Prove that $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic as groups

Question

I'm working on the following problem:

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

Here is my attempt at a solution:

If $\mathbb{Z} \cong \mathbb{Q}$, then there must exist a bijective homomorphism $\varphi: \mathbb{Z} \to \mathbb{Q}$. Consider $\varphi(x) = x$. Clearly, this is a homomorphism as it is the identity map. Moreover, it is also clearly injective. However, it fails to be surjective since there exist elements of $\mathbb{Q}$ that are not mapped to by $\varphi$ (like $3/2$ or $1/4$). Hence, since $\varphi$ is an injective homomorphism, but not surjective, we see that there cannot exist a bijection between $\mathbb{Z}$ and $\mathbb{Q}$. Thus, $\mathbb{Z} \not \cong \mathbb{Q}$.

Could anyone critique this solution? Am I on the right path?

1 can represent as 1/2+1/2 but there is no 1/2 in integers. consider isomorphism from integers to rationals. — Mizi, Dec 19 '14 at 04:50
Note that $\Bbb Q$ is an abelian group that is divisible, but $\Bbb Z$ is not. — Pedro, Dec 19 '14 at 04:50

score 14 · Answer 1 · answered Dec 19 '14 at 04:27

14

Your solution is incorrect. You have shown that one particular homomorphism is not an isomorphism.

To show that they are not isomorphic you need to show that there is no possible isomorphism whatsoever.

HINT: One of these groups is cyclic.

answered Dec 19 '14 at 04:27

Asaf Karagila

393,674

score 5 · Answer 2 · answered Dec 19 '14 at 04:30

Your solution proves that there is an injective, but not surjective homomorphism from $\mathbb Z\rightarrow \mathbb Q$. This is not the statement you wished to prove - it certainly doesn't prove that $\mathbb Z$ and $\mathbb Q$ are not isomorphic, because I can say the same about $f(x)=x+x$ mapping $\mathbb Z\rightarrow\mathbb Z$ - but that certainly doesn't mean that $\mathbb Z$ and $\mathbb Z$ are not isomorphic (because they obviously are).

A nice simple solution would be to notice there is no element $x$ in $\mathbb Z$ such that $x+x=1$, but for every $y$ in $\mathbb Q$, there is a $x$ such that $x+x=y$. The groups are therefore not isomorphic.

score 3 · Answer 3 · answered Dec 19 '14 at 04:44

3

Another thought: Let $\phi:\Bbb Z\to\Bbb Q$ be a homomorphism, and let $\phi(1)=a$. Because, $\Bbb Z$ is cyclic, $\phi$ is completely determined.

$$\phi(n)=na$$

It's easy now to show that $\phi$ is not surjective.

answered Dec 19 '14 at 04:44

Tim Raczkowski

20,046

score 0 · Answer 4 · answered Dec 19 '14 at 07:04

0

$\mathbb Q$ has a subgroup $N:=\mathbb{Z}$ such that $\mathbb{Q}/N$ is infinite

answered Dec 19 '14 at 07:04

Blah

5,374

Prove that $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic as groups

4 Answers4