Talk:Group/Examples/inv x = 1 - x

The set $S$

In that book of algebra and many other, the example most similar to that one described in this page is the one in which $S=\{x\in\R : 0\le x <1\}$. I think this group it's usually called "the reals modulo 1".

Be that as it may, but that is not the group which is being defined in this page, or indeed in 1971: Allan Clark: Elements of Abstract Algebra: $\S 26 \mu$. Maybe there's a mistake in the book, but it distinctly says:

$S = \left\{{x \in \R: 0 < x < 1}\right\}$ and so $0 \notin S$

The inverse of $x \in S$ is $1 - x$, NOT $1-x-\operatorname{floor}(1-x)$.

If this is a mistake in Clark, then fair enough, it needs to be announced as such. (This will have the positive effect of providing a source for confused students - like me - who can not make this example work.)

What we ought to do is start a new page called Real Numbers under Addition Modulo 1 form Group in which this concept is developed formally as a group (and I think some of the work has already been done there, somewhere in Analysis, possibly), perhaps linking from / to this result, with some sort of comment to the effect that it is suspected that Clark may have made a mistake and may have meant this group.--prime mover 01:59, 20 August 2011 (CDT)

I found the group you were talking about. Check the book of Alan Clark, the exercise just above this one it's the same group but with a tranlation $x\to x+1$ and a dilatation of $x\to 2x$--Dan232 16:18, 5 December 2011 (CST)

...or a dilation of $x \to x/2$. That would put the numbers in the correct range, I'd still want to check it was actually a group still. I'd need to think about it, but not tonight. --prime mover 16:38, 5 December 2011 (CST)

The suggested shift does not constitute a group homomorphism and so it can't be the operation sought after. --Lord_Farin 22:40, 23 July 2012 (UTC)

One would say an error was made but to prove that no group exists with these properties is probably very hard. --Lord_Farin 22:59, 23 July 2012 (UTC)

The proof is up; some tidying and linking is still to be done. I think that it is also valuable to determine directly that the operation $\circ$ determines a group (the identity element and inverses are done, only associativity essentially remains). --Lord_Farin (talk) 19:38, 9 October 2012 (UTC)

Good job. --prime mover (talk) 22:04, 9 October 2012 (UTC)