# Galois Theory Simplified

Galois discovered Quintic Equation has no radical (expressed with +,-,*,/, nth root) solutions, but his new Math “Group Theory” also explains:
$x^{5} - 1 = 0 \text { has radical solution}$
but
$x^{5} -x -1 = 0 \text{ has no radical solution}$

Why ?

$x^{5} - 1 = 0 \text { has 5 solutions: } \displaystyle x = e^{\frac{ik\pi}{5}}$
$\text{where k } \in \{0,1,2,3,4\}$
which can be expressed in
$x= cos \frac{k\pi}{5} + i.sin \frac{k\pi}{5}$
hence in {+,-,*,/, √ }
ie
$x_0 = e^{\frac{i.0\pi}{5}}=1$
$x_1 = e^{\frac{i\pi}{5}}$
$x_2 = e^{\frac{2i\pi}{5}}$
$x_3 = e^{\frac{3i\pi}{5}}$
$x_4 = e^{\frac{4i\pi}{5}}$
$x_5 = e^{\frac{5i\pi}{5}}=1=x_0$

=>
$\text {Permutation of solutions }{x_j} \text { forms a Cyclic Group: } \{x_0,x_1,x_2,x_3,x_4\}$

Theorem: All Cyclic Groups are Solvable
=>
$x^{5} -1 = 0 \text { has radical solutions.}$

However,
$x^{5} -x -1 = 0 \text{ has no radical solution }$
because the permutation of solutions is A5 (Alternating Group) which is Simple
ie no Normal Subgroup
=> no Symmetry!