Is $\mathbb{Q}/\mathbb{Z}$ isomorphic to $\mathbb{Q}$?

Question

My guess is no. Does the first isomorphism theorem have anything to do with this?

Any hints appreciated, thanks.

Isomorphic as what? They are isomorphic as plain sets, but not as rings (because $\Bbb Q/\Bbb Z$ is not a ring to begin with). Please always be clear about such polymorphic (no pun intended) terms as "isomorphic". I see this question is tagged group-theory, but tags should not be considered part of the question body. — Marc van Leeuwen, Apr 17 '13 at 09:36
Yes, I meant group isomorphism. I haven't studied ring theory yet so I wasn't aware of the different connotations of the term. I apologize. — misi, Apr 18 '13 at 02:44
OK I see. In any case you will see later that there are many theories, algebraic and other ones, that have their notions of "isomorphic". When posing a question here it is important that there should be no doubt about what the question is about precisely. In this case I really should have guessed, since $\Bbb Q/\Bbb Z$ really cannot be much more than a group (but $\Bbb Q$ can be, and usually is, more than just that). — Marc van Leeuwen, Apr 18 '13 at 05:23

score 35 · Accepted Answer · answered Apr 17 '13 at 08:08

35

HINT: What is the order of $\frac12+\Bbb Z$ in $\Bbb Q/\Bbb Z$? Does that match anything in $\Bbb Q$?

answered Apr 17 '13 at 08:08

Brian M. Scott

631,399

5

2? And there's no element of order 2 in $\mathbb{Q}$ because all elements in $\mathbb{Q}$ have order 1 or $\infty$? – misi Apr 17 '13 at 08:16
@misi: That’s right, on both counts. – Brian M. Scott Apr 17 '13 at 08:18

score 13 · Answer 2 · answered Apr 17 '13 at 08:20

13

Hint: $\Bbb Q$ is an ordered group, in particular $0<\frac12<1$. But in $\Bbb{Q/Z}$ we have that $$\left(\frac12+\Bbb Z\right)+\left(\frac12+\Bbb Z\right)=0+\Bbb Z=1+\Bbb Z.$$

answered Apr 17 '13 at 08:20

Asaf Karagila

405,794

I get why $(\frac{1}{2} + \mathbb{Z}) + (\frac{1}{2} + \mathbb{Z}) = 1 + \mathbb{Z}$. But why is $ 0 + \mathbb{Z} = 1 + \mathbb{Z}$? – CrypticParadigm Apr 15 '18 at 22:58
1

Okay, nvm. I think I see. $ 0 + \mathbb{Z} = {...,-2,-1,0,1,2,...}$ and $1 + \mathbb{Z} = {...,-1,0,1,2,3,...}$, right? – CrypticParadigm Apr 15 '18 at 23:04

score 12 · Answer 3 · edited Apr 20 '13 at 18:27

Just for fun, a general argument:

Notice that both $\mathbb{Q} / \mathbb{Z}$ and $\mathbb{Q}$ are divisible. But any divisible group can be written as the direct sum of subgroups isomorphic to $\mathbb{Q}$ or $\mathbb{Z}[p^{\infty}]$ for some prime $p$; then, you just have to compare the factors.

For $\mathbb{Q}$, there is nothing to do.

Let $A_p$ be the subgroup of $\mathbb{Q}/ \mathbb{Z}$ consisting of elements of order a power of $p$. Then $\mathbb{Q}/ \mathbb{Z} = \bigoplus\limits_{p \in \mathbb{P}} A_p$. But clearly, $A_p= \left\{ \frac{n}{p^k} \mid n \in \mathbb{Z},k \in \mathbb{N} \right\}$ is isomorphic to $\mathbb{Z}[p^{\infty}]$ (take $n/p^k \mapsto \exp( i2n\pi/p^k )$ for the isomorphism), so $\mathbb{Q}/\mathbb{Z} \simeq \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]$.

Reference: Infinite Abelian Groups, I. Kaplansky.

Is $\mathbb{Q}/\mathbb{Z}$ isomorphic to $\mathbb{Q}$?

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