Every infinite cyclic group is isomorphic to the additive group of Z

The Mathematician — Wed, 26 Apr 2023 13:00:51 +0000

We will prove that every infinite cyclic group is isomorphic to the additive group of Z.

Define the following mapping:

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

where

\langle x \rangle

is the infinite cyclic group. Then this map is isomorphic.

Proof. To prove that the map

f

is isomorphic, we need to show that

f

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

Well defined. The map is already well defined since there is no ambiguity in the representations of elements in the domain. To be more clear with an example,

2 = \frac{2}{1} = \frac{4}{2}

etc, but

\frac{2}{1},\frac{4}{2} \not \in \mathbb{Z}

. So all elements in

\mathbb{Z}

are distinct. Same holds for

\langle x \rangle

since it is an infinite cyclic group.

Injectivity. All elements are distinct from what we have seen before, so if

x^a \neq x^b

, then

a \neq b

.

Surjectivity. The element

x^k

\langle x \rangle

is the image of

k

under

f

, so

f

is surjective.

Homomorphism. To prove that

f

is a homomorphism, we have done that here. Now we have shown that every infinite cyclic group is isomorphic to the additive group of Z.

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Every infinite cyclic group is isomorphic to the additive group of Z