I wanted to prove that the theory of ordered abelian group has a strict order property. I know by the theory of Kikyo and Shellah we have:
a theory is unstable iff it has SOP or NIP
and by the : https://www.jstor.org/stable/1999281?seq=1
the ordered abelian group does not have IP but is there any way to prove SOP directly?
On
If you know about imaginary elements, it is not very difficult to see that the strict order property of $T$ is exactly equivalent to the existence of a $T^{\mathrm{eq}}$-definable partial order with an infinite chain (in a saturated model; in an arbitrary model, there may not be an infinite chain, but there are arbitrarily long finite chains).
In particular, if $T$ defines a partial order with an infinite chain, it has SOP, and in particular, if it defines an infinite linear order with an infinite chain, then it has the SOP.
All this follows from considering the formulas $x\leq p$, where $p$ is an element of the infinite chain.
The ordered abelian groups have an infinite linear order built in, so they certainly do have the SOP.
It is immediate from the definition that the formula $x\leq y$ witnesses SOP in the theory of ordered abelian groups. For example, in $\mathbb{Z}$, look at the family of sets defined by $x\leq n$ for various $n$. These sets form an infinite chain of proper containments.
In fact, the definition of SOP is a direct generalization of the property of having a definable linear order (which is infinite in some model). More precisely, a theory has SOP iff it has a definable preorder with infinite strict chains (in some model).
Also, a correction. You wrote "I know by the theory of Kikyo and Shellah we have:"
This is a theorem of Shelah. It appeared in Classification Theory in 1978. There is also a well-known theorem of Kikyo and Shelah (which appeared in their only joint paper, in 2002), but that theorem says that if $T$ is an SOP theory, then the theory of models of $T$ expanded by an automorphism has no model companion.