Is this proof for $(\mathbb{Z},+) \ncong (\mathbb{Q},+)$ valid?

Question

In an introductory cryptography course, our teacher demonstrated a proof for $(\mathbb{Z},+) \ncong (\mathbb{Q},+)$. I'm not convinced, even though the statement may be correct (I don't know).

Earlier in the course, we had seen the isomorphism of the Klein-4 group as $(\{0,1\}^2,+)$ and the 2D symmetry group of a rectangle not being a square (using identity, vertical reflection, horizontal reflection and 180 degree rotation).

Now, our teacher made a list of different equations and whether or not they could be solved in the sets $\mathbb{Z}^+, \mathbb{Z}, \mathbb{Q}$ and $\mathbb{R}$. This to show that some equations can be solved in certain sets but not in others (like $x=3$ being solvable in all four sets, but $3x=5$ only in the latter two and $x^2=-1$ in none of the considered sets).

Then, she stated that if we have some equation using an operation $\odot$ that is solvable in $A$ but not in $B$, then $(A,\odot) \ncong (B,\odot)$ (supposing of course that $(A,\odot)$ and $(B,\odot)$ are groups). This already is a rather vague description, isn't it?

She then proposed the equation $x+x=3$ which is solvable in $\mathbb{Q}$ but not in $\mathbb{Z}$. While that is obviously correct, I'm not sure if this argument is sound and if it actually proves the two groups not being isomorph.

I have difficulties to express my concerns. Somehow, I would expect the constant $3$ in that equation to be some number from the particular set instead of some constant. I'm not convinced by the argument because if we'd map all odd numbers from $\mathbb{Q}$ to even numbers in $\mathbb{Z}$, the equation would in fact be solvable in $\mathbb{Z}$.

What I'd like to know is:

• Is the proof given by my teacher valid (and why)?
• If not, are these groups still isomorph (out of curiosity)?

Answer 1

The reasoning is not correct. If there were an isomorphism $\mathbb Z \simeq \mathbb Q$ there's no apriori reason why it should send $1$ to $1$. Being able to solve the equation $2x = 3$ in $\mathbb Q$ would certainly then imply that there is an equation in $\mathbb Z$ of the form $2x = a$ which is also solvable, where $a \in \mathbb Z$ corresponds to $5 \in \mathbb Q$ under the isomorphism, but if the isomorphism doesn't send $1$ to $1$ then how would you know that $a$ should still be $5 \in \mathbb Z$?

This line of reasoning does prove that the natural inclusion $\mathbb Z \hookrightarrow \mathbb Q$ is not an isomorphism. But that's already obvious.

Ok, so that reasoning is incorrect, does that mean $\mathbb Z$ and $\mathbb Q$ are isomorphic? No. A similar but this time correct argument is proposed by Zhen Lin in the comments: For any $a \in \mathbb Q$ the equation $2x = a$ has a solution, but it's not true that for any $a \in \mathbb Z$ the equation $2x = a$ has a solution.

Side note: You may wonder why I'm all of a sudden assuming $2$ goes to $2$? I'm not. The $2$ in that equation just indicates that $x$ is the result of applying the group operation to the pair $(x, x)$. Writing $2x = a$ is just notation for $x + x = a$, while $x$ and $a$ are elements of the group, $2$ is notation and is not meant to be a group element.

Answer 2

This has been basically covered in the comments, but not to leave it unanswered: the argument works to show that $\mathbb{Q}$ and $\mathbb{Z}$ are not isomorphic as rings, because $1$ is preserved by homomorphisms of rings, and thus $3=1+1+1$ is as well. So an isomorphism, indeed, any homomorphism, of rings preserves solutions of equations like $2x=3$.

In an abelian group, "$3$" has no interpretation preserved by morphisms, so it's group-theoretically meaningless to point out that an equation like $2x=3$ has or lacks a solution. The only meaningful equations in abelian groups involving a specific constant are those involving the identity, essentially those of the form $nx=0$. So in an abelian group the argument from the solvability of a single equation works to distinguish groups with a specific order of torsion from those without, e.g. $\mathbb{Z}/m\mathbb{Z}$ from $\mathbb{Z}/n\mathbb{Z}$ and from $\mathbb{Z}$ itself, but otherwise we have to take families of equations, as was done in the comments.

Answer 3

I would call it an outline of a proof rather than a proof. But it is valid. Let $\varphi$ be an isomorphism from $\mathbb{Q}$ to $\mathbb{Z}$. Let $\varphi(r)=3$, and let $\varphi(r/2)=a$. Since $\varphi(r/2+r/2)=\varphi(r/2)+\varphi(r/2)$, we have $a+a=3$, which is impossible in $\mathbb{Z}$.