**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.6**

Prove that the additive groups $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic.

Solution: First we prove a lemma.

**Lemma.** If $\varphi : G \rightarrow H$ is a surjective group homomorphism and $G$ is generated by one element, then $H$ is generated by one element.

Proof: Let $h \in H$. Since $\varphi$ is surjective, there exists $g \in G$ with $\varphi(g) = h$. We can write $g = s^k$ for some $k \in \mathbb{Z}$, where $s$ is a generator of $G$. Now $$h = \varphi(g) = \varphi(s^k) = \varphi(s)^k.$$ Note that $\varphi(s)$ is a fixed element of $H$; thus $\varphi(s)$ is a generator of $H$.

Now suppose an isomorphism $\varphi : \mathbb{Z} \rightarrow \mathbb{Q} $ exists. We know that $\mathbb{Z}$ is generated by one element, so that by the lemma $\mathbb{Q}$ is also generated by one element; in particular, $\varphi(1)$. Suppose $\varphi(1) = \dfrac{m}{n}$. Now choose $\dfrac{1}{q} \in \mathbb{Q}$ such that $q$ does not divide $n$. Then we have $\dfrac{1}{q} = k \cdot \dfrac{m}{n}$ for some integer $k$, hence $n = kqm$. But then $q|n$, a contradiction. So no isomorphism $\varphi$ exists.

## Axel

21 Jul 2020Why can't we employ the same idea as in problem #5 (Cantor's theorem)? Isn't it also true that Q is the power set of Z?

## Linearity

22 Jul 2020Fact: $\mathbb Q$ is countable.

Proof. Let us define the mapping $\phi: \mathbb Q \to \Z \times \N$ as follows: $$\forall \dfrac p q \in \mathbb Q: \phi \left({\dfrac p q}\right) = \left({p, q}\right)$$ where $p$ and $q$ are relatively prime.

Then $\phi$ is clearly injective. Since Cartesian Product of Countable Sets is Countable, we have that $\Z \times \N$ is countably infinite.

The result follows directly from Domain of Injection to Countable Set is Countable.