TITLE: A binary operation on $G$ satisfies $a\cdot(b\cdot c)=(a\cdot b)\cdot (a\cdot c)$ and has an identity, is it a group?
QUESTION [6 upvotes]: A binary operation on $G$ satisfies $a\cdot(b\cdot c)=(a\cdot b)\cdot (a\cdot c)$ and has an identity, is it associative? Is it a group?
I don't think it is associative but can't produce a convincing argument.
Any help would be appreciated!
Thanks,

REPLY [11 votes]: Let $e$ denote the identity element.  Note that $$xy=x(ey)=(xe)(xy)=x(xy)$$ and $$xy=x(ye)=(xy)(xe)=(xy)x$$ for any $x,y\in G$.  Now we have $$a(bc)=(ab)(ac)=((ab)a)((ab)c)=(ab)((ab)c)=(ab)c,$$ proving associativity (the first two equalities are the given axiom, and the second two come from $(xy)x=xy$ and $x(xy)=xy$ for appropriate choices of $x$ and $y$).
However, $G$ does not have to be a group.  For instance, $G$ could be any totally ordered set with a greatest element, with the operation given by $ab=\min(a,b)$; this is not a group unless $G$ has only one element, since there are no inverses.  Or more generally, $G$ could be any partially ordered set in which any finite subset has a greatest lower bound, with $ab$ being the greatest lower bound of $a$ and $b$.  In fact, every commutative example is of this form (explicitly, you can define $a\leq b$ iff $ab=a$ and then $ab$ will be the greatest lower bound of $a$ and $b$ with respect to this order).
For an example that is not commutative, consider $G=\{e,a,b\}$ with the operation given by $xy=x$ unless $x=e$ in which case $xy=y$.