Is $\Bbb R/\Bbb Z$ isomorphic to $\Bbb R/2\Bbb Z$?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $[z]=zH=\{z,-z\}$.

Answer 4

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 $G \cong \mathbb{R}/\mathbb{Z}$.

Answer 5

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 $$ \mathbb{R}/k\mathbb{Z}\cong\mathbb{R}/\mathbb{Z} $$