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A fun result of Conway shows that the length of terms in the Look-and-say sequence approach a constant ratio which is the root of a certain degree 71 polynomial, which can be found at that link.

What is the Galois group generated by the roots of this polynomial? One would expect it is $S_{71}$; is this tractable with computer algebra systems, or any other techniques?

Beren Gunsolus
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2 Answers2

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Write the polynomial as $C(x)$. It is irreducible mod $5$, so it's irreducible over $\mathbf Q$. Therefore the Galois group of $C(x)$ over $\mathbf Q$ is a transitive subgroup of $S_{71}$. Since $71$ is prime, we will use the following result: for prime $p$, a transitive subgroup of $S_p$ containing some $p$-cycle and some transposition must be $S_p$.

A theorem of Dedekind says that if a monic irreducible polynomial $f(x)$ in $\mathbf Z[x]$ factors modulo some prime $p$ as $\pi_1(x)\cdots \pi_r(x)$ for distinct monic irreducibles $\pi_i(x)$, then the Galois group of $f(x)$ over $\mathbf Q$ contains a permutation on the roots with cycle type $(d_1, \ldots, d_r)$, where $d_i = \deg \pi_i(x)$.

So from $C(x) \bmod 5$ being irreducible, the Galois group of $C(x)$ over $\mathbf Q$ contains a $71$-cycle.

The monic irreducible factorization of $C(x) \bmod 43$ has distinct factors with degrees
$1, 1, 2, 5, 17$, and $45$, so the Galois group contains an element $\sigma$ that permutes the roots with cycle type $(1,1,2,5,17,45)$. Therefore $\sigma^{17\cdot 45}$ is a $2$-cycle (transposition). So the Galois group contains a $71$-cycle and transposition, proving the group is $S_{71}$.

KCd
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    Dang it, I was 43 seconds too late! – Greg Martin Mar 14 '23 at 04:43
  • Nice (to both solvers!), I kinda wish that this polynomial wasn't prime degree so this problem would be harder ;) – Beren Gunsolus Mar 17 '23 at 01:38
  • If it had degree $n \geq 2$ and Galois group $S_n$, which is the typical situation, then it probably would not be hard to verify that: a transitive subgroup of $S_n$ that contains a transposition and a $p$-cycle for some prime $p > n/2$ must be $S_n$. Factoring the polynomial modulo enough primes would probably lead to the discovery of such permutations in the Galois group quickly in the same way as above. What's hard to do is compute Galois groups that are smaller than $S_n$ and $A_n$. – KCd Mar 17 '23 at 04:59
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A computer can check that $C(x)$ is irreducible modulo $5$ (which implies that $C(x)$ is irreducible over the integers as well, if we didn't already know that). By a famous theorem of Dedekind, this means that the Galois group of $C(x)$ contains a $71$-cycle. (Since $71$ is prime I think this fact also follows from the general transitivity of Galois groups.)

If we (get our computer to) factor $C(x)$ modulo $43$, the result is $(x-5)(x+4)p_2(x)q_5(x)r_{17}(x)s_{45}(x)$, where the subscript of each polynomial indicates the degree of that polynomial. Consequently by Dedekind's theorem again, the Galois group of $C(x)$ contains a permutation $\sigma\in S_{71}$ that (has two fixed points and) is the product of a $2$-cycle, a $5$-cycle, a $17$-cycle, and a $45$-cycle that are all disjoint. It follows that $\sigma^{17\cdot45}$, which is also in the Galois group of $C(x)$, is simply a $2$-cycle, that is, a transposition.

We are now done, since any subgroup of $S_{71}$ that contains both a $71$-cycle and a transposition must be all of $S_{71}$ (again since $71$ is prime).

Greg Martin
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  • If you're going to index the higher-degree irreducible factors by their degree, you might as well write them as $p_2$, $p_5$, $p_{17}$, and $p_{45}$ instead of using a different letter for each one. – KCd Mar 14 '23 at 04:48