I have one probably basic question, but still it bothers me how to show it.
Namely, if two groups are isomorphic (i.e. there is a bijective group homomorphism between them) and if one of them is infinite cyclic (precisely, in my case it is discussed about $(\mathbb Z,+)$), does the other group necessary have to be infinite cyclic?
Precisely, I'm in algebraic topology and I know that the fundamental group of cycle $\mathbb S^1$ is isomorphic to integers, but does this imply that the fundamental group of cycle is infinite cyclic as well?
Any sketch of the proof would be welcome.
Best Answer
Indeed, this is true and is known as the fundamental theorem of finitely generated abelian groups..
In particular, we know that every finitely generated abelian group is of the form:
$$\mathbb Z^n \oplus \mathbb Z_p\oplus\cdots \oplus \mathbb Z_q$$
so if your group is infinite and cyclic, then it is abelian, of rank $1$ and torsion-free, meaning that it is indeed the same thing as $\mathbb Z$.
Edit: on the other hand, this now feels like a lot of overkill, despite being the way that I think about it. Here is a proof sketch:
Let $G$ be infinite cyclic $\langle a^n \mid n \in \mathbb Z \rangle $. Suppose further that $\phi: \mathbb Z \to G$ is the map $n \mapsto a^n$.
You should check that this is indeed a homomorphism. Surjectivity is clear. Injectivity follows from the fact that if $a^n=a^m$ for $ n >m$, then $a^n (a^m)^{-1}=e \implies a^{n-m}=e$ while the order of $a$ was supposed to be infinite.
Hence, this is an isomorphism.
Edit 2: Let me try to address your question in the comments. Let $\pi_1(S^1)$ be the group consisting of loop classes, equipped with the usual multiplication. We already have that there exists an isomorphism $$\rho: \pi_1(S^1) \to (\mathbb Z,+)$$
and we have further that
$$\phi:\mathbb Z \to G$$ is an isomorphism when $G$ is infinite cyclic. By the composition of isomorphisms,
$$\phi \circ \rho:\pi_1(S^1) \to G$$ is an isomorphism as well. Hence, we can conclude that it is infinite cyclic.
This seems like a more general problem, isomorphism is a transitive property, if $A \cong B \cong C$, then $A \cong C$ as well.
Edit 3: We claim that $(\mathbb Z,+)$ is an infinite cyclic group. To see this, consider $1 \in \mathbb Z$. Clearly, every element $n \in \mathbb N$ can be written as $\underbrace{1+1+\dots+1}_{n \, \mathrm{times}}$ and each inverse can be given by $-1$, which is the additive inverse of $1 \in \mathbb Z$. In other words, $1$ generates the group and has infinite order. Perhaps the notation is confusing, in additive notation:
A group $(G,+)$ is said to be infinite cyclic if $$G=\langle n \cdot a \mid n \in \mathbb Z\rangle.$$
Hence if you already know that $\pi_1(S^1) \cong \mathbb Z$, then it must be infinite cyclic, since $\mathbb Z$ already was.