What group is $\mathbb{R}/\mathbb{Z}$ isomorphic to? This is not so much a plea of ignorance, but rather me trying to see whether intuitively I actually understand what is going on in group theory. The question asks
What group is $\mathbb{R}/\mathbb{Z}$ isomorphic to?
Thinking about it like a real line which is periodic mod 1, I simply said.
The real numbers mod 1 under addition.
Is this correct? And am I really allowed to be asking such low level questions on MSE? 
 A: Using the map 
$$\begin{cases} \mathbb{R}\to S^1=\{z\in\Bbb C : |z|=1\} \\x\mapsto e^{2\pi i x}\end{cases}$$
we see this is a surjective group homomorphism--the group operation on $S^1$ is multiplication and $e^{z+w}=e^ze^w$--with kernel $\Bbb Z$, hence, by the first isomorphism theorem, we get $S^1$ is isomorphic to $\Bbb R/\Bbb Z$
A: Look at the definition of $\mathbb R/\mathbb Z$: Two elements $a,b\in\mathbb R$ are in the same coset if and only if $a-b\in\mathbb Z$. Now look at the definition of $a\equiv b$ mod $1$, it says that $a=b+k\cdot 1$ for some $k\in\mathbb Z$, which is equivalent to $a-b\in\mathbb Z$. So yes, you can think of $\mathbb R/\mathbb Z$ as "$\mathbb R$ modulo 1".
A: As a divisible group, $G:=\mathbb{R} / \mathbb{Z}$ may be written as a direct sum of Prüfer groups and copies of $\mathbb{Q}$. First we have $$G \simeq \mathrm{Tor}(G) \oplus G/ \mathrm{Tor}(G),$$ where $\mathrm{Tor}(G)$ is the torsion subgroup of $G$. According to one of my previous answer, $$\mathrm{Tor}(G) \simeq \mathbb{Q} / \mathbb{Z} \simeq \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z} [p^{\infty}].$$ On the other hand, $$G/ \mathrm{Tor}(G) \simeq \mathbb{R} / \mathbb{Q}$$ is a $\mathbb{Q}$-vector space of dimension $2^{\aleph_0}$. Finally, $$\mathbb{R} / \mathbb{Z} \simeq \mathbb{Q}^{(\mathbb{R})} \oplus \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z} [p^{\infty}].$$
