# Are there proper $1/2$ convex subsets of finite groups?

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,

1. Can there be any $1/2$ convex proper subsets of finite groups? if not, why?
2. What are the other notions of convexity on finite groups ?
• For 1., any cyclic group of odd order – user175968 Sep 23 '15 at 6:03
• A generalization of the definition that suggests itself would be: Any product of an $n$th power and an $m$th power is an $(n+m)$th power ... Specifically, such a group would be divisible ($1^{n-1}a=x^n$) – Hagen von Eitzen Sep 23 '15 at 6:34
• @frank000: Thank you for your response. But, I'm so sorry. I have to edit my question. What do you think about edited question. And I really appreciate if you can give a full answer. – Bumblebee Sep 25 '15 at 7:10
• If $K\le G$ is a subgroup of odd order, it seems to me that any coset of $K$ is $1/2$-convex. Just to show that a $1/2$-convex subset need not be a subgroup. Well, hardly suprising that a translation would preserve a notion of convexity :-/ – Jyrki Lahtonen Sep 28 '15 at 6:07
• @JyrkiLahtonen: Thank you for your valuable comment. But considering cosets of $A_3$ in $S_3$ I see that not every coset of a $1/2$ convex set is not convex. – Bumblebee Sep 30 '15 at 4:20

## 1 Answer

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 $m,n$.