Isomorphism between Galois Groups and triangulated cover groups of form π (X)

Though we discussed Galois groups before, it's interesting to go back and explore further an area that overlaps abstract and homological algebra. In this brief blog, I explore the isomorphism between a Galois group of form E(n) = 2 πi/n and triangulated cover groups of the form π(x).

As depicted in the accompanying above, a0, a1, ….etc. denote vertices ("edge covers") of the pentagon X. They also represent the points for five roots of unity in the associated Galois group. For a refresher on the roots of unity, see:

http://brane-space.blogspot.com/2010/03/more-complex-roots.html

In terms of the matching process between a0, a1 etc and the roots of unity we have:

a0 <-> 1

a1 <-> Z = exp(2πi/5)

a2 <-> Z2 = exp(4πi/5)

a3 <-> Z3 = exp(6πi/5)

a4 <-> Z4 = exp(8πi/5)

which leads to the table shown in the lower part of the graphic, where a group is formed with elements: 1, Z, Z2, Z3, Z4, matching a0, a1, a2, a3, a4.

One application of this branch of abstract algebra is the construction of regular polygons, or “n-gons” such as the pentagon shown in the graphic (leaving out the a0, a1 etc.)

The procedure for such construction using ruler and straight edge is known, but one can also ascertain in advance which n-gons are “constructible” using only rule and straight edge, and which are not. It turns out such regular polygons are only possible provided the extension field has degree of 2. (I.e. the regular n-gon is only constructible if (Z + 1/Z) = 2 cos(2π/n).


Problem:

Prove this is the case, if:

The Z roots are determined by: cos(2π/n) + isin(2π/n), and

The 1/Z roots are determined by: cos(2π/n) - isin(2π/n).