Let $f$ be an irreducible (over $\mathbb{Q}$) polynomial in $\mathbb{Z}[x]$, $\deg (f)=3,4,5$. The Galois group of an irreducible polynomial $f\in \mathbb{Z}[x]$ acts transitively on distinct roots in $\mathbb{C}$ of $f$. Hence $Gal(f)\leq S_m$, and acts transitively on distinct $m$-roots, where $m\leq n$ for $n=\deg (f)$.

$m=3$, $S_3,A_3=\mathbb{Z}_3$

$m=4$, $S_4,A_4,D_4,\mathbb{Z}_4,K_4$ ($K_4$ is the Klein four group)

$m=5$, $S_5,A_5,D_5,\mathbb{Z}_5,Fr_5$ ($Fr_5$ is a Frobenius group)

For example, by Eisenstein criterion, the following polynomials are irreducible, but:

Determine over $\mathbb{Q}$, $Gal(x^3-3x+1)=\mathbb{Z}_3$, not $S_3$? Why?

Determine over $\mathbb{Q}$, $Gal(x^4+3x+3)=D_4$, not $A_4,K_4, S_4$? Why?

Also by determine coefficients, $x^4+8x+12$ is irreducible. Why $Gal(X^4+8x+12)=A_4$, over $\mathbb{Q}$, not $S_4,K_4, \mathbb{Z}_4$?

Is there any method to determine the Galois group?

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    $\begingroup$ As told by Mr.Igor rivin, your subgroups of $S_n$ are all transitive... I would suggest you to try proving that $\textbf{galois group of an Irreducible polynomial is transitive}$ ( I am sure it would not be much difficult) and second part is to see that these are $\textbf{transitive subgroups of $S_n$}$ for respective $n$.... $\endgroup$
    – user87543
    Dec 17, 2013 at 2:49

1 Answer 1


The Galois group of an irreducible polynomial is transitive, so your list is the list of transitive permutation groups on $3, 4, 5$ elements.

And no, we cannot determine the Galois group if we have the number of real roots (in general), and no, there is no quick method to determine the Galois group, though for most polynomials of degree $n$ it is $S_n.$ Confirming or denying that it is $S_n$ is fast, as described at length in this quite recent preprint.

  • $\begingroup$ The paper is quite interesting.. I do not have necessary background to understand the paper completely but I could understand some parts and i would try to work on that... Thank you :) $\endgroup$
    – user87543
    Dec 17, 2013 at 2:54
  • $\begingroup$ @PraphullaKoushik My pleasure... $\endgroup$
    – Igor Rivin
    Dec 17, 2013 at 3:41

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