$G$ is an abelian group, and a permutation $T$. Suppose that $x-T(x)\neq y-T(y)$ for all $x\neq y$. Show that $p(x)=x-T(x)$ is also a permutation Let $G=(A,+)$ be an abelian group, T a permutation on $A$. Assume that $x - T(x) \neq y - T(y)$ for all distinct $x,y \in A$. Show that $p(x)=x - T(x)$ is a permutation on $A$.
For that I would show that $p: A\rightarrow A$ and $p$ bijective. The first property and injectivity are obvious. It remains to show that 
$\forall x \in A \ \exists z \in A : p(z)=x$.
I have no Idea how I should do that.
 A: The statement is false, here's a counterexample:
Let $G = (\Bbb Z,+)$, which is an abelian group. 
Define $T(x) := -x$, which is indeed a permutation on $\Bbb Z$. For any $x,y \in \Bbb Z$ with $x \neq y$, we have $$x-T(x) = 2x \neq 2y = y - T(y).$$ But $p(x) = x - T(x) = 2x$ is not a permutation on $\Bbb Z$ since $1$ is not in its image.
A: The only idea that I have is the following:
If G is a group that verifies the nullity+rank theorem, i.e. for all morphism $\psi: G\to G$ then $G$ can be written as $G\cong ker(\psi)\oplus im(\psi)$, then each injective morphism is surjective. 
In general your question is really complicated. For example in a Banach space $X$  if you have a morphism $T$ such that $||T||< 1$ then you have that $I-T$ is invertibile and its  inverse is 
$\sum_{k=0}^\infty T^n $ 
by Von Neumann lemma. So in general it is really difficult to find the inverse of $I-T$. 
Another problem is that there e exists some limited operator $T$ such that $1$ is in the residue specter of $T$ that means that $I-T$ could be injective but not surjective. So your assertion tell us that if $T$ is bijective than $1$ is not in the residue specter of $T$ but I don’t know if it is true. 
