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Let V be a finitely generated vector space with dimension(V) = $n \in \mathbb{N}>1$. Now let T: $ V \to V$ be a map such that $\forall \hat{v},\hat{w} \in V$, $\; T(\hat{v}+\hat{w}) \neq T(\hat{v})+T(\hat{w})$.So if $\chi(V,V)$ be the collection of mappings with the property that T has, then what I'm looking for is a subset $G \subset \chi(V,V) $ such that there is a function $\lambda: G \times G \to G$ and a function ${i}: G \to G $ with the following (4) properties:

  1. $\forall A,B \in G,\; \lambda(A,B) \in G \;$ (closure)

  2. $\exists E \in G$ such that $\forall A \in G,\; \lambda(A,E)=\lambda(E,A)=A \;$ (identity)

  3. $\forall A \in G,\; \exists A^{-1} \in G \;$ such that $\lambda(A, A^{-1})=\lambda( A^{-1},A)=E \; $ (inverse)

  4. $\forall A,B,C \in G, \; \lambda(A,\lambda(B,C))= \lambda(\lambda(A,B),C) \;$ (associativity)

And to clarify, ${i}(A) = A^{-1} \; (\forall A \in G)$. Can someone show an example(or two)of such a collection of operators(and a binary function) that satisfy these properties?

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Since you don't put any constraints on the way that $\lambda$ and $i$ interact with the operation on $V$, you can just choose any bijection between any $G\subset\chi(V,V)$ and any group. –  HJRW Jun 4 '13 at 5:40
    
Originally posted on math.SE: math.stackexchange.com/q/402350/264 –  Zev Chonoles Jun 4 '13 at 13:43
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1 Answer

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Fix $a\in V, a\ne 0$ and for $L\in {\rm Hom}(V,V)$ denote $L_a:x\to L(x)+a$. Let $G$ to be the set of all $L_a (L\in {\rm Hom}(V,V)$ and $$\lambda (L_a,M_a)(x) =L_a(x)+M_a(x)-a=(L+M)(x)+a.$$

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Just for clarification: Is Hom(V,V) the set of linear maps from V to V? –  Yog_Shohoth Jun 4 '13 at 20:31
    
An example of a nonlinear transformation on a finitely generated vector space that does not have the group property under addition or multiplication is the mapping $T_{\alpha}(\hat{v}) = [x^{2}y-\alpha^{2}, 2\alpha xy] \;$ where $\alpha$ is a nonzero scalar and (x,y) are the respective components of a vector in a 2-dimensional vector space. –  Yog_Shohoth Jun 4 '13 at 20:43
    
@Yog_Shohoth: Yes, of course. –  Boris Novikov Jun 4 '13 at 21:21
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