Let $L/\mathbb{Q}$ be a finite Galois extension with $[L:\mathbb{Q}]=n$ and let $G=\mathrm{Gal}(L/\mathbb{Q})$ be its Galois group. Then, we know from Galois theory that the order of $G$ is $n$. Since $G$ is the group of permutations of zeros of some polynomial $f$ why is not the order of $G$ always equal to $n!$? Or, in another words, when is $|G|=n!$ for $n>5$?

By Shafarevič theorem, we know for any finite solvable group $H$ there exist a Galois group which is isomorphic to $H $. I look for Galois group which is isomorphic to $\mathbb{Z}/n!\mathbb{Z}$. By Shafarevič theorem we know that this group exists because $\mathbb{Z}/n!\mathbb{Z}$ is abelian, but I want to determine it. I find just these two examples, but I am not sure if the Galois group of them is abelian whether any other examples?

  1. $[\mathbb{Q}(\sqrt[2]{p_{1}}, \sqrt[3]{p_{2}}, .......,\sqrt[n]{p_{n}}):\mathbb{Q}]=n!$, where $p_{i} $ is a prime number and $i=1,...,n$.

  2. $[\mathbb{Q}(\sqrt[n!]{2}): \mathbb{Q}]=n!$

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    $\begingroup$ If the degree $n$ of the splitting field of $f$ equals the degree of the polynomial $f$, then (for $n>2$) the Galois group can't be isomorphic to all of $S_n$ -- it will be isomorphic to subgroup of $S_n$. For example cyclotomic poly's over the rationals, like $x^4+x^3+x^2+x+1$, where the splitting field is generated by (any) one root and so has degree $4$, not $24$. $\endgroup$
    – Ned
    Commented Aug 9, 2020 at 15:19
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    $\begingroup$ Neither of the extensions in 1) or 2) is Galois when $n>2$. $\endgroup$ Commented Aug 9, 2020 at 16:55
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    $\begingroup$ @Abdo I have no idea about your example (1), but for (2), the splitting field will be generated by the real $n!$-th root of 2 AND a primitive $n!$-th root of unity, which will have degree strictly bigger than $n!$. The degree equation you wrote is right, but that is not a Galois (aka "normal") extension since it doesn't include the other roots. $\endgroup$
    – Ned
    Commented Aug 9, 2020 at 18:22
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    $\begingroup$ Abdo, one way is to find a prime number $p\equiv1\pmod{n!}$, say $p=1+k\,n!$. Then the Galois group of $\Bbb{Q}(\zeta_p)/\Bbb{Q}$ is cyclic of order $k\,n!$. Let $H$ be the subgroup of order $k$. Then the fixed field $L$ of $H$ is a degree $n!$ extension of $\Bbb{Q}$ such that $Gal(L/\Bbb{Q})$ is cyclic of order $n!$. $\endgroup$ Commented Aug 9, 2020 at 19:32
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    $\begingroup$ @Abdo you missed the “k”. There are infinitely many by Dirichlet. $\endgroup$
    – user208649
    Commented Aug 9, 2020 at 21:42

1 Answer 1


I like this question because it speaks to a natural misunderstanding: why can't we always send any root to any other one? I think the best way to absorb this idea is by example.

Example 1: Let $K = \mathbb{Q}[\sqrt{2},\sqrt{3}]$ and notice that $K/\mathbb{Q}$ has degree four and is the splitting field of the polynomial $(x^2 - 2)(x^2 - 3)$. Thus, $K/\mathbb{Q}$ is a Galois extension of degree four so, as you mention, $| \mathrm{Gal}(K/\mathbb{Q}) | = 4$. In fact, this Galois group has generators $\sqrt{2} \mapsto -\sqrt{2}$ and $\sqrt{3} \mapsto -\sqrt{3}$ and is thus isomorphic to the Klein four group. Now to your question: why isn't the Galois group just $S_4$? Here two ways to think about it:

