When do roots of a polynomial form a group?

[MostlyHarmless]

I've been studying for my final exam, and came across this homework problem (that has already been solved, and graded.):

"Show that the Galois group of $f(x)=x^3-1$ over ℚ, is cyclic of order 2."

I had a question related to this problem, but not about this problem exactly. What follows is the line of thought that lead me to my question.

The polynomial has roots, 1, $\lambda = \frac{-1+i\sqrt{3}}{2}$, and $\bar\lambda = \frac{-1-i\sqrt{3}}{2}$ over ℂ.

In this case, $\lambda^2 = \bar\lambda$. So, the conjugate of a root is the same as the root squared (which isn't always true, weird?). Further, $\lambda^3 = \lambda\cdot\bar\lambda=\bar\lambda\cdot\lambda = 1$. So, from that the set of roots, has inverses and the identity, is closed under multiplication, and is generated by $\lambda$. Thus, the set of roots is cyclic of order 3.

I was trying to think of the conditions under which this happens, and I thought maybe if the Galois group is cyclic, but it seems like 1 wouldn't always be a root of a polynomial whose Galois group is cyclic, so then the roots wouldn't even form a group.

Is there something "interesting" going on here? Like, when do the roots form a group? Cyclic group?

 

[pasmith]

The only finite multiplicative subgroups of \mathbb{C} are {0} and \{e^{2n\pi i /N} : n = 0, 1, \dots, N-1\} for strictly positive integer N. These groups are cyclic. They correspond to the roots of z = 0 and z^N - 1 = 0 respectively.

The only finite multiplicative subgroups of \mathbb{R} are {0}, {1} and {-1, 1}, which are also the only finite multiplicative subgroups of \mathbb{Q}. These groups are cyclic. They correspond to the roots of x = 0, x - 1 = 0 and x^2 - 1 = 0 respectively.

 

[mathwonk]

to say the same thing again, if you know group theory, you know that in a group of order n, every element gives 1 when raised to then nth power, hence in any group of order n, the elements satisfy the equation X^n -1 = 0. But if the group is abelian (roots of a polynomial over a field belong to the field and that has commutative multiplication), and not cyclic, i.e. no element has order n, then all of them also satisfy X^r = 1 for some smaller number r than n. Thus they would all satisfy the equation X^r - 1 = 0. But in a field you cannot hVE MORE SOLUTIONS TO AN EQUATION THAN THE DEGREE. oops.
so i think we have proved that in a field no equation can have solutions forming a group that is not cyclic. I think we have alkso proved that all finite multiplicative subgroups of a field are cyclic. of course we have assumed something about the structure of finite abelian groups, namely that the maximal order occurring for any element, annihilates the whole group. can you prove that?

hint: if a has order r and b has order s, what is the order of ab? then assume a has maximal order, and prove s divides r. deduce pasmith's claim about classifying all finite subgroups of R and C.