Metric on a group

Question

Is there a non-abelian finite group $G$ with the property: If metric $d$ on $G$ is left invariant then is also right invariant?

Answer 1

It seems that the answer is negative and holds the following.

Definition 1. A group $G$ is uniform if each left invariant metric on $G$ is right invariant.

Proposition 1. A group $G$ is uniform iff $y^{-1}xy\in\{x,x^{-1}\}$ for each $x,y\in G$.

Proof. Sufficiency. Let $d$ be a left invariant metric on $G$. If $x\in G$ then $d(e,x^{-1})=d(x,e)=d(e,x)$. Now let $y$, $z$, and $t$ be arbitrary elements of $G$. Then $d(zy,ty)=d(e,y^{-1}z^{-1}ty)=d(e,z^{-1}t)=$ $d(z,t)$. Hence the metric $d$ is right invariant.

Necessity. Let $G$ be a uniform group. Suppose that there exist elements $x,y\in G$ such that $y^{-1}xy\not\in\{x,x^{-1}\}$. Define a function $f:G\to\mathbb R$ by putting $f(e)=0$, $f(x)=f(x^{-1})=2$ and $f(z)=1$ for all $z\in G\backslash \{e,x,x^{-1}\}$. Then a function $d:G\times G\to\mathbb R$ such that $d(z,t)=f(z^{-1}t)$ is a left invariant metric on $G$. Suppose that $d$ is right invaraint. Then $2=f(x)=d(e,x)=d(y,xy)=$ $d(e, y^{-1}xy)=f(y^{-1}xy)=1$, a contradiction.$\square$

Now let $G$ be a uniform group. Suppose that $x,y\in G$ and $xy\not=yx$. Then $y^{-1}xy=x^{-1}$ and $x^{-1}yx=y^{-1}$. Then $xyx=y$ and $yxy=x$. Hence $x=y(xy)=y(yx^{-1})$. Therefore $x^2=y^2$.

Let $Z$ be the center of the group $G$ and $x$ be an arbitrary element of $G$. Since for each $y\in G$ we have $xy=yx$ or $x^2=y^2$, we see that $x^2y=yx^2$. Thus $x^2\in Z$.

Suppose that $x,y\in G,$ $xy\not=yx$ and $z\in Z$. Then $(xz)y\not=y(xz)$ and as above we can show that $(xz)y(xz)=y$. But we also have that $xyx=y$. Thus $z^2=e$ for each $z\in Z$. Hence $x^4=e$ for each $x\in G$.

If $G$ is finite, then both $Z$ and $G/Z$ are finite abelian groups of an exponent $2$.

Lemma 1. [Serge Lang, “Algebra”, I.6]. Let $H$ be a finite abelian group, $|H|=m$, and $p$ be a prime number such that $p|m$. Then $H$ contains a subgroup $H’$ such that $|H’|=p$.

Lemma 1 implies that $|G|$ is a power of 2 for each finite uniform group $G$

Example 1. It seems that the group of quarternions $\mathbb Q_8$ satisfies the conditions of Proposition 1 (I have not checked it very accurately). Therefore $\mathbb Q_8$ should be a uniform non-abelian finite group.