# Example of a ring satisfying this variant definition of “symmetric” on nilpotent elements

I want an example to show that if $a,b$ are nilpotent elements of a ring $R$ with 1 and if $c$ is any element of $R$, then $abc=0\Rightarrow acb=0$ but $cab=0$ does not imply $acb=0$.

This is unlike symmetric ring, where we know that if $a,b,c\in R$ and $abc=0$ implies that $acb=0$.

Please help me to find a ring where to search for an example or help me to show that if $abc=0 \Rightarrow acb=0$, then $cab=0 \Rightarrow acb=0$ for all nilpotent elements $a,b$ in $R$ and for all $c\in R$.

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Hi: I changed your title because it suggested you were looking for a special symmetric ring, but (as you noted in the body) no symmetric ring would be an example. – rschwieb Sep 29 '13 at 12:35
Where does this question come from? It sounds a little like something homemade, but of course I could be wrong :) Please let us know. – rschwieb Sep 29 '13 at 13:11
@ rschwieb, yes you are right. This question is homemade. I tried to define such ring, but couldn't be able to show that this types of rings are not right-left symmetric, though I have strong intuition that this is so. – Anupam Sep 29 '13 at 14:42
If you do not know if your question has a definite answer, you should say so in your post. Otherwise someone might waste a lot of time on it and be annoyed when you tell them later. If you warn them though, they'll know what they're getting themselves into :) – rschwieb Sep 29 '13 at 22:44
Thanks...I am changing the question a bit – Anupam Sep 30 '13 at 1:30