Why "An infinite cyclic group is isomorphic to the set of integers" and not "there is bijection $\varphi: \mathbb Z \to \langle g \rangle$"

Question

A group $(G,\circ)$ is finite if and only if it has finitely many subgroups.

Lemma: A group $(G,\circ)$ can be written as a union of cyclic subgroups.

Proof:

$$G=\bigcup_{g \in G}\left\{g \right\}$$

Every $g$ is in the cyclic group generated by itself,$\forall g \in G: g \in \langle g \rangle$ and hence $\forall g \in G: \left\{g \right\} \subseteq\langle g \rangle$,follows: $$G=\bigcup_{g \in G}\left\{g \right\} \subseteq \bigcup_{g \in G}\langle g \rangle \subseteq G$$

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$\Longrightarrow$ If $G$ is finite then the set of all subsets of $G$ is a finite set, a subgroup is a subset with extra structures which means the set of subgroups of $G$ is finite.

$\Longleftarrow$ If $G$ has an element $g$ of infinite order, then the cyclic group generated by $\langle g \rangle$ is an infinite cyclic group and isomorphic to $\mathbb Z$ and $n \mathbb Z$ is a subgroup of $\mathbb Z$ for every positive integer $n$, there are infinitely many such subgroups, so $\langle g \rangle$ has infinitely many subgroups, from which we conclude that $G$ has infinitely many subgroups,a contradiction.

From the lemma we know that $G$ can be written as the union of cyclic subgroups, in the previous section it's shown that every cyclic subgroup of $G$ is finite and by the assumption, there are finitely many such subgroups, so $G$ is a finite union of finite cyclic subgroups and hence is a finite group.

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The problem is that in all of the proofs one uses the fact that an infinite cyclic group is isomorphic to the set of integers, but I think it's enough to show that there is bijection $\varphi: \mathbb Z \to \langle g \rangle$ such that $ k \mapsto g^k$, and from here we conclude that that $\langle g \rangle$ has the same number of elements as $\mathbb Z$.

Answer 1

I think it's enough to show that there is bijection $\varphi: \mathbb Z \to \langle g \rangle$ such that $ k \mapsto g^k$

But that is what we mean when we say $\langle g\rangle$ is isomorphic to the integers. Your bijection is an isomorphism.

Also, note that in the context of group theory, we often want to use, as much as possible, group theoretic notions (a similar statement is true in general). Bijections between groups are, in general, not very useful in group theory, but isomorphisms are. So we phrase theorems and proofs, as much as possible, by using isomorphisms rather than bijections. A proof is easier to read when the number of notions to keep track of is smaller.

Answer 2

No, it's not sufficient. A simple bijection need not preserve group properties: for instance, $\mathbb{Z}/4\mathbb{Z}$ has a bijection with $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$, but one group is cyclic and the other isn't.

What the proof is actually using is that, up to isomorphisms, there is a single infinite cyclic group. Since an element of infinite order generates an infinite (cyclic) group, this group is isomorphic to $\mathbb{Z}$ and the bijection witnessing it does preserve group properties, in particular the subgroup structure. By just establishing a bijection, you cannot conclude that $\langle g\rangle$ has infinitely many subgroups.

You can easily avoid the isomorphism/bijection by proving directly that $\langle g^{n}\rangle$, for $n>0$, defines an infinite family of subgroups, because if $0<m<n$, then $g^{m}\notin g^{n}$.