  • One way is to leverage the fact that a $\mathbb{Q}$-automorphism must send roots of polynomials over $\mathbb{Q}$ to other roots of the same polynomial. In our example: take $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$ and consider the polynomial $x^2 - 2$. Then the fact that $\sigma$ is an automorphism fixing $\mathbb{Q}$ implies $\sigma(x^2 - 2) = \sigma(x)^2 - 2$, which in turn implies that $\sigma(x)$ is itself a root of $x^2 - 2$. That is, $\sigma(x) = \sqrt{2}$ or $\sigma(x) = -\sqrt{2}$. There are no other options. For $\mathrm{Gal}(K/\mathbb{Q})$ to equal $S_4$, we would need automorphisms which can permute any pair of roots, but that cannot happen here!

  • A less precise way of thinking about this is to ponder automorphisms conceptually: automorphisms --- as isomorphisms --- should preserve the structure of the field. In other words, once you apply an automorphism $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$, you should wind up with a field $\sigma(K)$ algebraically indistinguishable from $K$. So $\sqrt{2}$ and $\sigma(\sqrt{2})$ should have exactly the same algebraic properties. This works for $\sqrt{2}$ and $-\sqrt{2}$ (the ordering of the real numbers and the fact that $-\sqrt{2} < \sqrt{2}$ is an analytic, not algebraic, concern), but does not work for $\sqrt{2}$ and $\sqrt{3}$ because --- for example --- $(\sqrt{2})^2 = 2 \neq 3 = (\sqrt{3})^2$ (if $2 = 3$ then $0 = 1$ and we're dealing with the "zero ring").

This example teaches the following lesson: the Galois group is all about preserving the symmetry of a field, but the amount of symmetry is necessarily limited. Elements of a field will often have subtle/nuanced interactions, and the Galois group picks up on that! If $\mathrm{Gal}(K/\mathbb{Q})$ were always a full symmetric group, then that would suggest that we always have a maximum amount of symmetry (but we don't).

Exercise: Analyze the situation for the field $K = \mathbb{Q}[e^{2 \pi i/4}]$.

Example 2: Let $\zeta_3 = e^{2\pi i/3}$ be a primitive third root of unity and let $K = \mathbb{Q}[2^{1/3}, \zeta_3]$, an extension of degree six (take this on faith for now and then prove it later once you understand this example). Then $\mathrm{Aut}(K/\mathbb{Q})$ contains an element $\sigma$ which sends $\zeta_3$ to $\zeta_3^2$ and fixes everything unrelated to $\zeta_3$ ($\sigma$ is also known as complex conjugation); to make perfectly clear what $\sigma$ does, here are some examples of its action:

  • $\sigma(\zeta_3^2) = \sigma(\zeta_3)^2 = \zeta_3^4 = \zeta_3$,
  • $\sigma(2^{1/3}) = 2^{1/3}$,
  • $\sigma(\zeta_3 + 2^{1/3}) = \zeta_3^2 + 2^{1/3}$, and
  • $\sigma(q) = q$ for all $q \in \mathbb{Q}$.

Note that the first bullet implies that $\sigma^2$ is the identity map. (You should pause here to prove that $\sigma$ indeed defines a $\mathbb{Q}$-automorphism of $K$.) Similarly, $\mathrm{Aut}(K/\mathbb{Q})$ contains an element $\tau$ which sends $2^{1/3}$ to $\zeta_3 2^{1/3}$ and fixes everything unrelated to $2^{1/3}$ (as before, prove that $\tau$ is an automorphism, compute some examples of its action, and show that $\tau^3$ is the identity.) Then $\sigma$ and $\tau$ generate all of $\mathrm{Aut}(K/\mathbb{Q})$ (why?) so $\mathrm{Aut}(K/\mathbb{Q}) = < \sigma, \tau >$. We thus have a natural isomorphism of groups \begin{align*} \mathrm{Aut}(K/\mathbb{Q}) &\to S_3 \\ \sigma &\mapsto (2\, 3) \\ \tau &\mapsto (1\, 2\, 3) \end{align*} so in fact $K/\mathbb{Q}$ is a Galois extension with Galois group $S_3$. How does this example relate to the discussion of permuting roots of a polynomial? Without going into too many details, we have $K = \mathbb{Q}[2^{1/3}, \zeta_3] = \mathbb{Q}[2^{1/3} + \zeta_3]$ where $2^{1/3} + \zeta_3$ has minimum polynomial $$p(x) = x^6 + 3 x^5 + 6 x^4 + 3 x^3 + 9 x + 9$$ over $\mathbb{Q}$. The roots of $p(x)$ are

