Constructible regular n-gons and Euler totient function ($\phi(\cdots)$) We had seen the following Theorem in class:

Theorem: The regular n-gon is constructable $\iff$ $\phi(n)=2^l$ for $l \in \mathbb{N}$. $\phi(\cdot)$ is the Euler totient function.

In the lecture the professor proved only ($\implies$) but not the other direction. How can I see why the right side implies the left side?
Thanks a lot in advance.
 A: Constructing a regular n-gon is equivalent to constructing the angle $\frac{2\pi}{n}$.
Assuming you know or can show that a totally real Galois extension of degree $2^r$ is constructible, you consider the cyclotomic field of $n^{th}$ roots of unity.  The Galois group of $\Bbb{Q}_{\zeta_n}$ has order $\phi(n)$.  Then $\zeta_n+\bar{\zeta_n} = 2 \cos \frac{2\pi}{n}$ is a real subfield of degree $2^l$, and we're done.
A: The other direction requires a bit of group theory. As sharding4 explained, the extension $\Bbb{Q}(\cos(2\pi /n))$ is Galois of degree $\phi(n)/2$. Therefore, assuming $\phi(n)=2^\ell$, the corresponding Galois group $G$ is a 2-group. A basic property of $2$-groups is that they are solvable, more precisely we have a sequence of subgroups
$$
\{1_G\}=G_k\le G_{k-1}\le\cdots\le G_1\le G_0=G
$$
such that $[G_i:G_{i+1}]=2$. By Galois correspondence we then have a tower of intermediate fields
$$
\Bbb{Q}=K_0\subset K_1\subset \cdots\subset K_{k-1}\subset K_k=\Bbb{Q}(\cos(2\pi /n)),
$$
where $K_i$ is the fixed field of $G_i$, and $[K_{i+1}:K_i]=2$.
As all the extensions are quadratic, an induction on $i$ shows that all the elements of all the fields $K_i$, $\cos(2\pi/n)$ in particular, are constructible.
