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(I know the existence of Galois group of $x^4-2$, so please don't start tagging this question as duplicate)

The degree of the field extension is 8, so the only group that the Galois group can be isomorphic to is $D_4$, the symmetry group of the square.

Now, I know about the definition of $D_4$, which is $D_4= \langle a,b\mid a^4=1,b^2=1,ba=a^{-1}b\rangle$. Why then are the elements of $D_4$ $a, a^2, a^3, b, ab, a^2b, a^3b$ and id? Would it make any sense to make a Cayley table of the automorphisms and $D_4$? How do I know that all the automorphisms are in $D_4$?

Furthermore, how can I know the Galois group is isomorphic to $D_4$ in another way?

I know it's a lot of questions, but could someone please answer something, it might clear up a lot for me. :)

Community
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Dr. Heinz Doofenshmirtz
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  • The question, "How can I know the Galois group is isomorphic to D4 in another way?" really does seem to be a duplicate of the indicated question, and so perhaps it should be excised. The question "Why then are the elements of $D_4$ $a, a^2, a^3, b, ab, a^2b, a^3b$ and id?" also is a duplicate (or near-duplicate) of a question that has been asked in the last few days, and so probably should be removed, too. – Travis Willse Oct 31 '15 at 22:55
  • As for the remaining questions, one could certainly identify elements $a, ab,$ etc., of $D_4$ with explicit automorphisms in $\operatorname{Gal}(x^4 - 2)$, and indeed this is probably instructive if you haven't done that sort of exercise before. But this identification is not canonical---it's only defined up to automorphism of $D_4$. (Here, $\operatorname{Aut}(D_4) \cong D_4$, but NB the analogous statement is not true for most dihedral groups.) – Travis Willse Oct 31 '15 at 23:00
  • It would (IMO) help to have a square handy for pictures to better visualize $D_4$.. this has always helped me out.. it’s the rigid Motions of a square. Which is rotations and symmetries. – MyMathYourMath Sep 29 '24 at 17:04
  • Related: https://math.stackexchange.com/questions/4909800/galois-group-of-xn-p-d-n – Nic Jan 27 '25 at 21:06

2 Answers2

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One of the many questions in your post is why the elements of $D_4$ are those 8 expressions in $a$ and $b$? From your definition of $D_4$ (as presentation) it is clear that elements are expressions of the type: $$a^{k_1}ba^{k_2}b\ldots ba_{k_n}\qquad(*)$$ where $k_j$'s are integers between 1 to 3. Allowing $k_1$ and $k_n$ to be zero takes care of generating expressions starting/ending in $b$.

Now the condition $ba= a^{-1}b$ can be used straighten all the expressions: that is, if $b$ happens to be ahead of a we can make the $a$ to come ahead, but at a price. The should be replaced by its inverse, which is the same as $a^3$. Thus all the $a$'s can migrate to the left leaving all the $b$'s to the right. As $b^2$ is is identity, an expressions of the type $(*)$ with en even number of $b$'s will simplify to a mere power of $a$, and expressions with an odd number of $b$'s end as a power of $a$ followed by a single $b$.

P Vanchinathan
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Since $x^4-2$ is irreducible, we can adjoining any root $\alpha $ and get an extension of degree $4.$

To get the other roots and make sure we have the splitting field we adjoin a primitive fourth root of unity $\zeta, $ which is another extension of degree $2.$

The field extension is finally of order $2\cdot 4=8.$

Playing with the automorphisms that permute the roots we find that the Galois group is non-abelian.

There's only two non-abelian groups of order $8.$

But since more than one of the automorphisms have order two, the quaternions are ruled out.

suckling pig
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