  • $2^{1/3} + \zeta_3$
  • $\zeta_3 2^{1/3} + \zeta_3$
  • $\zeta_3^2 2^{1/3} + \zeta_3$
  • $2^{1/3} + \zeta_3^2$
  • $\zeta_3 2^{1/3} + \zeta_3^2$
  • $\zeta_3^2 2^{1/3} + \zeta_3^2$

so that $\sigma$ and $\tau$ together shuffle around all roots of $p(x)$! In some sense, this example therefore has the maximum possible amount of symmetry.

Now, to the extensions 1) and 2) in your question: as Jyrki notes in their comment on your question, neither extension is Galois (over $\mathbb{Q}$). I hope that the examples I've explained help you to see why.

EDIT: As pointed out in the comments, I neglected to address the construction of a Galois group isomorphic to $\mathbb{Z}/n!\mathbb{Z}$, the cyclic group of order $n!$. I think the easiest way to do this is to apply the fundamental theorem of Galois theory to a cyclic Galois extension. Indeed, let $n!$ divide $d$, where $d$ is the degree of some cyclic Galois extension $L/K$. Then there is a cyclic subgroup of $\mathrm{Gal}(L/K)$ of degree $n!$ and the fixed field of this subgroup will be the desired Galois extension (note that we are using that cyclic groups are abelian here to get that the intermediate extension is Galois). There are many options for the cyclic Galois extension $L/K$ (e.g. extensions of finite fields, whose Galois groups are comparatively straightforward).

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    $\begingroup$ You should be careful of your characterization of the "common" misconception. For an irreducible polynomial, it is true that you can send any root to any other, and this leads to the actual misconception that you can simultaneously do this for all the roots. $\endgroup$
    – user208649
    Commented Aug 9, 2020 at 18:11
  • $\begingroup$ @roxas3582 yes i understand your explanation .but I am looking for Galois extension which is isomorphic to $\mathbb{Z}/n!\mathbb{Z}$ do you know any example ? $\endgroup$
    – Abdo
    Commented Aug 9, 2020 at 18:25
  • $\begingroup$ @roxas3582 can you explain why is neither of the extension 1) and 2) are not Galois $\endgroup$
    – Abdo
    Commented Aug 9, 2020 at 18:34
  • $\begingroup$ @Abdo I edited my answer to address $\mathbb{Z}/n!\mathbb{Z}$. As for why 1) and 2) aren't Galois, I suggest that you think about the possible $\mathbb{Q}$-automorphisms for those extensions and compare that to example 2 in my answer. $\endgroup$
    – xion3582
    Commented Aug 9, 2020 at 22:24
  • $\begingroup$ @Abdo I'm not sure what "it" refers to, but either way that polynomial is not irreducible (so it cannot be a minimum polynomial). But I think you're on the right track. I suggest that you read a bit about either 1) the Galois theory of finite field extensions, or 2) the Galois theory of cyclotomic extensions. Either one will help you identify a Galois group isomorphic to $\mathbb{Z}/n!\mathbb{Z}$ for any $n$. $\endgroup$
    – xion3582
    Commented Aug 10, 2020 at 14:38

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