I'm trying to apply the result of Jordan's theorem (cited below) to find the Galois group for a given polynomial. My goal is to provide an example where Jordan's theorem is useful, so the polynomial I'm using will have to be one whose Galois group is difficult to calculate without using Jordan's theorem. I chose the polynomial $f(x)=x^7-7x-1$, but any other examples that you think will better serve my goal are welcome.

First we have to show that the conditions for Jordan's theorem are met:

  • $f$ is irreducible and hence its Galois group $G$ is caontained in $S_7$.
  • $G$ is primitive acting on the spilitting field of $f$.
  • $G$ contains a $p$-cycle ($p<7-2=5$).

Proving the conditions are met:

  • $f$ is irreducible since $f(x)= x^7+x+1(mod 2)$ which is irreducible .
  • $G$ containes a permutations of cycle type (2,5) because $f(x)=(x^2+2x+2)(x^5+x^4+2x^3+2x+2)(mod3)$, and hence $G$ contains a 2-cycle ((2,5), whose 5th power is a 2-cycle).
  • $G$ is primitive acting on the roots of $f$ because G is transitive and $f$ has 7 roots. (by the way, is it ok to say "$G$ acts on the roots of $f$" in this context?)

Therefore, By jordan's theorem $G$ is $S_n$ or $A_n$. Since $disc(f)$ is not square we get that $G=S_7$.

Is this proof correct?

Jordan's theorem:

Let $G$ be a primitive permutation subgroup of the symmetric group $S_n$. If $G$ contains a $p$-cycle for some prime number $p<n−2$, then $G$ is either the whole symmetric group $S_n$ or the alternating group $A_n$. (They are in Section 3.3 of the book on Permutation groups by Dixon and Mortimer)

  • 1
    $\begingroup$ That is all correct. Nice. The statement on $G$ being primitive could be a little clearer. Maybe "$G$ is a primitive subgroup of $S_7$ since $G$ is the Galois group (hence acts transitively on the roots) of a $7^{th}$ degree polynomial" or such, $\endgroup$ – sharding4 Jun 11 '17 at 2:54
  • $\begingroup$ If the degree of $f$ is a prime $p$, you need only to know that $G$, as a subgroup of $S_p$, contains a $p$-cycle and a transposition, because $S_p$ is generated by such permutations. $\endgroup$ – nguyen quang do Jun 11 '17 at 10:14

Your Answer

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

Browse other questions tagged or ask your own question.