1

Let $S$ be a subset of a finite abelian group of even order $G$ such that $x+y\not\in S$ for all $x,y\in S$, and the identity element $0\not\in S$. Define $-S=\{-x|~x\in S\}$ and $S-S=S+(-S)=\{x-y|~x,y\in S\}$. Suppose $S \cap (-S)=\varnothing$. Prove or disprove that $S-S$ cannot contain an involution in $G$.

I tried proving it. Here is my attempt. Suppose $x-y$ is an involution in $S-S$ for $x,y\in S$, with $x\neq y$. Then $2x=2y$. How do I obtain a contradiction? [I understand that $2x=2y$ does not necessarily imply that $x=y$; for instance, in $\mathbb{Z}_4=\{0,1,2,3\}$, we have that $2(1)=2\equiv 2(3)\mod 4$; but $1\neq 3$.]

Daniel Fischer
  • 206,697
LTX
  • 41
  • Your idea is right: as long as $S$ is nonempty, $S-S$ always contains an involution, namely the identity $0$ (or $1$, if you prefer, although this is not standard for additive notation). – Luiz Cordeiro Dec 19 '16 at 21:03
  • As far as I know, in group theory an involution is an element of order two. Zero has order one. – DonAntonio Dec 19 '16 at 21:04
  • @Luiz: Except that in a group-theoretic context involution generally means element of order $2$, in which case $S={1}$ and $G=\Bbb Z/4\Bbb Z$ is an example in which $S-S$ contains no involution. – Brian M. Scott Dec 19 '16 at 21:05
  • @DonAntonio, @BrianMScott Oh, right. Anyway we can work his example out just to fix the ``problem'' that $-1=3$ in $\mathbb{Z}_4$. See my answer below. – Luiz Cordeiro Dec 19 '16 at 21:11
  • @ Luiz, it is helpful. Thanks. – LTX Dec 19 '16 at 21:13
  • 2
    Don't delete your question just after you received an answer. That is rude. – Daniel Fischer Dec 19 '16 at 21:27

1 Answers1

1

There can be involutions. Take $G=\mathbb{Z}_4^2$ and $S=\left\{(1,0),(3,2)\right\}$. Then $(1,0)-(3,2)=(2,2)$ is an involution.

Luiz Cordeiro
  • 18,513