# $G$ 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.

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.

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.