📂Abstract Algebra

Proving That All Cyclic groups are Isomorphic to the Integer group

Theorem ¹

If a cyclic group $\left< a \right>$ is a finite group, then $\left< a \right> \simeq \mathbb{Z}_{n}$, and if it is an infinite group, then $\left< a \right> \simeq \mathbb{Z}$.

Explanation

With this theorem, the exploration of cyclic groups is virtually complete. The fact that groups, which were only abstract, suddenly fall into the domain of number theory means that there is quite a lot that can be done. Conversely, it should also be possible to solve problems of number theory using the theories of group theory.

Proof

We divide the proof into cases where $m \in \mathbf{n}$ satisfies $a^m = e$ and cases where it does not. This effectively divides it into cases where $\left< a \right>$ is finite and cases where it is infinite.

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Case 1. For some natural number $m$, $a^m = e$

Let the smallest natural number that satisfies $a^m = e$ be $n$. Considering natural number $s = nq + r$ for $0 \le r < n$, $$a^{s} = a^{nq + r} = (a^{n})^{q} a^{r} = e^{q} a^{r} = a^{r}$$ If we assume that there exist two smaller, distinct natural numbers $h > k$ such that $a^{h} = a^{k}$, then $a^{h-k} = e$ and since $(h-k) < n$, $n$ cannot be the smallest natural number that satisfies $a^m = e$, which is a contradiction. Therefore, $a^{0} = e$ and $a , a^2 , \cdots , a^{n-1}$ must all be distinct elements. Now, if we define the function $\phi : \left< a \right> \to \mathbb{Z}_{n}$ as $\phi (a^{i}) = i$, it becomes bijective, $$\phi ( a^{i} a^{j} ) \equiv \phi ( a^{i + j } ) \equiv i + j \equiv \phi (a^{i}) + \phi (a^{j}) \pmod{n}$$ and thus satisfies $$\left< a \right> \simeq \mathbb{Z}_{n}$$

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Case 2. For all natural numbers $m$, $a^m \ne e$

Assuming that there exist two distinct natural numbers $h > k$ such that $a^{h} = a^{k}$, $$a^{h} a^{-k} = a^{h - k } = e$$ thus a contradiction arises since $( h - k) \in \mathbb{N}$ exists, and therefore if there exists a natural number $m$ that satisfies $a^m = e$, it must be unique. Now, if we define the function $\phi : \left< a \right> \to \mathbb{Z}$ as $\phi (a^{i}) = i$, it becomes bijective, $$\phi ( a^{i} a^{j} ) = \phi ( a^{i + j } ) = i + j = \phi (a^{i}) + \phi (a^{j})$$ and thus satisfies $$\left< a \right> \simeq \mathbb{Z}$$

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In the end, cyclic groups, whether finite or infinite, are isomorphic to the group of integers.

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