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Assume you've got a set $S$ and an associative multiplication $S\times S\to S$. Let's further assume that for some given $a\in S$ there exists an element $b\in S$ such that $a^2b = aba = ba^2 = a$. Now define two elements $n=ab$ and $i=ab^2$ (note that $n$ resembles somewhat a neutral element for $a$, and $i$ resembles somewhat an inverse element, however they may be different for different elements, and $S$ might not have a neutral element at all). My question is: Do those elements $n$ and $i$ have established names? Also, does there exist a name for those elements $a$ for which such a $b$ (and thus an $n$ and an $i$) exists?

BTW, note that while there can be different elements $b$ with this property for a given $a$, $n$ and $i$ do not depend on the choice of $b$ (but in general will depend on the given $a$).

(I'm sorry for the bad title, but I can't think of a better one; feel free to improve it.)

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Well, a semigroup is regular if for every $a$ there exists a $b$ such that $aba=a$. The element $b$ is called a "pseudoinverse" of $a$. Your constructions seem related, so I would look in that context to see if they have a name. –  Arturo Magidin Jul 3 '12 at 17:17
@ArturoMagidin: Thanks, knowing a name for $b$ is also nice (so the above definition of $b$ could have been shortened to "$b$ is a pseudoinverse of $a$ which commutes with it"). Googling for it came up with the corresponding term "pseudo-invertible" for $a$ which answers my last question. –  celtschk Jul 3 '12 at 17:38
I've now also found in the Wikipedia article that $bab$ (which in my case is the same as $ab^2$) is simply called inverse of a. –  celtschk Jul 3 '12 at 17:51
That's not quite accurate. If $aba=a$ and $bab=b$, then $b$ is called an inverse of $a$. –  Arturo Magidin Jul 3 '12 at 17:51
@ArturoMagidin: And for $bab=ab^2=i$, we have $iai=(ab^2)a(ab^2)=(ab)^3b=b$, and therefore $ab^2=bab=i$ is an inverse ($b$ isn't, but I didn't claim that). Maybe I would have been less confusing if I had used the name $i$ in my previous comment instead of spelling it out. (Note that my conditions imply that $a$ and $b$ commute.) –  celtschk Jul 3 '12 at 19:28

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