Cyclic Groups Same Order Isomorphic

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Theorem

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

Let $G_1 = \left \langle {a} \right \rangle, G_2 = \left \langle {b} \right \rangle$.

Then, by the definition of a cyclic group, $\left|{a}\right| = \left|{b}\right| = k$.

Also, by definition, $G_1 = \left\{{a^0, a^1, \ldots, a^{k-1}}\right\}$ and $G_2 = \left\{{b^0, b^1, \ldots, b^{k-1}}\right\}$.

Let us set up the obvious bijection $\phi: G_1 \to G_2: \phi \left({a^n}\right) = b^n$. We would now like to show that $\phi$ is an isomorphism.

Note that $\phi \left({a^n}\right) = b^n$ 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 the Power of Remainder, $a^n = a^r, b^n = b^r$.

Thus, $\phi \left({a^n}\right) = \phi \left({a^r}\right) = b^r = b^n$.

Since $G_1 = \left \langle {a} \right \rangle$, $\exists s, t \in \Z: x = a^s, y = a^t$.

Thus:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \phi \left({x y}\right)$	$=$	$\displaystyle \phi \left({a^s a^t}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \phi \left({a^{s + t} }\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle b^{s + t}$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle b^s b^t$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \phi \left({a^s}\right) \phi \left({a^t}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \phi \left({x}\right) \phi \left({y}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $

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.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \phi \left({x y}\right)\)	\(=\)	\(\displaystyle \phi \left({a^s a^t}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({a^{s + t} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle b^{s + t}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle b^s b^t\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({a^s}\right) \phi \left({a^t}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({x}\right) \phi \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)