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

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

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 9:36
• 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 2:44
• 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 5:23

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

• 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 8:16
• @misi: That’s right, on both counts. – Brian M. Scott Apr 17 '13 at 8:18

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.

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

• I get why $(\frac{1}{2} + \mathbb{Z}) + (\frac{1}{2} + \mathbb{Z}) = 1 + \mathbb{Z}$. But why is $0 + \mathbb{Z} = 1 + \mathbb{Z}$? – physicsmajor Apr 15 '18 at 22:58
• Okay, nvm. I think I see. $0 + \mathbb{Z} = \{...,-2,-1,0,1,2,...\}$ and $1 + \mathbb{Z} = \{...,-1,0,1,2,3,...\}$, right? – physicsmajor Apr 15 '18 at 23:04