# lattice-ordered group

I am just starting to read the lattice-ordered group. I was trying to prove one of the elementary properties. Give an $\ell$-group $G$, for all $a, b$ and $c$ in $G$, $a + \textit{sup} (b, c) = \textit{sup} ((a + b),(a + c))$. I have done the following.

$b \leq \textit{sup} (b, c)$ and $c \leq \textit{sup} (b, c)$. Then $a + b \leq \textit{sup} (b, c) + a$ and $a + c \leq \textit{sup} (b, c) + a$. So, $\textit{sup}((a + b), (a + c)) \leq a + \textit{sup} (b, c)$. How to show $a + \textit{sup} (b, c) \leq \textit{sup} ((a + b),(a + c))$ ?

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Hint: In $a+\sup(b,c) \leq \sup((a+b),(a+c))$ you can move the $a$ to the right side (because you are in a group). – boumol Jan 21 '13 at 20:13
A lattice-ordered group need not be Abelian, so from $b\le b\lor c$ you should get $a+b\le a+(b\lor c)$ rather than $a+b\le(b\lor c)+a$. If you intend the group to be Abelian (as suggested by your use of $+$), please say so explicitly. – Brian M. Scott Jan 21 '13 at 20:16
The group $G$ is abelian here. – Sharma Sharma Jan 21 '13 at 21:15

To avoid any implication that the group is Abelian, I will write it multiplicatively. In that notation, and keeping track of the order of elements in a product, you have $b,c\le b\lor c$, so $ab\le a(b\lor c)$ and $ac\le a(b\lor c)$, and hence $ab\lor ac\le a(b\lor c)$.

HINT: One way to get the opposite inequality is to show that if $ab\lor ac\le x$, then $a(b\lor c)\le x$. If $ab\lor ac\le x$, then $ab\le x$ so $b\le a^{-1}x$. Similarly, ... ? (The conclusion is spoiler-protected; mouse-over to see it if you get stuck.)

Similarly, $c\le a^{-1}x$, and therefore $b\lor c\le a^{-1}x$. Now multiply on the left by $a$.

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@ Brian, Thank you very much. – Sharma Sharma Jan 21 '13 at 21:28
@Sharma: You’re very welcome. – Brian M. Scott Jan 21 '13 at 21:29