Isomorphism to certain Galois group and cyclic groups

[PsychonautQQ]

Homework Statement

Let c be a pth root of unit where p is prime. Then the Galois group G(Q(c):Q) is isomorphic to Z_p*. Show that if there is some m that divides p-1, then there is an extension K of Q such that G(K:Q) is isomorphic to Z_q*

Homework Equations

The Attempt at a Solution

I suspect that K=Q(c^m) is the field extension that we are looking for.

Then the basis elements for Q(c^m):Q are {1,c^m,c^2m,...,c^(r)m) where r+1 is the order of <c^m> a subgroup of <c>. In the Galois group of Q(c^m):Q there will exist a unique element g such that g(1)=c^(im)
for each 0,...,r possible exponents of the basis elements.

Am I on the right track here? Thanks!

 

[fresh_42]

Homework Statement

Let c be a pth root of unit where p is prime. Then the Galois group G(Q(c):Q) is isomorphic to Z_p*. Show that if there is some m that divides p-1, then there is an extension K of Q such that G(K:Q) is isomorphic to Z_q*

Can you clarify the roles of $m$ and $q$ here? Is is $m\,\cdot\, q= p-1$ or $m=q$?

Homework Equations

Which theorems have already been proven in this context? It looks like you may use the correspondence between Galois extensions and automorphism groups, but you haven't mentioned it.

The Attempt at a Solution

I suspect that K=Q(c^m) is the field extension that we are looking for.

Then the basis elements for Q(c^m):Q are {1,c^m,c^2m,...,c^(r)m) where r+1 is the order of <c^m> a subgroup of <c>. In the Galois group of Q(c^m):Q there will exist a unique element g such that g(1)=c^(im)
for each 0,...,r possible exponents of the basis elements.

Am I on the right track here? Thanks!

So it is actually a unique element $g_i$ depending on $i$! Which theorem have you used here?

See, if you assume the availability of all those theorems, then there will be nothing left to prove.