Galois group of a degree 6 polynomial

Solution 1:

This is a summary of my calculations based on the comments by David E. Speyer. I used a different description of the Galois group, but that doesn't matter.

The basic tool in use (see the earlier comments) is Dedekind's theorem relating factorizations of $$ f(x)= x^6 + 2 x^5 + 3 x^4 + 4 x^3 + 5 x^2 + 6 x + 7 $$ modulo various (unramified) primes to the cycle structure of elements of the Galois group $G$ as permutations of the roots (here six, so $G\le S_6$). As explained by the OP

• Modulo $p=3$ $f(x)$ is irreducible, so there is a 6-cycle in $G$. In particular $G$ is transitive and $f(x)$ is also irreducible over $\Bbb{Q}$.
• Modulo $p=11$ $f(x)$ splits as a product of a linear and a quintic, implying that $G$ contains a 5-cycle. Therefore the point stabilizer of $G$ among the set of roots acts transitively among the remaining roots, and $G$ is doubly transitive.
• The testing I did revealed that modulo $p=19$ $f(x)$ splits as a product of two linear factors and an irreducible quartic. Therefore $G$ contains a 4-cycle, and we can also conclude that $G$ is a triply transitive subgroup of $S_6$.
• More testing only gave cycle structures that are powers of the already listed, so this suggested that $G$ might be a transitive copy of $S_5$ inside $S_6$. It turns out that this is true. Let's record the fact that we already know $G$ to have order at least $120$.

I constructed that copy of $S_5$ from its conjugation action on its six Sylow $5$-groups: $P_1=\langle(12345)\rangle$, $P_2=\langle(12354)\rangle$, $P_3=\langle(12435)\rangle$, $P_4=\langle(12453)\rangle$, $P_5=\langle(12534)\rangle$ and $P_6=\langle(12543)\rangle$.

The group $S_5$ is generated by $\alpha=(12345)\in P_1$ and $\beta=(45)$. Conjugation action by $\alpha$ permutes the subscripts of the $P_i$s according to $a=(24653)$ whereas conjugation by $\beta$ permutes the Sylow-$5$s according to $b=(12)(35)(46)$. It follows that a desired copy of $S_5$ is the group $$ \tilde{G}=\langle(24653),(12)(35)(46)\rangle. $$ Using a suitable CAS (I used Mathematica) it is then straight forward to generate a list of elements of $\tilde{G}$.

David E. Speyer's idea is that the polynomial $$ P(x_1,x_2,\ldots,x_6)=\sum_{g\in \tilde{G}}x_{g(1)}^2x_{g(2)}^2x_{g(3)}x_{g(4)} $$ is invariant under $\tilde{G}$ but isn't invariant under $S_6$. Because $\tilde{G}$ is a maximal subgroup of $S_6$ we can use $P$ to distinguish the Galois groups $\tilde{G}$ and $S_6$. Namely, if $P$ evaluated at a carefully chosen permutation of the roots of $f(x)$ produces an integer, then that equation must be respected by the Galois group $G$, implying $G=\tilde{G}$.

Mathematica kindly gave me approximate zeros, and those are (rounded to only 4 decimals to save space) $$ \begin{aligned} z_1\approx-1.3079-0.5933i,&&z_2=\overline{z_1},\\ z_3\approx-0.4025-1.3417i,&&z_4=\overline{z_3},\\ z_5\approx\hphantom{-}0.7104-1.1068i,&&z_6=\overline{z_5}. \end{aligned} $$ Which permutation of these roots should we use? Because $\tilde{G}$ is triply transitive, there is no need to try anything other than $x_1=z_1,x_2=z_2,x_3=z_3$. Simply try out the different orderings of $z_4,z_5,z_6$. It turns out that $$ P(z_1,z_2,z_3,z_6,z_5,z_4)=264 $$ an integer to the precision I had available.

