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.

  • $\begingroup$ 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

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 $n>2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.