Question

Let \( \mathbb{C}^* = \mathbb{C} \setminus \{0\} \) denote the group of non-zero complex numbers under multiplication. Suppose \[ Y_n = \{ z \in \mathbb{C} \mid z^n = 1 \}, n \in \mathbb{N}. \] Which of the following is (are) subgroup(s) of \( \mathbb{C}^* \)?

• A. \( \bigcup_{n=1}^{100} Y_n \)
• B. \( \bigcup_{n=1}^{\infty} Y_{2^n} \)
• C. \( \bigcup_{n=100}^{\infty} Y_n \)
• D. \( \bigcup_{n=1}^{\infty} Y_n \)

Hint:

To check if a union of sets forms a subgroup, ensure the set is closed under the group operation.

Answer 1

The Correct Option is B, C, D

Given: $Y_n = {z \in \mathbb{C} \mid z^n = 1}$ for $n \in \mathbb{N}$

These are the $n$-th roots of unity, which form a cyclic subgroup of $\mathbb{C}^*$ of order $n$.

Key observations:

$Y_n$ is a subgroup of $\mathbb{C}^*$ for each $n$

$Y_m \subseteq Y_n$ if and only if $m \mid n$

For a union of subgroups to be a subgroup, special conditions must hold

Option (A): $\bigcup_{n=1}^{100} Y_n$

This is a finite union of subgroups.

For a union of subgroups to be a subgroup, one must contain all the others (i.e., the union must equal one of the subgroups).

Since $Y_1 \subseteq Y_2 \subseteq Y_4 \subseteq Y_8 \subseteq \cdots$ is NOT a chain covering all terms, and $Y_3, Y_5, Y_7,$ etc. are not nested, we have:

For example: $Y_2 = {1, -1}$ and $Y_3 = {1, e^{2\pi i/3}, e^{4\pi i/3}}$

Neither contains the other, so their union is not a subgroup.

To verify: Take $-1 \in Y_2$ and $e^{2\pi i/3} \in Y_3$.

Then $-1 \cdot e^{2\pi i/3} = -e^{2\pi i/3} = e^{i\pi} \cdot e^{2\pi i/3} = e^{i(3\pi + 2\pi)/3} = e^{5\pi i/3}$

For this to be in $\bigcup_{n=1}^{100} Y_n$, we need $(e^{5\pi i/3})^n = 1$ for some $n \leq 100$.

This gives $e^{5\pi i n/3} = 1$, so $5\pi n/3 = 2\pi k$ for integer $k$, meaning $5n = 6k$, so $n$ must be a multiple of 6.

Since $6 \leq 100$, this element IS in the union. Let me reconsider.

Actually, $\bigcup_{n=1}^{100} Y_n = Y_{100}$ if $100$ is divisible by all other numbers up to 100, which it's not.

Actually, $\bigcup_{n=1}^{100} Y_n = Y_{\text{lcm}(1,2,...,100)}$ only if we're lucky, but that's not how unions work.

The union $\bigcup_{n=1}^{100} Y_n$ contains all elements whose order divides some $n \leq 100$. This is NOT closed under multiplication.

Option (A) is NOT a subgroup 

Option (B): $\bigcup_{n=1}^{\infty} Y_{2^n}$

We have: $Y_2 \subseteq Y_4 \subseteq Y_8 \subseteq Y_{16} \subseteq \cdots$

This is a nested chain of subgroups!

The union of a nested chain of subgroups is always a subgroup.

$$\bigcup_{n=1}^{\infty} Y_{2^n} = {z \in \mathbb{C} \mid z^{2^n} = 1 \text{ for some } n}$$

This is the set of all $2^n$-th roots of unity for all $n$, which forms a subgroup (the group of all roots of unity whose order is a power of 2).

Option (B) IS a subgroup 

Option (C): $\bigcup_{n=100}^{\infty} Y_n$

Since $Y_{100} \subseteq Y_{200} \subseteq Y_{300} \subseteq \cdots$ is NOT true (e.g., $Y_{101}$ is not contained in $Y_{100}$), this is not a nested chain.

However, let's check if it's still a subgroup.

$$\bigcup_{n=100}^{\infty} Y_n = {z \in \mathbb{C} \mid z^n = 1 \text{ for some } n \geq 100}$$

Take $z_1 \in Y_{100}$ and $z_2 \in Y_{101}$. Then $z_1^{100} = 1$ and $z_2^{101} = 1$.

For closure: $(z_1 z_2)^m = 1$ for some $m \geq 100$?

We have $(z_1 z_2)^{\text{lcm}(100, 101)} = 1$, and $\text{lcm}(100, 101) = 10100 \geq 100$.

So the product is in the union. Similarly for inverses.

Option (C) IS a subgroup 

Option (D): $\bigcup_{n=1}^{\infty} Y_n$

This is the set of ALL roots of unity:

$$\bigcup_{n=1}^{\infty} Y_n = {z \in \mathbb{C} \mid z^n = 1 \text{ for some } n \in \mathbb{N}}$$

Identity: $1 \in Y_1$ ✓

Closure: If $z_1^{n_1} = 1$ and $z_2^{n_2} = 1$, then $(z_1z_2)^{\text{lcm}(n_1,n_2)} = 1$ ✓

Inverses: If $z^n = 1$, then $(z^{-1})^n = 1$ ✓

Option (D) IS a subgroup 

Answer: (B), (C), and (D)