Suppose $B$ is a finite subset of $A$. $A =$ {$e^{2πij/n},0 ≤ j < n , n ≥ 1$}. Show $B$ is contained in some cyclic subgroup of $A$. I need to give a good generator of this cyclic subgroup of $A$ and show that every element of $B$ is a power of the generator.
I need to find a generator of $A$ and show it works for any subset $B$.
- For $a ∈ A$, $<a>$ is the cyclic subgroup generated by $a$. If $A$ contains some element $a$ where $A = <a>$, then $A$ is a cyclic group. Here, $a$ is a generator of $A$.
I'm not really sure what cyclic subgroup I need. I was thinking about $H =$ {$e^{i\pi\ q}, q∈Z$} maybe?
So $<e^{i\pi\ q}>$ would generate the cyclic subgroup of $A$? And all elements of $B$ would be powers of the generator? I need to define $q$ by giving some formula for it or some conditions to show it works for any $B$. I was thinking something related to $gcd$ maybe? I'm not really sure if it's the right generator though.
Notice that $H=\{-1,+1\}$, so it does not work for arbitrary $B$.
Also $A$ is an infinite group each of whose elements has finite order, therefore it is impossible for $A$ to be a cyclic group, and it therefore makes no sense to talk about a single generator of $A$.
Your intuition about gcd is aiming approximately in the right direction.
But to get anywhere you have to turn that intuition into mathematics.
And the very first step of doing that is to write out explicit expressions for the elements of the finite set $B$, so that you have some actual mathematical expressions to do some arithmetic with.
Step 1: Write the elements of the finite set $B$ in the explicit form $$B = \bigl\{\underbrace{\exp\left(2 \pi i \frac{j_1}{n_1}\right)}_{b_1}, \underbrace{\exp\left(2 \pi i \frac{j_2}{n_2}\right)}_{b_2}, ... \underbrace{\exp\left(2 \pi i \frac{j_M}{n_M}\right)}_{b_M} \bigr\} $$ for some integers $n_1,...,n_M \ge 0$ and some integers $j_1,...,j_M$ such that $0 \le j_m \le n_m$ for all $m=1,...,M$. $$
Step 2: Find one element of $A$, which we shall express in the explicit form $a = \exp\left(2 \pi i \frac{j}{n} \right)$, such that the cyclic subgroup of $A$ generated by $a$ contains each of $b_1,...,b_M$.
And at this point, I'll give you a hint and stop: You can find a value of $a$ having the even more restrictive form $a = \exp\left(2 \pi i \frac{1}{n} \right)$, in fact all you have to do is to find a single fraction $\frac{1}{n}$ such that each of the fractions $\frac{j_1}{n_1},\ldots,\frac{j_M}{n_M}$ is an integer multiple of $\frac{1}{n}$.