Pretty much just the title. In a finite non-abelian group $(G,+)$, is it true that for any $a,b,c,d\in G\setminus\{e\}$, $a+b=c+d$ if and only if $b+a=d+c$? I have tried using proof by contradiction (i.e. assuming $a+b\not= c+d$ but $b+a=d+c$), but was unable to make any progress.
I have been asked to provide more context, so here goes. My professor assigned the following problem:
Let $G$ be a finite multiplicative group with identity 1 and let $D\subseteq G$. Prove that the following are equivalent: (1) For any $g\in G\setminus\{1\}$, there are exactly $\lambda$ distinct $(x,y)\in D\times D$ s.t. $g=xy^{-1}$. (2) For any $g\in G\setminus\{1\}$, there are exactly $\lambda$ distinct $(x,y)\in D\times D$ s.t. $g=x^{-1}y$.
There's no assigned book for the class, so I have no idea where he got the problem. We'd been studying difference sets, but until this problem, we'd only considered them in abelian groups. My thought was that if I can prove the statement I made in the title, I can prove that the statement is true for difference sets with $\lambda=1$, then use that to prove the statement for general $\lambda$.
No. For instance, taking $d$ to be the identity element and $c=a+b$, this would say that $a+b=c=b+a$ for all $a,b\in G$, so that $G$ is abelian.