# basic group theory questions

• Dec 5th 2008, 07:31 AM
nocturnal
basic group theory questions
taken from D&F, question 1.1.7:

Let $\displaystyle G = \{x \in \mathbb{R} | 0 \leq x < 1\}~\text{and for}~x,y \in G, \text{let }~x * y~\text{be the fractional part of } x + y$ (i.e. $\displaystyle x * y = x+y - [x+y]$). Show that $\displaystyle G$ is an abelian group.

﻿The question asks for verification of the group axioms, although I'm a little confused with showing the existence of inverses. One can say that for an arbitrary $\displaystyle a \in G, a * (-a) = a + (-a) + [a-a] = 0$, which is the identity. But notice that I said $\displaystyle -a$, and that's confusing me. Is $\displaystyle (-a) \in G$? Is it literally the negative of $\displaystyle a$? I would presume as much since I actually took that into consideration when I added $\displaystyle a + (-a)$. But $\displaystyle -a$ is not part of $\displaystyle G$ by construction. So how do I show the existence of inverses?
• Dec 5th 2008, 07:49 AM
Plato
You are correct, G contains no negatives.
But consider this: $\displaystyle x \in G\backslash \left\{ 0 \right\} \Rightarrow \quad x * \left( {1 - x} \right) = ?$
• Dec 5th 2008, 07:55 AM
nocturnal
Isn't it x + 1 - x + [x + 1 - x] = 1 + [1] = 2? But $\displaystyle 2 \notin G\backslash \left\{ 0 \right\}$!
• Dec 5th 2008, 08:02 AM
Plato
Quote:

Originally Posted by nocturnal
Isn't it x + 1 - x -[x + 1 - x] = 1 - [1] = 0? But $\displaystyle 2 \notin G\backslash \left\{ 0 \right\}$!

You missed a sign.
• Dec 5th 2008, 08:06 AM
Gamma
Another look
I have been taking topology this semester and was thinking this same problem could be approached in a more topological way by looking at the bijection from $\displaystyle f:[0,1)\rightarrow S^1$ (the unit circle) by $\displaystyle f(x)=e^{i2\pi x}$ and then this is obviously an abelian group with the operation rotations of x degrees about the origin of the complex plane and you can check to see that this would preserve addition easier this way, at least in my opinion. It is not a homeomorphism (the inverse function is not continuous at $\displaystyle 1+0i$), but I don't think that will affect this in terms of it still being and isomorphism.
• Dec 5th 2008, 08:10 AM
nocturnal
Haha thanks Plato. I'm trying to reach your train of thought;

Ordinarily, we have two cases for a * b:

(Case 1) if 0 <= (a+b) < 1, then we have a*b = a + b, and
(Case 2) if 1 <= (a+b) < 2, a*b = a+b-1.

Suppose a*b = 0. Then if 0 <= a+b < 1, a + b = 0, and b = -a. Now if 1 <= a+b < 2, then a + b - 1 = 0, so b = 1 - a.

So I guess we have two possibilities for the inverse. Clearly the first one is absurd. The second one, though, will always hold. Yay! Thanks Plato. That is what you were getting at, right?
• Dec 5th 2008, 08:15 AM
nocturnal
That looks neat Gamma! Unfortunately I'm only a semester away from Topology so I'm not getting everything in your post, but that will hopefully be temporary -- thanks for sharing!
• Dec 5th 2008, 08:18 AM
Plato
Quote:

Originally Posted by nocturnal
Ordinarily, we have two cases for a * b: WHAT?
Just one case a*b = a + b - [[a+b]]

For that operation, 0 is the operational identity.
If $\displaystyle x \not= 0$ then $\displaystyle (1 - x)$ is its operational inverse.