Let $f_n(x)=x^n-x^{n-1}-....-x-1$ be the polynomial where $n$ is even. Now, $f_n(x)$ is an irreducible polynomial over $\mathbb{Q}$, so Galois group of $f_n(x)$, say Gal$(f_n)$ is a transitive subgroup of $S_n$. Also, Gal($f_n$) contains a transposition. Now, I need to show that Gal$(f_n)$ is generated by transpositions. How do I show that?


  • $\begingroup$ How do you know, that $f_n(x)$ is irreducible? $\endgroup$ – user276611 Jul 22 '17 at 9:16
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    $\begingroup$ The polynomial is irreducible, may be for similar reasons like here, using the reciprocal. It does not matter that $n$ is even. $\endgroup$ – Dietrich Burde Jul 22 '17 at 11:31
  • $\begingroup$ If you assume all the results that I mentioned, can you prove that Gal($f_n$) is generated by transpositions? $\endgroup$ – uuuuuuuuuu Jul 22 '17 at 12:45
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    $\begingroup$ This is the same as claiming that $\textrm{Gal}(f_n)=S_n$: math.stackexchange.com/questions/1993832/… $\endgroup$ – user138530 Jul 22 '17 at 19:13
  • $\begingroup$ @ChristianRemling yes I am actually trying to prove that only $\endgroup$ – uuuuuuuuuu Jul 22 '17 at 19:18

Not a full answer. Basically I want to explain why the ultimate target here is out of reach of the listed observations:

  • $Gal(f_n)$ is a transitive subgroup of $S_n$ because $f_n$ is irreducible,
  • $Gal(f_n)$ contains a transposition.

I don't know how the OP concluded these. Leaving that. But let's record the following

Fact. For all even $n$, $n=2k, k>1$ there exists a proper transitive subgroup $G$ of $S_n$ such that $G$ contains a transposition.

Proof. The wreath product $G=S_2\wr S_k$ has these properties. We can also describe it as the group of signed permutations. Let $I_k=\{\pm x_1, \pm x_2,\ldots,\pm x_k\}$ be distinct variables, and then $$ G=\{\sigma:I_k\to I_k\mid \text{$\sigma$ is bijective, and $\sigma(-x_i)=-\sigma(x_i)$ for all $i$} \}. $$ Clearly $G$ contains transpositions like the permutation that swaps $+x_1$ and $-x_1$, and has the remaining elements of $I_k$ as fixed points. On the other hand $G$ is easily seen to act transitively on $I_k$.

A few remarks about the Galois groups. Let $G_n$ be $Gal(f_n)$ when viewed as a subgroup of $S_n$, the permutations of roots. The polynomial $f_n, n=2k,$ is easily seen to have exactly two real roots. Therefore complex conjugation is a product of $(k-1)$ transpositions $\in G_n$. After all, the remaining $2(k-1)$ zeros are $k-1$ conjugate pairs. I include what can be (easily, if you have Mathematica or some such tool) added by using Dedekind's theorem relating the cycle structure of elements $G_n$ to the factorization of $f_n$ modulo primes.


  • Modulo $p=2$ the polynomial $f_4$ is irreducible (so $f_4$ is also irreducible over $\Bbb{Q}$). Therefore $G_4$ contains a 4-cycle.
  • Modulo $p=3$ the polynomial $f_4$ has a linear factor $x-1$ and an irreducible cubic factor. Therefore $G_4$ contains a 3-cycle.
  • This is enough to conclude that $G_4=S_4$. Lagrange's theorem tells us that the Galois group has order $12$ or $24$. The former can occur only if the Galois group were $A_4$, but that doesn't contain $4$-cycles.


  • Modulo $p=3$ $f_6$ is irreducible, so $f_6$ is irreducible over $\Bbb{Q}$ as well, and the group $G_6$ contains $6$-cycles.
  • Modulo $p=17$ $f_6$ is a product of a linear factor and a quintic factor. Therefore $G_6$ contains a $5$-cycle. Together with the previous bullet this implies that $G_6$ is doubly transitive. After all, at least one point stabilizer is transitive, hence all of them are (by transitivity).
  • Modulo $p=5$ $f_6$ has two linear factors and one irreducible quartic. Therefore at least one stabilizer of a pair of roots contain a $4$-cycle. We can conclude that $G_6$ is thus 3-transitive.
  • Modulo $p=59$ $f_6$ has three linear factors and one irreducible cubic. Together with the previous bullet this shows that a two-point-stabilizer subgroup of $G_6$ contains a $4$-cycle and a $3$-cycle. As in the case $n=4$ this implies that the two-point-stabilizer must be all of $S_4$. This implies that $G_6=S_6$.

The order of $S_6$ is $720$. The group contains $\binom 62=15$ transpositions, so one element out of $48$ is a transposition. By Chebotarev's density result asymptotically one prime $p$ out of forty-eight has the property that modulo $p$ $f_6$ should have four linear factors and a single irreducible quadratic factor. I checked the $55$ smallest primes up to $p=257$ until I found one such prime. The smallest prime giving a factorization of type $2+2+1+1$ is $p=241$, and that should be three times as frequent (and is the conjugacy class of complex conjugation). My sample is small so the observed probabilities are probably well within the known error bounds :-)

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  • $\begingroup$ Sorry about not being able to say more. With paper and pencil work all I could say about a more general $n=2k$ would be a few things about the factorization of $f_n$ modulo $p=2$. Even in the case $n=4$ that was not enough to conclude, so I brute forced these two cases. $\endgroup$ – Jyrki Lahtonen Jul 23 '17 at 13:54

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