Show that a group G is cyclic and infinite

Question

I want to show, that for a group $(G,\circ)$:

$ \mathbb{Z}$ is isomorphic to $G$ $\Rightarrow G$ is cyclic and infinite

Therefore, to my understanding, I will have to show that there exists $$\exists a \in G: \langle a\rangle = \{ a^{k}: k \in \mathbb{Z} \} $$

From the left side of the statement I can assume that $G$ is well-defined, isomorphic and therefore also homomorphic to $G$ as well as bijective. At this point unfortunately, I don't really know where to continue. Am I supposed to construct a concrete example to prove it or is it possible to generally show this?

Answer 1

From an isomorphism $f\colon \mathbf Z\longrightarrow G$, you can easily check that $a=f(1)$ is a generator of $G$. Suppose the group law on $G$ is denoted multiplicatively::

• An element $g\in G$ is the image of an integer $n$: $$g=f(n)=f(\underbrace{1+1+\dots+1}_{n \text{ terms}})=f(1)^n=a^n.$$
• As a group isomorphism is a bijection, $G$ has the same cardinal as $\mathbf Z$, i.e. $\aleph_0$.

Answer 2

$\bf{Hint}$: An isomorphism $\phi: \mathbb{Z} \to G$ must send $1$ to the generator of $G$. Just use the homomorphism property of the isomorphism to show this. Use surjectivity of the iso to show that every element in $G$ is $\phi(1)^k$ for some integer $k$. The bijection piece of $\phi$ guarantees that both sets have the came cardinality.

Answer 3

Since $\mathbb{Z}\cong G,$ there is an induced isomorphism,call $\varphi:\mathbb{Z}\longrightarrow G.$

Then$$\langle\varphi(1)\rangle=\varphi(\langle1\rangle)=\varphi(\mathbb{Z})=G$$ ,since $\varphi $ is surjective and $\varphi(1)$ is an element in $G$,so $G$ is cyclic.

If G is finite then there is some positive number $m$ such that $\bigg(\varphi(1)\bigg)^m=e_{G}$, then it contradicts the fact that $ker\varphi=\langle e\rangle$.

when G is finite, there is some detail I omitted.But I think it would be an easy exercise for you.