Is $\Bbb R/\Bbb Z$ isomorphic to $\Bbb R/2\Bbb Z$? I see that both ${\Bbb R}/{\Bbb Z}$ and ${\Bbb R}/{2\Bbb Z}$ are isomorphic to $S^1$. But when I apply the third isomorphism theorem I get
${\Bbb R}/{\Bbb Z}\simeq\frac {{\Bbb R}/{2\Bbb Z}}{{\Bbb Z}/{2\Bbb Z}}$ i.e. ${\Bbb R}/{\Bbb Z}\simeq\frac {{\Bbb R}/{2\Bbb Z}}{\Bbb Z_2}$.
So if the claim in the title is true, I get ${\Bbb R}/{\Bbb Z}\simeq\frac {{\Bbb R}/{\Bbb Z}}{\Bbb Z_2}$, which seems incorrect to me.
Since $\Bbb R$ is abelian, all the quotient sets are groups.
 A: Indeed, if you consider $S^1$ as the multiplicative group $\{ z\in\mathbb C : \lvert z\rvert = 1\}$, then $H=\{+1,-1\}$ is a subgroup isomorphic to $\mathbb Z/2\mathbb Z$ and you have an isomorphism
\begin{align*}
S^1/H &\longrightarrow S^1, \\
[z] &\longmapsto z^2,
\end{align*}
where $[z]=zH=\{z,-z\}$.
A: Elements of $\mathbb{R}/2 \mathbb{Z}$ are of the form $x + 2\mathbb{Z}$ for $x \in \mathbb{R}$ and a complete set of representatives is given by choosing $x$ from $[0,2)$, although many other choices are possible.
Elements of $\mathbb{Z}/2 \mathbb{Z}$ are of the form $z + 2\mathbb{Z}$ for $z \in \mathbb{Z}$ and a complete set of representatives is given by choosing $z$ from $\{0,1\}$.
Elements of $G = \frac{\mathbb{R}/2 \mathbb{Z}}{\mathbb{Z}/2 \mathbb{Z}}$ are apparently, then, of the form $x + 2\mathbb{Z} + \mathbb{Z}/2\mathbb{Z}$.  We may find all the elements of the equivalence class containing $x$ by  \begin{align*}
\bigcup_{z \in \mathbb{Z}/2\mathbb{Z}} & x + 2\mathbb{Z} + z  \\
    &= \left( x + 2\mathbb{Z} + (0 + 2\mathbb{Z}) \right) \cup \left( x + 2\mathbb{Z} + (1 + 2\mathbb{Z}) \right)  \\
    &= \left( x + 2\mathbb{Z} \right) \cup \left( x + 1 + 2\mathbb{Z} \right)  \\
    &= x + (2\mathbb{Z} \cup (1 + 2\mathbb{Z}) )  \\
    &= x + \mathbb{Z}  \text{.}
\end{align*}
That is, the equivalence classes of the quotient defining $G$ are the same as those in $\mathbb{R}/\mathbb{Z}$ (and addition goes through without complications).  So $G \cong \mathbb{R}/\mathbb{Z}$.
A: It actually is correct!  Taking $S^1$ and identifying antipodes gives $S^1$ again.  This explains why $\mathbb{R}P^1$ is homeomorphic to $S^1$, and what you have is the group-theoretic version of this.  
You have to be careful about how you are realizing $\mathbb{Z}_2$ as a subgroup of the circle.  The sensible thing is as the multiplicative group $\{\pm 1\}$, and the cosets are then antipodal pairs, which gives my first paragraph above.
A: This is not a problem at all.
Consider the surjective group morphism $f: S^1 \to S^1$
$$f(z)=z^2$$
The kernel is $\{ 1 ; -1\} \cong \Bbb Z_2$ and by the first isomorphism theorem
$$S^1 \cong S^1 / \{ 1 ; -1\}$$ which is exactly what you found.
A: You're approaching the problem from the wrong side, in my opinion. The problem you're facing is due to how you identify $\mathbb{Z}/2\mathbb{Z}$ with a subgroup of $\mathbb{R}/\mathbb{Z}$ after applying the isomorphism.
There's a simpler way to proceed. Consider the (additive) group homomorphism $\mu\colon\mathbb{R}\to\mathbb{R}$ defined by $\mu(x)=x/2$. This is of course an isomorphism. Compose it with the canonical projection $\pi\colon\mathbb{R}\to\mathbb{R}/\mathbb{Z}$.
Then $\pi\circ\mu$ is surjective and
$$
\ker(\pi\circ\mu)=\{x\in\mathbb{R}:\mu(x)\in\mathbb{Z}\}=2\mathbb{Z}
$$
so the (first) homomorphism theorem provides
$$
\mathbb{R}/2\mathbb{Z}=
\mathbb{R}/\ker(\pi\circ\mu)\cong
\mathbb{R}/\mathbb{Z}
$$
If instead of $\mu(x)=x/2$ we use $\mu(x)=x/k$, where $k$ is any nonzero real, we get similarly that
$$
\mathbb{R}/k\mathbb{Z}\cong\mathbb{R}/\mathbb{Z}
$$
