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Partially ordered group $G$ is a lattice ordered if and only if for every $a\in G$ there is a least upper bound $a\vee e$ in $G$. [L.Fuchs, Partially Ordered Algebraic Systems, 1963].

Addendum: Thanking to the comments of Mark Sapir and Tobias Schlemmer one can prove:

Proposition. Let $S(\cdot,\leq)$ be a group with a lattice order and the condition: $x\leq 1\implies xz\leq z, \ zx\leq z$, and similarly for $x\ge 1$.

1. If $S(\cdot,\leq)$ is a partially ordered group, then it is a lattice ordered group.

2. If $S(\cdot,\leq)$ is not a partially ordered group, then there is such a linear (hence lattice) exstension $\preceq$ of $\leq$ that $(x\vee y)z\ne xz\vee yz$ for some $x,y,z$ (and thus your condition is not sufficient).

Proof. 1) Since there is $a\vee e$, $S$ is lattice ordered (see above).

2. If $S(\cdot,\leq)$ is not a partially ordered, then $x< y$, but $xz\not< yz$ or $zx\not< zy$ for some $x,y,z$. Let, for example, $xz\not< yz$. By Szpilrajn theorem we can extend $\leq$ up to linear $\preceq$ such that $yz\prec xz$. Then $(x\vee y)z=yz\ne xz=xz\vee yz$.

Partially ordered group $G$ is a lattice ordered if and only if for every $a\in G$ there is a least upper bound $a\vee e$ in $G$. [L.Fuchs, Partially Ordered Algebraic Systems, 1963].

Addendum: Thanking to the comments of Mark Sapir and Tobias Schlemmer one can prove:

Proposition. Let $S(\cdot,\leq)$ be a group with a lattice order and the condition: $x\leq 1\implies xz\leq z, \ zx\leq z$, and similarly for $x\ge 1$.

1. If $S(\cdot,\leq)$ is a partially ordered group, then it is a lattice ordered group.

2. If $S(\cdot,\leq)$ is not a partially ordered group, then there is such a linear (hence lattice) exstension $\preceq$ of $\leq$ that $(x\vee y)z\ne xz\vee yz$ for some $x,y,z$.

Proof. 1) Since there is $a\vee e$, $S$ is lattice ordered (see above).

2. If $S(\cdot,\leq)$ is not a partially ordered, then $x< y$, but $xz\not< yz$ or $zx\not< zy$ for some $x,y,z$. Let, for example, $xz\not< yz$. By Szpilrajn theorem we can extend $\leq$ up to linear $\preceq$ such that $yz\prec xz$. Then $(x\vee y)z=yz\ne xz=xz\vee yz$.

Partially ordered group $G$ is a lattice ordered if and only if for every $a\in G$ there is a least upper bound $a\vee e$ in $G$. [L.Fuchs, Partially Ordered Algebraic Systems, 1963].

Addendum: Thanking to the comment of Mark Sapir one can prove:

Proposition. Let $S(\cdot,\leq)$ be a group with a lattice order and the condition: $x\leq 1\implies xz\leq z$ and similarly for $x\ge 1$.

1. If $S(\cdot,\leq)$ is a partially ordered group, then it is a lattice ordered group.

2. If $S(\cdot,\leq)$ is not a partially ordered group, then there is such a linear (hence lattice) exstension $\preceq$ of $\leq$ that $(x\vee y)z=xz\vee yz$ for some $x,y,z$ (and thus your condition is not sufficient).

Proof. 1) Since there is $a\vee e$, $S$ is lattice ordered (see above).

2. If $S(\cdot,\leq)$ is not a partially ordered, then $x< y$, but $xz\not< yz$ for some $x,y,z$. By Szpilrajn theorem we can extend $\leq$ up to linear $\preceq$ such that $yz\prec xz$. Then $(x\vee y)z=yz\ne xz=xz\vee yz$.

Partially ordered group $G$ is a lattice ordered if and only if for every $a\in G$ there is a least upper bound $a\vee e$ in $G$. [L.Fuchs, Partially Ordered Algebraic Systems, 1963].

Addendum: Thanking to the comments of Mark Sapir and Tobias Schlemmer one can prove:

Proposition. Let $S(\cdot,\leq)$ be a group with a lattice order and the condition: $x\leq 1\implies xz\leq z, \ zx\leq z$, and similarly for $x\ge 1$.

1. If $S(\cdot,\leq)$ is a partially ordered group, then it is a lattice ordered group.

2. If $S(\cdot,\leq)$ is not a partially ordered group, then there is such a linear (hence lattice) exstension $\preceq$ of $\leq$ that $(x\vee y)z\ne xz\vee yz$ for some $x,y,z$ (and thus your condition is not sufficient).

Proof. 1) Since there is $a\vee e$, $S$ is lattice ordered (see above).

2. If $S(\cdot,\leq)$ is not a partially ordered, then $x< y$, but $xz\not< yz$ or $zx\not< zy$ for some $x,y,z$. Let, for example, $xz\not< yz$. By Szpilrajn theorem we can extend $\leq$ up to linear $\preceq$ such that $yz\prec xz$. Then $(x\vee y)z=yz\ne xz=xz\vee yz$.

Partially ordered group $G$ is a lattice ordered if and only if for every $a\in G$ there is a least upper bound $a\vee e$ in $G$. [L.Fuchs, Partially Ordered Algebraic Systems, 1963]

Partially ordered group $G$ is a lattice ordered if and only if for every $a\in G$ there is a least upper bound $a\vee e$ in $G$. [L.Fuchs, Partially Ordered Algebraic Systems, 1963].

Addendum: Thanking to the comment of Mark Sapir one can prove:

Proposition. Let $S(\cdot,\leq)$ be a group with a lattice order and the condition: $x\leq 1\implies xz\leq z$ and similarly for $x\ge 1$.

1. If $S(\cdot,\leq)$ is a partially ordered group, then it is a lattice ordered group.

2. If $S(\cdot,\leq)$ is not a partially ordered group, then there is such a linear (hence lattice) exstension $\preceq$ of $\leq$ that $(x\vee y)z=xz\vee yz$ for some $x,y,z$ (and thus your condition is not sufficient).

Proof. 1) Since there is $a\vee e$, $S$ is lattice ordered (see above).

2. If $S(\cdot,\leq)$ is not a partially ordered, then $x< y$, but $xz\not< yz$ for some $x,y,z$. By Szpilrajn theorem we can extend $\leq$ up to linear $\preceq$ such that $yz\prec xz$. Then $(x\vee y)z=yz\ne xz=xz\vee yz$.