# Cyclic Groups of Same Order are Isomorphic

## Theorem

Two cyclic groups of the same order are isomorphic to each other.

## Proof

Let $G_1$ and $G_2$ be cyclic groups, both of finite order $k$.

Let $G_1 = \gen a, G_2 = \gen b$.

Then, by the definition of a cyclic group:

- $\order a = \order b = k$

Also, by definition:

- $G_1 = \set {a^0, a^1, \ldots, a^{k - 1} }$

and:

- $G_2 = \set {b^0, b^1, \ldots, b^{k - 1} }$

Let us set up the obvious bijection:

- $\phi: G_1 \to G_2: \map \phi {a^n} = b^n$

The next task is to show that $\phi$ is an isomorphism.

Note that $\map \phi {a^n} = b^n$ holds for all $n \in \Z$, not just where $0 \le n < k$, as follows:

Let $n \in \Z: n = q k + r, 0 \le r < k$, by the Division Theorem.

Then, by Element to Power of Remainder:

- $a^n = a^r, b^n = b^r$

Thus:

- $\map \phi {a^n} = \map \phi {a^r} = b^r = b^n$

Now let $x, y \in G_1$.

Since $G_1 = \gen a$, it follows that:

- $\exists s, t \in \Z: x = a^s, y = a^t$

Thus:

\(\displaystyle \map \phi {x y}\) | \(=\) | \(\displaystyle \map \phi {a^s a^t}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {a^{s + t} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b^{s + t}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b^s b^t\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {a^s} \map \phi {a^t}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi x \map \phi y\) |

So $\phi$ is a homomorphism.

As $\phi$ is bijective, $\phi$ is an isomorphism from $G_1$ to $G_2$.

Thus $G_1 \cong G_2$ and the result is proved.

$\blacksquare$

## Sources

- 1964: Walter Ledermann:
*Introduction to the Theory of Finite Groups*(5th ed.) ... (previous) ... (next): $\S 9$: Cyclic Groups - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.7$: Theorem $9$ - 1978: John S. Rose:
*A Course on Group Theory*... (previous) ... (next): $0$: Some Conventions and some Basic Facts: Exercise $2$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 46$. Isomorphic groups: Worked Example - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $4$: Subgroups: Example $4.8$