At this point I also tested that "complex conjugation" (based on the positions of the conjugate pairs in the list of variables), i.e. the permutation $(12)(36)(45)$ is, indeed, an element of $\tilde{G}$. This added to my confidence :-)

This could still be a false alarm in the sense that the actual value might just happen to be extremely close to $264$. David's suggestion was to look at the polynomial $$ H(T)=\prod_{\sigma\in Sym\{4,5,6\}}(T-P(z_1,z_2,z_3,z_{\sigma(4)},z_{\sigma(5)},z_{\sigma(6)})). $$ This is known to be invariant under $S_6$, and hence absolutely guaranteed to have integer coefficients. The idea is to verify that $H(264)=0$, which we can do with exact integer arithmetic alone. An expansion (still using approximate zeros, but now "legally" allowed to round the coefficients of $H(T)$ to the obvious nearest integer) gives $$ H(T)=T^6-240 T^5-101440 T^4+24410112 T^3+2093608960 T^2-447570968576 T-1492648329216, $$ and we can readily check that $H(264)=0$.

Solution 2:

Verification of the integer $P$:

? r = polroots(X^6+2*X^5+3*X^4+4*X^3+5*X^2+6*X+7)
%27 = [
-1.3078697439524358868574947207136826382 - 0.59329470741458755880701799527984032602*I, 
-1.3078697439524358868574947207136826382 + 0.59329470741458755880701799527984032602*I, 
0.71037886931271562450142054081955002103 - 1.1068452983838490198383021953838865547*I, 
0.71037886931271562450142054081955002103 + 1.1068452983838490198383021953838865547*I, 
-0.40250912536027973764392582010586738286 - 1.3416668277593834410394603953456211331*I, 
-0.40250912536027973764392582010586738286 + 1.3416668277593834410394603953456211331*I]~
? x = [r[1],r[2],r[3],r[6],r[4],r[5]]
%28 = [-1.3078697439524358868574947207136826382 - 0.59329470741458755880701799527984032602*I, 
-1.3078697439524358868574947207136826382 + 0.59329470741458755880701799527984032602*I, 
0.71037886931271562450142054081955002103 - 1.1068452983838490198383021953838865547*I, 
-0.40250912536027973764392582010586738286 + 1.3416668277593834410394603953456211331*I, 
0.71037886931271562450142054081955002103 + 1.1068452983838490198383021953838865547*I, 
-0.40250912536027973764392582010586738286 - 1.3416668277593834410394603953456211331*I]
? 4*x[6]^2*x[1]^2*x[2]*x[3] + 4*x[6]*x[1]*x[2]^2*x[3]^2 + 4*x[6]^2*x[1]*x[2]^2*x[4] + 4*x[1]^2*x[2]^2*x[3]*x[4] + 4*x[6]*x[1]^2*x[3]^2*x[4] + 4*x[6]^2*x[2]*x[3]^2*x[4] + 4*x[6]*x[1]^2*x[2]*x[4]^2 + 4*x[6]^2*x[1]*x[3]*x[4]^2 + 4*x[6]*x[2]^2*x[3]*x[4]^2 + 4*x[1]*x[2]*x[3]^2*x[4]^2 + 4*x[6]*x[1]^2*x[2]^2*x[5] + 4*x[6]^2*x[2]^2*x[3]*x[5] + 4*x[6]^2*x[1]*x[3]^2*x[5] + 4*x[1]^2*x[2]*x[3]^2*x[5] + 4*x[6]^2*x[1]^2*x[4]*x[5] + 4*x[2]^2*x[3]^2*x[4]*x[5] + 4*x[6]^2*x[2]*x[4]^2*x[5] + 4*x[1]*x[2]^2*x[4]^2*x[5] + 4*x[1]^2*x[3]*x[4]^2*x[5] + 4*x[6]*x[3]^2*x[4]^2*x[5] + 4*x[6]^2*x[1]*x[2]*x[5]^2 + 4*x[6]*x[1]^2*x[3]*x[5]^2 + 4*x[1]*x[2]^2*x[3]*x[5]^2 + 4*x[6]*x[2]*x[3]^2*x[5]^2 + 4*x[1]^2*x[2]*x[4]*x[5]^2 + 4*x[6]*x[2]^2*x[4]*x[5]^2 + 4*x[6]^2*x[3]*x[4]*x[5]^2 + 4*x[1]*x[3]^2*x[4]*x[5]^2 + 4*x[6]*x[1]*x[4]^2*x[5]^2 + 4*x[2]*x[3]*x[4]^2*x[5]^2
%29 = 264.00000000000000000000000000000000000 + 0.E-37*I