# Infinite cyclic group isomorphism.

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.

• All cyclic groups are isomorphic to either $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$ for a natural number $n$. – Student Mar 6 '17 at 21:20

## 1 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.

• Thanks. I understand what you are trying to say, but isn't this (in some sense) the reversed statement, i.e. for a given infinite cyclic group say $(G,*)$, one can construct isomorphism between $G$ and $\mathbb Z$. But I think I have the opposite problem; I already know that my (fundamental) group $\pi_1(\mathbb S^1,*)$ is isomorphic to $(\mathbb Z,+)$ where for $(\mathbb Z,+)$ "infinite cyclic" behaviour is known and what I am trying to show (if possible) is that the first group must be infinite cyclic as well. – edward_scissorhands Mar 7 '17 at 8:39
• It's unclear what you are confused about to me. What I have just demonstrated is that it does not matter whether you say that the fundamental group is infinite cyclic of is the additive group on the integers, since the two of them are the same group, up to symbolic manipulation. – Andres Mejia Mar 7 '17 at 8:44
• the fundamental group is infinite cyclic. I included another argument. However, I think that this argument is backwards in the way of intuition. The thing to understand is that $\pi_1(S^1)$ is a free group given by a single generator, so it is in fact cyclic, so it is abelian, and hence is $\mathbb Z$. In particular, any free group modulo its commutator subgroup is again $\mathbb Z^n$ for some $n$. To see an example where it is a free group (infinite) on two generators, but not $\mathbb Z$, consider $\pi_1(S^1 \vee S^1)$. – Andres Mejia Mar 7 '17 at 8:54
• If it is isomorphic to $(\mathbb Z,+)$ then it is infinite cyclic, since $\mathbb Z$ is itself infinite cyclic. For example, it is generated by a single element $1$ that has infinite order. – Andres Mejia Mar 7 '17 at 8:57
• @Eurydice see my edits – Andres Mejia Mar 7 '17 at 9:01