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Let $f$ be an irreducible separable polynomial of degree $n$ over a field $k$. Let $G$ be the Galois group of the splitting field of $f$ over $k$. Let $G$ be simple. Then, $G$ is a subgroup of $A_n$.

Since $G$ is simple, it has no proper normal subgroup. So no intermediate field between $k$ and the splitting field of $f$ can be normal over $k$.

To show subgroup of $A_n$ I need to show $G$ fixes $\sqrt{D}$ where $D$ is the discriminant. I am not sure how the proof should go.

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  • $\begingroup$ Some subgroup of $G$ fixes $\sqrt{D}$, including the identity. It's the kernel of the action of the Galois group on $k(\sqrt{D})$ so it's a normal subgroup. So, it is either trivial or the whole group. Can it be trivial? $\endgroup$ – probablystuck May 6 '18 at 17:08
  • $\begingroup$ I'm sorry but I don't quite follow how you infer the subgroup is normal. Can you please use the fundamental theorem of GT language? $\endgroup$ – Landon Carter May 6 '18 at 17:16
  • $\begingroup$ I've clarified my thinking (and taken a different approach) in the answer below, which should be straightforward to understand. The only bit of Galois Theory you need is that $G$ is a subgroup of $S_n$ - the rest follows from group theory. $\endgroup$ – probablystuck May 6 '18 at 17:31
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Suppose $G$ is simple. Then $G$ is isomorphic to a transitive subgroup of $S_n$ (you can see this by seperability, irreducibility and considering the action on the roots). Consider the homomorphism $\phi:G\rightarrow \{1,-1\}^\times$ given by taking the sign of each permutation. This homomorphism has kernel all even permutations in $G$. Hence, all even permutations in $G$ form a normal subgroup. If $G$ is nontrivial, then either:

Case 1: $\phi$ has trivial kernel and $G$ consists only of elements of order $2^k$. From this it is possible to show $G$ is either not simple or is isomorphic to $C_2$ - which is simple and not a subgroup of $A_2$ - the statement is false in this case: look for example at $x^2-2$.

Case 2: $\phi$ has nontrivial kernel. The kernel is a normal subgroup of $G$ and hence must be the whole of $G$.

In either case, $G$ lies in $A_n$ so long as $n>2$.

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