TITLE: $G$ 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
QUESTION [3 upvotes]: 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.

REPLY [7 votes]: 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.
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