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!

See also: From Durian to Group

Galois Theory, Third Edition (Chapman Hall/CRC Mathematics Series)

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