I have the groups $C_{n}=\{ z \in \mathbb{C} |z^n=1\}$ and $\mathbb{Z}_{n}=\{0,1,...,n-1\}$. I already proved that $C_n$ is a subgroup of $(\mathbb{C}\backslash\{0\},*) $. Now how can I show that these groups are isomorphic? Any hint or help would be welcome.
How to show Isomorphism between Cn and $\mathbb{Z}_{n}$
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0Send $1\in\mathbb Z_n$ to $cos(2\pi/n)+i*sin(2\pi/n)$. – 2017-01-31
2 Answers
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Since $C_n $ and $Z_n $ are two cyclic groups of order $n $, they are isomorphic.
Theorem: Two cyclic groups are isomorphic if and only if they have the same cardinal number.
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Any group homomorphism $\;\bar\varphi\colon \mathbf Z/n\mathbf Z\to G$ is induced by a group homomorphism $\;\varphi\colon \mathbf Z\to G$ which vanishes at $n$.
Explicitly, this means a homomorphism $\mathbf Z/n\mathbf Z\to C_n$ is determined by the choice of the image in $C_n$ of the congruence class $1+n\mathbf Z$. The vanishing condition is automatically satisfied in $C_n$.
This homomorphism will be an isomorphism if an only if we make the choice of a generator of the cyclic group $C_n$. If $u$ is a given generator, say $u=\mathrm e^{\tfrac{2i\pi}n}$, we have to choose one of the $u^k$, $k\wedge n=1$.