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I am trying to understand the proof of the Gauss-Wantzel Theorem given on page 601 of Abstract Algebra by Dummit and Foote.

Proposition 14.5.29 (Gauss-Wantzel): A regular $n$-gon can be constructed by a straightedge and compass if and only if $n = 2^k p_1 \cdots p_r$ is the product of a power of $2$ and distinct Fermat primes.

The proof they outline is as follows:

  1. Recall that $\alpha \in \mathbb{R}$ is constructible over $\mathbb{Q}$ if and only if the field $\mathbb{Q}(\alpha)$ is contained in a field $K$ obtained by a series of quadratic extensions: \begin{equation*} \mathbb{Q} = K_0 \subset \dots \subset K_m = K \end{equation*} with $[K_{i + 1} : K_i] = 2$ for all $i$.
  1. Constructing a regular $n$-gon is equivalent to constructing the $n$-th roots of unity $1, \zeta_ n, \zeta_n^2, \dots, \zeta_n^{n -1}$ since they form the vertices of a unit regular $n$-gon. The roots of unity $1, \zeta_n, \zeta_n^2, \dots, \zeta_n^{n -1}$ are constructible if and only if $\zeta_n$ is constructible since powers of constructible numbers are constructible (they form a field). Furthermore, $\zeta_n = e^{2\pi i/n} = \cos(\frac{2\pi}{n}) + i \sin(\frac{2\pi}{n})$ is constructible if and only if $\cos(\frac{2\pi}{n})$ and $\sin(\frac{2\pi}{n})$ are constructible and as $\sin(\frac{2\pi}{n}) = \sqrt{1 - \cos^2(\frac{2\pi}{n})}$, we see $\zeta_n$ is constructible if and only if $\cos(\frac{2\pi}{n})$ is constructible. Therefore, the regular $n$-gon is constructible if and only if $\cos(\frac{2\pi}{n})$ is constructible.
  1. Note that $\zeta_n$ satisfies $\zeta_n^2 - 2 \cos(2\pi/n) \zeta_n + 1 = 0$ over $\mathbb{Q}(\cos(2\pi/n))$. Since $\mathbb{Q}(\cos(2\pi/n))$ consists only of real numbers it follows that $[\mathbb{Q}(\zeta_n) : \mathbb{Q}(\cos(2\pi/n))] = 2$, so $[\mathbb{Q}(\cos(2\pi/n)) : \mathbb{Q}] = \phi(n)/2$.
  1. The regular $n$-gon is constructible if and only if $\phi(n)$ is a power of 2.
    (i) For one direction, if a chain of extensions as in (1) exists, then $\phi(n) = [\mathbb{Q}(\zeta_n) : \mathbb{Q}] = [\mathbb{Q}(\zeta_n) : \mathbb{Q}(\cos(2\pi/n) = K_m] [K_m : K_{m - 1}] \dots [K_1 : K_0 = \mathbb{Q}] = 2^{m + 1}$.
    (ii) Conversely if $\phi(n) = 2^m$ is a power of 2, then the Galois group $\mathrm{Gal}(\mathbb{Q}(\zeta_n)/ \mathbb{Q})$ is an abelian group of order a power of 2, so the same is true for the Galois group $\mathrm{Gal}(\mathbb{Q}(\cos(2\pi/n))/\mathbb{Q})$. It is easy to see by the fundamental theorem of abelian groups that the an abelian group $G$ of order $2^m$ has a chain of subgroups \begin{equation*} G = G_m > G_{m - 1} > \dots > G_{i + 1} > G_i > \dots > G_0 = 1 \end{equation*} with $[G_{i + 1} : G_i] = 2$ for all $i$. Applying this to the Galois group $G = \mathrm{Gal}(\mathbb{Q}(\cos(2\pi/n))/\mathbb{Q})$, and taking the fixed fields for the subgroups $G_i$, by the Fundamental Theorem of Galois Theory, we obtain the required sequence of quadratic extensions.
  1. $\phi(n)$ is a power of 2 if and only if $\phi(n)$ is a power of 2.

I am having trouble understanding 4(ii), the claim that by fundamental theorem of abelian groups that an abelian group $G$ of order $2^m$ has a chain of subgroups $G = G_m > \cdots G_0 = 1$ with $[G_{i + 1} : G_i] = 2$. How does this follow from the fundamental theorem of abelian groups?

Shaun
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user942
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    Please don't rely on pictures of text. They're not user-friendly. Besides, if you want people to put more effort into helping you, then you ought not to be shy of typing things up. – Shaun Apr 14 '21 at 16:27
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    See this post. – Dietrich Burde Apr 14 '21 at 16:42
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    To expand on Shaun's comment: images are not searchable (either within math.se, or through web searches) so your post will not come up if people are searching for this material. Also, some people use screen readers, and many of them cannot handle pictures that do not have alt-text that contains the material in the picture, which is most of them and is in fact the case here, so your post is inaccessible. Finally, posts should be self-contained. The picture can be used to help, but it should be possible to understand the question without having to be able to see the picture. Please re-type. – Arturo Magidin Apr 14 '21 at 17:25
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    @ArturoMagidin thank you for the helpful suggestions for improvements to the post. I have retyped the question. – user942 Apr 14 '21 at 17:54

1 Answers1

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They are using the fundamental theorem of abelian groups to write $G$ as a product of cyclic groups. Since $|G|$ is a power of $2$, all of these factor groups must be of the form $\def\Z{\mathbb Z}\Z/2^i\Z$ for some $i\ge 1$. In the case where $G=\Z/2^i\Z$, the following cha $$ \langle 0\rangle=\langle 2^i\rangle <\langle 2^{i-1}\rangle<\dots<\langle 2\rangle<\langle 1\rangle=\Z/2^i\Z $$ In words, the smallest nontrivial subgroup is generated by $2^{i-1}$, the second smallest by $2^{i-2}$, and so on.

Furthermore, given a chain $1<G_1<G_2<\dots<G_n=G$ for $G$, and a chain $1<H_1<\dots<H_m=H$ for $G$, we can construct a chain for $G\times H$ as follows: $$ 1\times 1<G_1\times 1<G_2\times 1<\dots<G_n\times 1<G_n\times H_1<\dots <G_n\times H_m $$ This allows you to obtain the sequence for any product of cyclic groups, which accounts for all possible $G$.

Mike Earnest
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