TITLE: Are there proper $1/2$ convex subsets of finite groups?
QUESTION [5 upvotes]: According to this paper by N. H. Bingham and A. J. Ostaszewski,

A subset $H$ of a group $G$ is said to be $1/2$ convex (or mid-point convex)
  if for each $x,y\in H$, there exist a unique element $z\in H$ satisfying the equation $z^2=xy$.

My questions are, 

Can there be any $1/2$ convex proper subsets of finite groups? if not, why?   
What are the other notions of convexity on finite groups ?

REPLY [4 votes]: As frank000 points out, any cyclic group of odd order is $1/2$-convex. So if you have a non-cyclic $1/2$-convex group and take an element of odd order, the subgroup generated will be a proper $1/2$-convex subgroup. For an example, let $n$ be an odd positive integer and consider $\mathbb G=Z_n^2$. This is a $1/2$-convex group, and the subgroup $H$ generated by $(1,0)$ is also $1/2$-convex.
As Hagen von Eitzen suggested in the comments, a natural generalization would be: given positive integers $m$ and $n$, a subgroup $H$ of $G$ is $(m,n)$-convex if for all $x,y\in H$, there exists a unique $z\in H$ such that $x^my^n=z^{m+n}$ (so in this new terminology, $1/2$-convexity is the same as $(1,1)$-convexity). For example, $\mathbb Z_k$ is $(m,n)$-convex if and only if $k$ and $m+n$ are coprime. In particular, this shows there is no finite cyclic group $G$ which is $(m,n)$-coprime for all $m,n$.