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

Solution 1:

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

Solution 2:

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

Solution 3:

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.