# how to prove $G$ is an abelian group under $*$ (called the real numbers mod 1)

Let $G = \{x \in \mathbb{R}~|~0\leq x < 1\}$ and for $x,y \in G$ let $x*y$ be the fractional part of $x+y$ i.e $x*y = x + y - [x + y]$ where $[a]$ is the greatest integer less than or equal to $a$. I need help proving $*$ is a well defined binary operation on $G$ and that $G$ is an abelian group under $*$.

To prove that $*$ is well defined, I rely on the assumption that +,- is well defined in the set of real numbers. evaluating the brackets [] is also well-defined in the set of real numbers. (is this an okay assumption?)

To prove $*$ is associative, (I imagine because +,- is associative, so will *?) I show that $(x*y)*z = x*(y*z)$ $$(x*y)*z =(x+y-[x+y])*z = x+y-[x+y]+z - [x+y-[x+y]+z]$$ $$x*(y*z) = x*(y+z - [y+z]) = x + y + z - [y+z] - [x+y+z - [y+z]]$$ rearranging $$(x*y)*z = x+y+z-[x+y]-[x+y+z-[x+y]]$$ $$x*(y*z) = x+y+z-[y+z]-[x+y+z-[y+z]]$$ I am not so sure that these two are equal. I don't see how distributing the $-$ to remove the terms $[x+y]$ and $[y+z]$ is fair (ex: $x = y = \frac{1}{2}$ so $x*y = .5+.5-[.5]-[.5]$ yields a different answer than $x*y = .5+.5-[.5+.5]$

The identity element would be $0$, the inverse would have to be $-x$ which isn't in $G$. What am I doing wrong?

edit: commutativity would be proven by showing $x*y = y*x$ which is easy to show $x+y - [x+y] = y+x - [y+x]$ right?

• For a proof see for example here. May 13, 2016 at 19:35
• I'm actually using that to check my solutions for the exercises in D&F. I'd like hints/corrections for my proof before I check the solution though. May 13, 2016 at 19:36
• I think the best way to think about this group is as rotations in the plane. $x\in G$ signifies rotation of $360x$ degrees. May 13, 2016 at 19:58
• @CameronBuie Thank you for your help. The inverse has to be $y = 1-x$ since $x+y$ is to be the integer 1. I forgot to reply and thank you :p May 16, 2016 at 17:16
• also, for the part about associativity in case anyone else is wondering, the bracketed section inside of the bracketed section can be treated almost like a constant in an integral. It won't change the value of the bracketed section since it's an integer in itself so it's okay to take it out to cancel terms. If it were single values that were not integers, however, it would not be okay to cancel out. i.e $[x+[a+b]] = [x]+[a+b]$ but not $[x+a+b] = [x]+[a+b]$ May 16, 2016 at 17:21

To prove $$∗$$ is associative

Let $$a,b,c\in G$$. \begin{align*} (a*b)*c &= (a+b-[a+b])*c \\ &= a+b+c-[a+b]-\left[a+b+c-[a+b]\right] \end{align*} Similarly, \begin{align*} a*(b*c) &= a+b+c-[b+c]-\left[a+b+c-[b+c]\right] \end{align*}

Given $$0\le a,b,c < 1, [a+b]$$ and $$[b+c]$$ must be either zero or one individually, for a total of four possible cases. In the cases when both are zeros or ones, clearly $$(a*b)*c = a*(b*c)$$.

Consider the case when $$[a+b]=1$$ and $$[b+c]=0$$. Necessarily $$1 \le a+b < 2, b+c< 1$$ and $$1\le a+b+c< 2\implies [a+b+c] = 1$$ and $$[a+b+c-1] = 0$$. Therefore, $$(a*b)* c=a+b+c-1-[a+b+c-1]=a+b+c-1$$, whereas $$a*(b* c)=a+b+c-[a+b+c]=a+b+c-1$$; i.e. $$(a* b)* c = a*(b* c)$$.

Similar argument applies when $$[a+b]=0$$ and $$[b+c]=1$$. This covers all four cases.

Hence $$G$$ is associative over $$*$$.

You can think of your group $(G, *)$ as $(\mathbb{R}, +)/\mathbb{Z}$: That is, the reals under addition modded out by the integers.

This clearly works because we are essentially saying that is $x, y \in \mathbb{R}$, $x - y \in \mathbb{Z}$, then $x = \mathbb{Z} + y$ in $\mathbb{R} /\mathbb{Z}$. This has the same effect as “truncation”, but is now a nice subgroup Hence,

$$(G, *) \simeq (\mathbb{R}/\mathbb{Z}, +)$$

To imagine this, realize that $\mathbb{R}$ just contains shifted copies of $[0, 1)$ all laid down one next to the other. So, we simply “collapse” all these copies together for this construction to work out.

Now, for this to be a group, we need the kernel of the map $$\phi: (\mathbb{R}, +) \to (\mathbb{R}, +) /\mathbb{Z}$$

to be a normal subgroup. Since the kernel of a quotient map is the quotienting subgroup itself, we simply need to show that $\mathbb{Z}$ is a normal subgroup of $\mathbb{R}$. This is obvious in abelian groups.

• As you're defining it, $\;[0,1)\;$ is not even a subgroup of $\;(\Bbb R,+)\;$ , so that map $\;\phi\;$, whatever it is, is not a group homomorphism and its kernel is irrelevant, in my opinion. Also, if you're taking the usual group $\;(\Bbb R,+)\;$ , then this is an abelian group, so any subgroup is normal. May 13, 2016 at 19:51
• Whoops, you're totally right. What I meant to use was $\mathbb{R} /\ \mathbb{Z}$ which does the same thing (but is actually a subgroup) May 13, 2016 at 19:58
• @Joanpemo - the proof should be correct now. Could you retract your downvote if it is? May 13, 2016 at 20:04
• Sorry but can you prove it is an abelian group by showing that $(G,*)$ satisfies the abelian group axioms instead of treating it as a different group (no idea what kernels, quotient sub groups, etc. are) I'm still a beginner to group theory so I don't have access to your methods in my knowledge yet. May 13, 2016 at 20:19
• @SiddharthBhat I've never downvoted a posted answer for a tiny mistake and even less after a few minutes. Sorry, it wasn't me. May 13, 2016 at 20:30