# identity element for partial orders

I am given (possibly total) order $\phi$ (represented as a directed acyclic graph), how can I construct an identity element $E_I$ for the order (i.e. $\phi$ $\otimes$ $E_I$=$\phi$)? let's assume $\otimes$ to be the product order ($(a_1,b_1)\succ (a_2,b_2)$ iff $a_1\succ a_2$ and $b_1\succ b_2$ ).

• One of them is an ordering of one set and the other is an ordering of a different set.... They can't be equal. What do you actually mean? – dfeuer Jun 21 '13 at 5:21
• I maybe mistaken but I just have an order and want to construct an identity element. – seteropere Jun 21 '13 at 5:38
• You're going to have to explain what you mean by that, and you're going to have to explain it better than you already did. I'm not trying to be harsh, but an ordering of elements can't possibly equal an ordering of ordered pairs of those elements, although it can sometimes be isomorphic to it. – dfeuer Jun 21 '13 at 5:39

We take all partially ordered sets and mod out by the relation $\sim$ that is given by $A\sim B$ whenever $A$ and $B$ are order-isomorphic. Next we want to consider this as an algebraic structure and see that it forms a monoid. First of all, the definition of product you mention neatly descends to one on isomorphism classes, so there's no problem there. Also, associativity doesn't seem to give much of a problem.
As to the identity element: notice that all ordered singleton sets are order-isomorphic, the only possible order being $a\succ a$ if the set is $\{a\}$. Therefore, there is only one isomorphism class corresponding to these sets. If we take this class as $E_I$, it is the identity element. Why? Just choose a representative singleton set and see what happens. In the product order you mention, the condition $b_1\succ b_2$ is automatically fulfilled, since $b_1=b_2$ and you get $(a_1,b_1)\succ(a_2,b_2)\iff a_1\succ a_2$. The product therefore is order-isomorphic to your original order.