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I was trying to show that every finite subgroup of $\mathbb C^{\times}$ is equal to $G_n$ (the nth roots of unity) for some $n \in \mathbb N$ without invoking Lagrange's theorem, I got stuck at one point, I would appreciate some help to complete my solution.

Let $H \neq \{1\}$ be a finite subgroup of $\mathbb C^{\times}$. For any $x \in H \setminus \{1\}$ it is easy to show that there is a minimum $a$ natural such that $x^a=1$ and for all $k$, $x^k \in \{1,x,...,x^{a-1}\}$ with all these elements distinct. Since $H$ is finite, then $H=\{x_1,...,x_m\}$, consider $H'=\{x_{n_1},...,x_{n_r}\}$ all the elements in $H$ such that $x_{n_k} \neq x_{n_j}^m$ for all $k \neq j$ and all $m \in \mathbb Z$. For each $n_j$, $\exists$ $\alpha_{n_j}$ with ${x_{n_j}}^{\alpha_{n_j}}=1$ and $\alpha_{n_j}$ the minimum natural number that satisfies this.

I want to show $H=G_{lcm(\alpha_{n_1},...,\alpha_{n_r})}$

Let $h \in H$, then $h=x_{n_1}^{\beta_{n_1}}...x_{n_r}^{\beta_{n_r}}$, $0\leq \beta_{n_i} <\alpha_{n_i}$

If $t={lcm(\alpha_{n_1},...,\alpha_{n_r})}$, it is easy to verify that $h^t=1$, so $H \subset G_t$. I got stuck trying to show the other inclusion (or that $|H|=t$). I would appreciate any corections or/and suggestions. Thanks in advance.

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  • $\begingroup$ What are you defining $G_n$ to be here? $\endgroup$ – Ramified_Minds May 1 '15 at 23:36
  • $\begingroup$ The nth roots of unity $\endgroup$ – user156441 May 1 '15 at 23:54
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Here's one way you could go: Since you have already found that $H \subset G_t$, and you presumably know that $G_t$ is cyclic, generated by $e^{2\pi i/t}$, use the fact that every subgroup of $G_t$ is cyclic, equal to $G_{t'}$ for some $t' \mid t$. So if $H \subsetneq G_t$, then $H = G_{t'}$ and $x_{n_j}^{t'} = 1$ for every $j$.

Derive a contradiction with the fact that the order of each $x_{n_j}$ is $\alpha_j$ and $t = lcm(\alpha_{n_1}, \ldots, \alpha_{n_r})$; namely, $\alpha_{n_j} \mid t'$ for every $j$. You don't need Lagrange's theorem for this fact; you simply need the fact that the powers of $x_{n_j}$ cycle with period $\alpha_{n_j}$.

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Think about the structure theorem for finite abelian groups:

Every finite abelian group is isomorphic to the product of cyclic groups with prime power degree, ie

$$H\cong\prod_{p\: \text{prime}}\prod_{i=0} \mathbb{Z}/p^{i}\mathbb{Z}$$

Where the $i$ are $0$ for all but finitely many $i$.

Then we can see that working under $\mathbb{C}^{\times}$, the group $G_{p^{i}}$ is isomorphic to $\mathbb{Z}/p^{i}\mathbb{Z}$.

So we can think about finite subgroups as the 'product' of the sets $\mu_{p^{i}}$, the $p^{i}$-th roots of unity. Since each $\mu_{p^{i}}$ contains $\mu_{p^{j}}$, for $j<i$, we need only consider the highest power of $p$. So take the product of sets of the $\mu_{p^i}$, for different primes $p$. Since each $p$ is coprime, we see that $\zeta_{p_1}\not\in\mu_p$ for any $p\neq p_1$. Thus, the product of all the sets must contain $\{1,\zeta_{p_1^i},\zeta_{p_1^i}^2,\ldots,\zeta_{p_1^i}^{p_1-1}\}$ and $\{\zeta_{p_2^j},\zeta_{p_2^j}^2,\ldots,\zeta_{p_2^j}^{p_2-1}\}$, and so on. This means that the product of these sets must be the $l$-th roots of unity, where $l=$lcm$(p_1^i,p_2^j,\ldots,p_n^k)$. But since these are all primes, this is simply $p_1^{a_1}p_2^{a^2}\ldots p_n^{a_n}$.

For an example, let's think about the possible subgroups of size $8$. An abelian group of size $8$ is either $C_8$, $C_4\times C_2$ or $C_2\times C_2\times C_2$. The second group does not have an analogue as a subgroup of $\mathbb{C}^{\times}$, because $G_2\subset G_4$, so when we take the 'set product' we simply get $G_4$ again, and similarly with $C_2^3$. Thus we are left with $G_8$.

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