Any two cyclic groups of the same order are isomorphic

Define the following map:

\begin{align*} f: \langle x \rangle &\longrightarrow \langle y \rangle \\ x^k &\longmapsto y^k \end{align*}

where

\langle x \rangle

and

\langle y \rangle

are cyclic groups of order

n \in \mathbb{Z}^{+}

. Then the map

f

is isomorphic.

Proof. We need to prove four things for the map

f

, i.e.:

The map $f$ is well defined. To show this, we need to prove that if

x^r = x^s

, then

f(x^r) = f(x^s)

. We know that

x^{r-s} = 1

, which implies that

n \mid r-s

, i.e.,

r = tn + s

for some

t \in \mathbb{Z}

. Now it is a straight calculation that:

\begin{align*} f(x^r) = f(x^{tn + s}) = y^{tn + s} = y^{tn}y^s = y^s = f(x^s). \end{align*}

So the map is well defined.

The map $f$ is injective. In order to show that it is injective, it needs to follow that if

f(x^r) = f(x^s)

, then

x^r = x^s

. Notice that

y^r = f(x^r) = f(x^s) = y^s

, and therefore,

y^{r-s} = 1

. We apply the same trick as we did when we proved that

f

is well defined map:

r = tn + s

for some

t \in \mathbb{Z}

. Therefore, we get:

\begin{align*} x^r = x^{tn + s} = x^{tn}x^s = x^s, \end{align*}

since both cyclic group have order

n

. This proves the injectivity.

The map $f$ is surjective. For each

y^k \in \langle y \rangle

, there exists an

x^k \in \langle x \rangle

such that

f(x^k) = y^k

. Since both groups do have order

n

, we have already proved the surjectivity.

The map $f$ is a homomorphism. This one is pretty straightforward. Take

a,b \in \mathbb{Z}

. Then

\begin{align*} f(x^{a} x^{b}) = f(x^{a + b}) = y^{a + b} = y^{a}y^{b} = f(x^a)f(x^b), \end{align*}

which proves the homomorphism. We finally proved that any two cyclic groups of the same order are isomorphic