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:
\(\ds \map \phi {x y}\) | \(=\) | \(\ds \map \phi {a^s a^t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {a^{s + t} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b^{s + t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b^s b^t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {a^s} \map \phi {a^t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x \map \phi y\) |
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): Chapter $\text {I}$: The Group Concept: $\S 9$: 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 $4.8$