totally ordered abelian groups with many convex subgroups I am wondering if the following can be proved in ZFC: 
For every infinite cardinal $\kappa$, there exists a totally ordered abelian group G of cardinality $\kappa$ with at least $\kappa^+$ many convex subgroups.
I'm inclined to think the answer is "yes" since it can be shown in ZFC that for infinite $\kappa$, there is a chain of subsets of $\kappa$ of size $\kappa^+$. References would be appreciated as well as this question lies outside my research area. Thanks!
 A: The answer is yes. To see this, for any given infinite cardinal $\kappa$, let $\lambda$ denote the least cardinal such that $2^{\lambda} > \kappa$.
Let $2^{<\lambda}$ denote the set of maps $\alpha \rightarrow \{-1;1\}$, $\alpha < \lambda$, ordered lexicographically taking $-1$ to be strictly inferior to nothing, which is strictly inferior to $1$. It is a total order, and incidently it can be endowed with an interesting ordered field structure, seeing it as a subfield of $No$, the Field of surreal numbers.
Its Dedekind completion is $2^{<\lambda+1}$, which contains the set $\{-1;1\}^{\lambda}$, and has therefore a cardinal $\geq \kappa^+$. $2^{<\lambda}$ however, is the union $\bigcup \limits_{\alpha < \lambda} \{-1;1\}^{\alpha}$, and has cardinal $\sup \limits_{\alpha < \lambda} 2^{|\alpha|} = \kappa$ by minimality of $\lambda$ (if this supremum was $< \kappa$, then $2^{\lambda}$ would be $\leq \kappa$).
Now consider the group $G:=\mathbb{Z}^{(2^{<\lambda})}$ of maps $f: 2^{<\lambda} \rightarrow \mathbb{Z}$ whose support $s(f):=f^{-1}(\mathbb{Z} - \{0\})$ is finite. For $0 \neq f \in \mathbb{Z}^{(2^{<\lambda})}$, define $d^{\circ} f := \max(s(f))$, and for $f\neq g \in \mathbb{Z}^{(2^{<\lambda})}$ say that $f \prec g$ when $0 < (g-f)(d^{\circ}(g-f))$.
$(G,\prec)$ is an ordered group of cardinal $\max(|\mathbb{Z}|,|2^{<\lambda}|) = \kappa$.
For any element $x \in 2^{<\lambda+1}$, consider the subgroup $G_x:= \mathbb{Z}^{(]-\infty;x[_{2^{<\lambda}})}$. $G_x$ is a convex subgroup of $G$. Indeed, for $0 \prec f \in G_x$, and $0 \prec g \prec f$, $s(g)$ cannot contain an element $y \in 2^{<\lambda}$ strictly superior to $x$, otherwise $d^{\circ} g$ would be out of $s(f)$ and so $(f-g)(d^{\circ}(f-g))$ would be $-g(d^{\circ}g)$, which is strictly negative because $g$ is strictly positive: this contradicts $g \prec f$. So $g \in \mathbb{Z}^{(]-\infty;x[_{2^{<\lambda}})}$ . 
This and the stability of $\mathbb{Z}^{(]-\infty;x[_{2^{<\lambda}})}$ by taking opposites suffices to prove that it is convex.
By density of $2^{<\lambda}$ in its Dedekind completion, for any two $x < z \in 2^{<\lambda + 1}$, there is $y \in 2^{<\lambda}$ such that $x < y < z$, and so $\chi_{\{y\}} \in G_z - G_x$, so $2^{<\lambda+1} \ni s \longmapsto G_s$ is injective in the set of convex subgroups of $G$.
Therefore, $G$ is an ordered group of cardinal $\kappa$ with at least $\kappa^+$ convex subgroups.
