On the theory of $2$-ordered groups

1. On the order on a straight line $l_{e, \alpha}$.
Let $\langle G, \cdot, \zeta\rangle$ is a non-degenerate $2$-ordered group, $\alpha\in G$, $o(\alpha)=2$, $l_{e, \alpha}=\{x\in G\mid \zeta(\alpha, e, x)=0\}$.
It is known that $l_{e,\alpha} \triangleleft G$. As $l_{e,\alpha}\ne G$, then $\exists c\in G(\zeta(c,\alpha,e)\ne0)$. Let $\zeta(c, \alpha, e)=1$.
Let: $x<y \Leftrightarrow \zeta_c(x,y)=\zeta(c, x, y)=1$.
It is known that the function $\zeta_c$ sets linear order on the line $l_{e,\alpha}$. Let us note that $\alpha<e$ regarding this order. As $\alpha\in l_{e,\alpha}$ then the group $\langle l_{e,\alpha},\cdot\rangle$ cannot be linearly ordered. Let us find a subgroup which is linearly ordered regarding to the specified order $\zeta_c$.
Theorem 1.1. Let $P=\{x\in l_{e,\alpha}\mid x\geqslant e\}$, $H=P\cup P^{-1}$. If $|P|\ne1$, then $\langle H, \cdot, \zeta_c\rangle$ is a linearly ordered group.
2. On the cardinality of the set of elements of order $n$ in $2$-ordered group
Let $n\in\mathbf{N}$ and $H=\{x\in G \mid x^n=e\}$. As $T(G) \subset Z(G)$, then $H < G$ and $H$ is an Abelian group. Consequently, $\langle H, \cdot, \zeta\rangle$ is a locally finite $2$-ordered group. Let $\zeta\not\equiv0$ on the set $H$.
Theorem 2.1. Let $\langle G, \cdot, \zeta\rangle$ be a non-degenerate $2$-ordered group, $n\in\mathbf{N}$ and $H=\{x\in G\mid x^n=e\}$. If $\zeta\not\equiv0$ on the set $H$, then $|H| \leqslant n$.