may wish to use the Fundamental Theorem a second time.]