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Let $G$ be the multiplicative group of all complex $2^n$-th roots of unity, where $n = 0 , 1, 2 , \ldots$. Then assess the following claims:

  1. Every proper subgroup of $G$ is finite.

  2. $G$ has a finite set of generators.

  3. $G$ is cyclic.

  4. Every finite subgroup of $G$ is cyclic.

  5. $G \cong \dfrac{\mathbb Z[1/2] }{\mathbb Z}$.

$G = \bigcup_{n=1}^{\infty} G_{2^n}$ , where $G_{2^n} = \{ e^{\frac{2\pi ik}{2^n}} \mid k = 0,1,2, \dots , n-1 \}$.

Suppose $G$ is generated by finitely many elements. Suppose $$ S = \left\lbrace\exp\left({2\pi i k_j\over 2^{n_j}}\right)\right\rbrace_{j=1}^m $$

and $G = \langle S \rangle $

Choose $N = \max \{ n_j\}_{j-1}^m$, then $ e^{\frac{2\pi i}{N+1}}$ is not generated by $S$. So $G$ is not a cyclic.

Thus (2) and (3) are false.

Please tell me about (1) and (4) and (5)

any help would be appreciated. Thank you

Community
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Struggler
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    Hint for (1): a proper subgroup must omit some element $x$ in some $G_{2^n}$; can that subgroup then contain any element in any $G_{2^N}$ with $N>n$? Hint for (4): given a finite subgroup $H$, consider the maximal $n$ such that $H$ has elements from $G_{2^n}$. Hint for (5): dream up a group homomorphism that has the potential to be an isomorphism, and check that it is. – Greg Martin Jun 16 '15 at 22:18
  • Also, check this question and the answer here if you get stuck. – Daniel Fischer Jun 16 '15 at 22:21
  • Thank you so much for sharing the link – Struggler Jun 17 '15 at 02:06
  • can anyone please tell me that for a fixed integer n, why a multiplicative group G of $ (complex)^{2^n} $ th roots of unity contains n elements ? – suchanda adhikari Feb 12 '19 at 18:16
  • I could not understand that for a fixed integer n how many elements does the multiplicative group G of $ (compex)^{2^n} $ th roots of unity contain ? I thought the number is 2^n because the multiplicative group of nth roots of unity contains n elements .please answer. – suchanda adhikari Feb 12 '19 at 20:43

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