1. G is cyclic
2.G is non abelian
3. order of G is 20
4. G has an element of order 4
2.G is non abelian
3. order of G is 20
4. G has an element of order 4
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First of all, the splitting fields equals Q[2^(1/5), ζ], where ζ is a primitive 5th root of unity.
It's easy to check that
[Q[2^(1/5), ζ]: Q] = [Q[2^(1/5), ζ]: Q[2^(1/5]] * [Q[2^(1/5)] : Q] = 4 * 5 = 20.
So, |G| = 20.
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For any subgroup H of G, there is a corresponding field, consisting of those elements of the splitting field which are fixed by any automorphism (element) in H (and vice versa).
Note that Q[ζ] is a subfield of the splitting field over Q, which is fixed by the (real) automorphisms
2^(1/5) --> (2^(1/5))^k for k = 1,2,3,4. So, this is a subgroup of G, which is cyclic of order 4.
Hence, G has an element of order 4.
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Finally, G is nonabelian (and hence non-cyclic):
To describe the maps, note that G being Galois implies that any such automorphism
must send 2^(1/5) to another zero of x^5 - 2, and send ζ to ζ^k (k = 1, 2, 3, 4), since Q[ζ] is a subfield of the splitting field and x^5 - 1 is its minimal polynomial.
So, we have the maps (defined on two indices i = 0, 1, 2, 3, 4 and j = 1, 2, 3, 4):
f_(i,j) (2^(1/5)) = 2^(1/5) ζ^i,
f_(i,j) (ζ) = ζ^j.
(As a bonus, f_(0, 1) is an element of order 4, since [f_(0, 1)]^4 = identity map,
and no smaller power does this.)
It can be checked that these elements satisfy f_(i,j) * f_(k,l) = f_(i+jk, jl).
From this, you should be able to see that this does not satisfy commutativity.
I hope this helps!
It's easy to check that
[Q[2^(1/5), ζ]: Q] = [Q[2^(1/5), ζ]: Q[2^(1/5]] * [Q[2^(1/5)] : Q] = 4 * 5 = 20.
So, |G| = 20.
------------
For any subgroup H of G, there is a corresponding field, consisting of those elements of the splitting field which are fixed by any automorphism (element) in H (and vice versa).
Note that Q[ζ] is a subfield of the splitting field over Q, which is fixed by the (real) automorphisms
2^(1/5) --> (2^(1/5))^k for k = 1,2,3,4. So, this is a subgroup of G, which is cyclic of order 4.
Hence, G has an element of order 4.
--------------
Finally, G is nonabelian (and hence non-cyclic):
To describe the maps, note that G being Galois implies that any such automorphism
must send 2^(1/5) to another zero of x^5 - 2, and send ζ to ζ^k (k = 1, 2, 3, 4), since Q[ζ] is a subfield of the splitting field and x^5 - 1 is its minimal polynomial.
So, we have the maps (defined on two indices i = 0, 1, 2, 3, 4 and j = 1, 2, 3, 4):
f_(i,j) (2^(1/5)) = 2^(1/5) ζ^i,
f_(i,j) (ζ) = ζ^j.
(As a bonus, f_(0, 1) is an element of order 4, since [f_(0, 1)]^4 = identity map,
and no smaller power does this.)
It can be checked that these elements satisfy f_(i,j) * f_(k,l) = f_(i+jk, jl).
From this, you should be able to see that this does not satisfy commutativity.
I hope this helps!