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Groups of real numbers

A group as we all know is a set A with a binary operation "*" defined on it such that for every a,b belongs to A a*b belongs to A, the operation is assocaitive, there exists an identity element and for every a in A there exists a unique p in A such that a*p=identity element. Can anyone provide me examples of groups where the operation defined on the set is not addition + or multiplication x.

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marked as duplicate by Arturo Magidin, t.b., Ross Millikan, Asaf Karagila, J. M. Sep 2 '11 at 8:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

-1: The question admits infinitely many answers (in a very strong sense!) and is answered in literally every course in group theory (e.g., any non commutative group does the trick). – PseudoNeo Aug 30 '11 at 18:25
Should this question get made into a community wiki, or have I misunderstood the term? – Doug Spoonwood Aug 30 '11 at 19:20
@PseudoNeo: That doesn't make it any less valuable a question, because this person might not have the opportunity to attend a course in group theory. I really don't think we should be picky about which questions we think too 'standard' to ask. It's highly conceivable that they might be learning group theory from a non-standard (or just plain bad) textbook and not know it. – Billy Aug 30 '11 at 19:31
If one interprets your question as "what topological/Lie group structures exist on subspaces of $\mathbb{R}^n$", it becomes very interesting! – Alexei Averchenko Aug 30 '11 at 20:34
@Billy- So, you think I.N.Herstein's algebra text is a bad one? – Primeczar Aug 31 '11 at 3:59

Here are some examples I've used (class examples, homework, tests, etc.):

Let $G = \{x \in \mathbb R: x \neq -1\}$ and define $*$ by $a*b=a+b+ab$.

Fix an integer $k$. Let $G = \mathbb Z$ and define $*$ by $a*b = a + b + k$.

Fix a real number $\beta > 0$. Let $G = \{x \in \mathbb R: x > 0 \}$ and define $*$ by $a*b = \beta ab$.

Let $G = \{x \in \mathbb R: -1 < x < 1 \}$ and define $*$ by $a*b = \frac{a+b}{1 + ab}$.

Let $G = \mathbb R$ and define $*$ by $a*b = \sqrt[3]{a^{3}+b^{3}}$.

More examples can be obtained by the following general methods:

Let $(G,\bullet)$ be a group and fix $g \in G$. Then $(G,*)$ is a group, where $*$ is the operation defined by $a*b = a \bullet g \bullet b$.

Let $(G,\bullet)$ be a group and $f:G \rightarrow G$ be a 1-1 and onto (i.e. bijective) function. Then $(G,*)$ is a group, where $*$ is the operation defined by $a*b = f^{-1}\left(f(a) \bullet f(b) \right)$.

If we go outside the realm of real numbers for the set $G$, then far more exotic examples are possible. For example, let $X$ be any set and let $P(X)$ be the set of subsets of $X$. Then $\left( P(X),\Delta \right)$ is a group, where $\Delta$ is the symmetric difference operation defined by $A \Delta B = (A-B) \cup (B-A)$. For another example, let

$$G = \left\{ x, \frac{x + 5}{3 - x}, \frac{x - 5}{x-1}, \frac{3x-5}{x+1} \right\},$$

whose elements are functions from ${\mathbb R} - \{-1,1,3\}$ into ${\mathbb R} - \{-1,1,3\}.$ Then $(G,*)$ is a group, where $*$ is the operation of composition of functions.

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Thanks Dave. I appreciate this contribution. – The Chaz 2.0 Sep 1 '11 at 21:23

Take a real number x not equal to 0 or 1. There is a group of order 6 generated by the operations which take $x\to \frac 1 x$ and $x \to 1-x$.

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I think it's important to note that the way you want to think about groups is not necessarily as number systems (a role better suited for rings), but as symmetries or in terms of actions.

To make this more mathematically precise, for any set $X$, you can form a group by looking at all one-to-one, onto maps from $X$ onto $X$, let's call this $\operatorname{Perm}(X)$. The binary operation is composition, the identity is the the map $x\mapsto x$, and the inverse is the compositional inverse. Lots of examples of groups can be found by considering subgroups of $\operatorname{Perm}(X)$, namely those that preserve additional "structure".

So for concrete examples, if $X=\{1,2,\dotsc,n\}$, then $\operatorname{Perm}(X)$ is the $n$-th symmetric group and is denoted $S_n$.

If $X=\mathbb{R}^n$, then we can look at the non-singular linear maps from $\mathbb{R}^n$ onto $\mathbb{R}^n$. This group is called $GL_n(\mathbb{R})$, and it's the subgroup of $\operatorname{Perm}(\mathbb{R}^n)$ that preserves the vector space structure of $\mathbb{R}^n$.

Neither of these (for $n>1$) are subgroups of $\mathbb{R}$ with addition or multiplication.

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Suppose the set has just one real number, it's not equal to 1 or 0, and closure holds for the binary operation *. In other words, (a*a)=a. Every such structure qualifies as a group, and they all come as isomorphic to each other.

Suppose the set has two distinct real numbers "a" and "b" and neither of them equals 1 or -1. Suppose the binary operation has this table:

   a  b
a  a  b
b  b  a

Every structure of this class qualifies as a group, and they all come as isomorphic.

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