# Prove that any subgroup of order $n$ of $\Bbb C^*$ is $A_n$

Is every finite subgroup of order $n$ of $\Bbb C^*=$ (set of all non-zero complex numbers) of the form $A_n=\{z:z^n=1\}$ is a subgroup of $C^{*}$.

Any element of $A_n$ is a solution of $z^n=1$.Now the solutions of $z^n=1$ for any $n\in \Bbb N$ are $e^{\frac{{2ki\pi}}{{n}}};1\le k\le n$ and hence the subgroup $A_n$ is generated by $a=e^{\frac{{2ki\pi}}{{n}}}$ and hence cyclic.

Conversely if $H$ is any subgroup of order $n$ say $H=\{g_1,g_2,\cdots ,g_n\}$ then $g_i^n=1$

$\implies g_i$ is a solution of $x^n=1\implies g_i=e^{\frac{2k\pi i}{n}}$ .

Since each $g_i$ is a solution and $g_i's$ are distinct and also $x^n-1$ has exactly $n$ solutions so $G=A_n$ .

Is my solution correct?

Please suggest edits.

## 1 Answer

Your proof of the converse is basically correct. You should probably mention how you know that $g_i^n=1$, though. You should also be careful to state your quantifiers: when you write $g_i=e^{2k\pi i/n}$, what you really mean is that for every $i$ there exists $k$ such that this is true.

The first part of your proof has some problems and I'm not sure what it is you are actually trying to prove. I'm guessing you're trying to prove that $A_n$ is indeed a subgroup of $\mathbb{C}^*$ of order $n$. In that case you can't say "the subgroup $A_n$" yet because you are still in the process of proving it. What you can instead say is that $A_n$ consists of all integer powers of $a$ (why?), and thus is equal to the subgroup generated by $a$ (in particular, it is a subgroup). Also, your definition $a=e^{2ki\pi/n}$ is incomplete: what is $k$?

• Since the order of the subgroup is $n$ so $g_i^n=1$ ;Isn't it right? – Learnmore Feb 24 '17 at 5:33
• That's correct. – Eric Wofsey Feb 24 '17 at 5:36