I'm trying to find the lattice of subgroups of $\mbox{Gal}\left(\mathbb Q\left(n^{\frac{1}{4}},\,i\right)\Big/\mathbb Q \right)$.

As far I found that this group (seen as a subgroup of $S_4$) has the following elements (disjoint cycles):

  • $\sigma_1 = \mbox{id}$
  • $\sigma_2 = (3\;\;4)$
  • $\sigma_3 = (1 \;\;2)\,(3\;\;4)$
  • $\sigma_4 = (1\;\;2)$
  • $\sigma_5 = (1\;\;3\;\;2\;\;4)$
  • $\sigma_6 = (1 \;\;3)\,(2\;\;4)$
  • $\sigma_7 = (1\;\;4\;\;2\;\;3)$
  • $\sigma_8 = (1 \;\;4)\,(2\;\;3)$

Then I try to find the subgroups generated by these elements:

  • $<\sigma_5> = <\sigma_7> = \{\sigma_1,\sigma_3,\sigma_5,\sigma_7\}$, the only subgroup of order $4$.
  • $<\sigma_i>=\{\sigma_1,\sigma_i\}$, for $i=2,3,4,6,8$ the other subgroups of order $2$.

However in my class notes I have that this lattice corresponds to $\mbox{Dih}_4$.

I don't know where is my mistake.

EDIT Sorry, $n$ is a square-free integer.

  • $\begingroup$ what is $n?$ .. $\endgroup$
    – Ram
    Jan 3, 2013 at 10:02
  • $\begingroup$ And it is the dihedral group, if your corners numbered 1 and 2 are diagonally opposite in the square (ditto 3 and 4). But something is wrong with the cycle structure of $\sigma_5$ and $\sigma_7$ in that case. How did you number the roots of $x^4-n=0$? Which root is $\#1$, which is $\#2$ et cetera??? $\endgroup$ Jan 3, 2013 at 10:05
  • $\begingroup$ @JyrkiLahtonen: I numbered the roots the following way: $\alpha_1 = n^{1/4},\alpha_2=-n^{1/4},\alpha_3 = i\,n^{1/4}, \alpha_4=-i\;n^{1/4}$. $\endgroup$
    – synack
    Jan 3, 2013 at 10:11

1 Answer 1


Following your notation $\alpha_1=n^{1/4}$, $\alpha_2=-n^{1/4}$, $\alpha_3=in^{1/4}$, $\alpha_4=-in^{1/4}$, where $n$ is a square-free integer larger than 1.

An element $\sigma$ of the Galois group is fully determined, once we know $\sigma(\alpha_1)$ and $\sigma(i)$. The former can be any one of the roots, and the latter can be either $i$ or $-i$. The general theory tells us (presumably covered in your class notes) that there are altogether eight elements in the Galois group, so all choices are possible (this is not always the case, when a field extension is described by a list of generating elements).

You have written down most of the automorphisms correctly. I spotted the following two mistakes, so I cover them in detail. [Edit: These errors have since been corrected by the OP. The following is thus somewhat obsolete now, but still serves as an example in how to get the permutation of roots given the action of the automorphism on the generators of the field extension.]

The choices $\sigma(\alpha_1)=\alpha_3$, $\sigma(i)=i$ mean that: $$ \sigma(\alpha_2)=\sigma(-\alpha_1)=-\sigma(\alpha_1)=-\alpha_3=\alpha_4, $$ $$ \sigma(\alpha_3)=\sigma(i\alpha_1)=\sigma(i)\sigma(\alpha_1)=i\alpha_3=\alpha_2,$$ and $$\sigma(\alpha_4)=\sigma(-i)\sigma(\alpha_1)=-i\alpha_3=\alpha_1.$$ Thus this automorphism permutes the roots according to the 4-cycle $$ \alpha_1\mapsto\alpha_3\mapsto\alpha_2\mapsto\alpha_4\mapsto\alpha_1 $$ or in other words $\sigma=(1 3 2 4)$. Thus $\sigma$ is your automorphism $\sigma_5$.

The other error was with your $\sigma_7$. It is actually the inverse of the above $\sigma=\sigma_5$. It is similarly determined from the data $\sigma_7(i)=i$, $\sigma_7(\alpha_1)=\alpha_4$, and the corresponding permutation is then $(1 4 2 3)$.

Note that these 8 permutations are the symmetries of the square, if the numbering of corners is such that $1$ and $2$ as well as $3$ and $4$ are diagonally opposite. Draw a picture where you plot the roots in the complex plane to see this! The permutation $(1 2)$ must map $\alpha_1\mapsto\alpha_3=-\alpha_1$, but it must also map $\alpha_2=i\alpha_1$ to itself, so it must map $i\mapsto -i$. The permutation $(3 4)$ that interchanges the two imaginary roots is just the usual complex conjugation.

A subgroup missing from your list is the Klein 4-group consisting of the permutations $$ V_4=\{\sigma_1,\sigma_3,\sigma_6,\sigma_8=\sigma_3\sigma_6\}=\langle \sigma_3,\sigma_6\rangle. $$

  • $\begingroup$ Thanks, I've corrected the errors. Your explanation helped me a lot. However, if I want to find the subgroups, how can I know when to stop because I have all of them? I think that there is still another subgroup of order 4 missing. $\endgroup$
    – synack
    Jan 3, 2013 at 10:35
  • 1
    $\begingroup$ In general that may be somewhat laborious. Start by listing the cyclic subgroups as you did. Add one generator at a time to any one of them. You may then get the full group (happens here more often than not), but you may also get something new. Do this for all of them and remove duplicates (there will be plenty). Repeat this with the freshly gotten subgroups. Practice will allow you to spot duplicates early. For example, above the subgroup generated by $\sigma_3$ and $\sigma_6$ contains $\sigma_8$, so it is pointless to check e.g. the subgroup generated by $\sigma_3$ and $\sigma_8$. $\endgroup$ Jan 3, 2013 at 10:41
  • 2
    $\begingroup$ +1 Okay, I've found another subgroup of order $4$, the one generated by $<\sigma_3,\sigma_4>=\{\sigma_1,\sigma_2,\sigma_3,\sigma_4\}$. $\endgroup$
    – synack
    Jan 3, 2013 at 10:50
  • $\begingroup$ @Kits89: Well done!! $\endgroup$ Jan 3, 2013 at 11:23

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