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Result: Fix $a \in \mathbb{R}$. Then $(\mathbb{R} \backslash \{a\}, *)$ is a group, where our group operation is defined by $x*y = (x-a)(y-a) + a$.

One consequence of this is the standard fact that the set of nonzero real numbers forms a group under regular multiplication. (Just pick $a = 0$.)

As another, slightly more interesting option, we could let $a = -1$. Then we have the group $(\mathbb{R} \backslash \{-1\}, *)$, where the group operation is defined by $x*y = x + y + xy$.

My question: Has anyone seen this result (or a generalization of it) elsewhere?

It may already exist as an exercise in an Abstract Algebra text, but I suspect this particular type of group is really a specific instance of something far more general. Any references or remarks on its generalization would be most welcomed.

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If memory serves, I’ve seen the case $a=1$ quite a few times as a problem in a textbook or (occasionally) contest. –  Brian M. Scott Sep 28 '12 at 5:41
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First, in your third line you surely meant "the set of positive integers", and second: how does this follow from choosing $\,a=0\,$ and then taking $\,\Bbb R -\{0\}\,$ in your construction? Added I think I get it: you surely meant "the non-zero real numbers", right? –  DonAntonio Sep 28 '12 at 5:41
    
Right, nonzero real numbers; sorry for the typo! –  Benjamin Dickman Sep 28 '12 at 6:03
    
I think these are called isotopes. –  i. m. soloveichik Sep 29 '12 at 14:26

1 Answer 1

up vote 3 down vote accepted

Each of these groups is isomorphic to $(\mathbb R\setminus \{0\},\cdot)$, via the isomorphism $\varphi(x)=x-a$. To see this, note that $\varphi$ is clearly bijective and $\varphi(x*y)=\varphi((x-a)(y-a)+a)=(x-a)(y-a)=\varphi(x)\varphi(y)$. Thus all their properties are the same as $(\mathbb R\setminus \{0\},\cdot)$ up to a translation by $a$.

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Thanks; in retrospect this is so clear that I'm embarrassed at having asked! –  Benjamin Dickman Oct 26 '12 at 23:41

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