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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 $\mathbb{R}/\mathbb{Z}$ isomorhpic 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?

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It is actually a circle. –  mesel Jul 12 at 20:12
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And don't worry about the level of this: there are plenty of arithmetic or pre-calculus questions which fit in well here: group theory is certainly sensible! –  Adam Hughes Jul 12 at 20:13

3 Answers 3

up vote 4 down vote accepted

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".

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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 $\Bbb R/\Bbb Z$

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I see. And would $S^1$ in turn be isomorphic to the reals mod 1? It seems so. –  Elie Bergman Jul 12 at 20:13
    
@ElieBergman: $[0,1)$ is a circle –  mesel Jul 12 at 20:14
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@mesel Nitpickery: only if you define neighborhoods correctly - $[0,1)$ isn't inherently a circle in any meaningful way. –  Steven Stadnicki Jul 12 at 20:25
    
@ElieBergman keep in mind, the way the question is worded "the reals modulo $1$" is essentially the definition of $\Bbb R/\Bbb Z$, so the intent may not be to just restate this, but to find a group which is not literally the same but still isomorphic and familiar. –  Adam Hughes Jul 12 at 20:39
    
@StevenStadnicki: As we see from the glass modulo $\mathbb Z$, $[0,1)$ is circle from this aspect since $0\cong 1$. (it also preserve topological structure as well) –  mesel Jul 12 at 23:49

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, $$\mathbb{R} / \mathbb{Z} \simeq \mathbb{Q}^{(\mathbb{R})} \oplus \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z} [p^{\infty}].$$

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Love divisible ones so love this post. +1 –  Babak S. Jul 13 at 10:19

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