4

If we let $$S = \{ e^{q\pi i} : q\in Q \} $$ Prove that for each $ n \ge 1$ there is a unique cyclic subgroup of order $n$ in $S$ and the union of these cyclic subgroups is $S$. Any help on this?

Timur Lame
  • 73
  • 5

2 Answers2

2

The elements of $S$ are complex numbers. The group operation is complex multiplication and the identity element is $1 = e^{2\pi i}$. So $x \in S$ lies in a copy of $\mathbb Z/ n \mathbb Z$ if and only if it is an $n$th root of unity. We have explicit formulas for the roots of unity. This will prove existence. From the fundamental theorem of algebra, we know there are exactly $n$ roots of unity. This will prove uniqueness.

manthanomen
  • 3,186
  • That there are at most $n$ roots of any polynomial is true of any field. It does not require the fundamental theorem of algebra. – lhf Nov 08 '17 at 10:52
1

$S_n = \{ e^{2 \pi i \frac{m}{n}} : m\in \{0...n-1\} \}$

$(S_n,\cdot)$ is cyclic of order n (you can show the isomorphism with $\mathbb Z_n$)

For the uniqueness use an argument on the generator of the subgroup (show that if $x\in S \ x^n=1$ then $x \in S_n$ )

1)$ \forall n \ S_n\subset S \implies \bigcup_{n=1}^{\infty}S_n \subset S$

2)$ x \in S\implies \ x=e^{i \pi q}, q \in Q \implies x=e^{i \pi \frac{m}{n}} \ x\in S_n$ for some n $\implies S\subset \bigcup_{n=1}^{\infty}S_n$

$1)+2)\implies S=\bigcup_{n=1}^{\infty}S_n$

polbos
  • 395
  • 2
  • 10