Any two cyclic groups of the same order are isomorphic

The Mathematician — Sat, 22 Apr 2023 13:00:14 +0000

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.:

it is well defined,
is injective,
is surjective,
and is a homomorphism.

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

The post Any two cyclic groups of the same order are isomorphic appeared first on Epsilonify.

Are cyclic groups abelian?

The Mathematician — Fri, 07 Oct 2022 13:00:32 +0000

All cyclic groups are abelian groups. It is not true that every abelian group is cyclic.

Proof. Take the cyclic group $G = \langle g \rangle$ . Take the elements $a,b \in G$ . We need to show that $ab = ba$ . Since $G$ is cyclic, we have that $a = g^i$ and $b = g^j$ for some $i,j \in \mathbb{Z}$ . This implies that $ab = g^ig^j = g^{i + j} = g^{j + i} = g^jg^i = ba$ .

This proves that cyclic groups are abelian groups.

The post Are cyclic groups abelian? appeared first on Epsilonify.

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Any two cyclic groups of the same order are isomorphic

Are cyclic groups abelian?