Does $\mathbb{Z} \hookrightarrow \mathbb{Z} \star \mathbb{Z} \star \mathbb{Z} \star \mathbb{Z}$?

Question

Probably very trivial, I'm pretty sure it does. Just want to make sure.

We note that $\mathbb{Z} \star \mathbb{Z}$ is the free group on two letters. In this case, we are consider the free group on 4 letters.

Answer 1

Let $\{a,b,c,d\}$ be a basis for $\mathbb Z \star \mathbb Z \star \mathbb Z \star \mathbb Z$ and consider $$ \iota \colon \mathbb Z \to \mathbb Z \star \mathbb Z \star \mathbb Z \star \mathbb Z, z \mapsto a^z, $$ where $$ a^z := \begin{cases} \underbrace{aa\ldots a}_{z \text{ times}} & \text{, if } z \ge 0 \\ \underbrace{a^{-1}a^{-1}\ldots a^{-1}}_{-z \text{ times}} & \text{, if } z < 0 \end{cases} $$

You need to verify that

1. $\iota$ is a group homomorphism and
2. $\iota$ is injective (this follows easily from 1.).

Answer 2

Subgroups of free groups are free, and in particular torsion-free. Thus the map $1 \to g$ is a well-defined injection $\mathbb{Z} \to \mathbb{Z}^{*4}$ for any $g\not = 1$.