# Are these exactly the abelian groups?

I'm thinking about the following condition on a group $G$.

$$(\forall A\subseteq G)(\forall g\in G)(\exists h\in G)\ Ag=hA.$$

Obviously every abelian group $G$ satisfies this condition. Are there any other groups that do? Can we give a familiar characterization for them? Can we give one if we confine the considerations to finite groups?

Certainly not all groups satisfy the condition. Let $G$ be the free group on $\{x,y,z\}.$ Let $A=\{x,y\}$ and $g=z.$ Then $$Ag=\{x,y\}z=\{xz,yz\}.$$ Suppose there is $h\in G$ such that $\{hx,hy\}=hA=\{xz,yz\}.$ Then either $$\begin{cases}hx=xz\\hy=yz\end{cases}$$

or $$\begin{cases}hx=yz\\hy=xz\end{cases}$$

From the first case we get $h=xzx^{-1}$ and $h=yzy^{-1}$, which is a contradiction. From the second case we get $h=yzx^{-1}$ and $h=xzy^{-1}$, which is also a contradiction.

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Isn't this same argument works generally: let $A:=\{x,y\}$ and $g$ and $h$ varies, so that we'll get $x$ and $y$ must commute? – Berci Feb 14 '13 at 17:52
@Berci But aren't I using the freeness of $G$ here? I need to know that $xzx^{-1}\neq yzy^{-1}$ and $yzx^{-1}\neq xzy^{-1}$ to have a contradiction... – Bartek Feb 14 '13 at 17:55
aha, ok, yes. Nice question anyway. – Berci Feb 14 '13 at 17:58

Suppose $G$ satisfies your condition. Let $x,y \in G$ be distinct elements of $G$ different from $1$ so that the listed elements of all sets in this answer are distinct. Pick $g=x$ and $A=\{1,x,y\}$ Then $Ag = hA$ implies

$$\{ x, xx, yx \} = \{ h, hx, hy \}$$

Case 1: $x=h$

Then $\{ xx,yx \} = \{ xx,xy \}$, and $yx = xy$

Case 2: $x = hx$

Then $h=1$ and $\{ xx, yx \} = \{ 1, y \}$.

If $yx = y$, then $x=1$ and $xy=yx$. Otherwise $yx=1$, and so $y=x'$ and $xy=1=yx$.

Case 3: $x=hy$

Then $h=xy'$ and $\{ xx, yx \} = \{xy',xy'x\}$

If $xx = xy'$, then $x=y'$ and thus $xy=1=yx$. Otherwise $xx = xy'x$ and thus $1=y'$, and again $xy=yx$.

In all cases where $x,y,1$ are distinct, we've shown $x$ and $y$ commute. Thus $G$ is an abelian group.

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Fantastic! Thank you very much! – Bartek Feb 14 '13 at 18:36
The proof implicitly uses: distinct elements stay so under scalings. Thus since elements of the bag/multiset $A$ are distinct, so too for the bags $\,Ag\,$ and $\,hA.$ Thus every "set" notated in the proof has distinct elements, which enables the decomposition processes used, $$\begin{eqnarray}\text{e.g the proof implicitly uses}\quad \{a,b,c\} = \{a,d,e\}\:&\Rightarrow&\{b,c\} = \{d,e\}\\ \text{which is false if nondistinct}\quad \{a,b,b\} = \{a,b,a\}\:&\not\Rightarrow&\{b,b\} = \{b,a\}\end{eqnarray}$$ Perhaps it would be helpful to the reader to mention this subtle yet crucial point. – Math Gems Feb 14 '13 at 20:49

Assume $ab\ne ba$. Let $A=\{1,a\}$, $g=b$. Then there is $h\in G$ such that $\{h,ha\}=\{b,ab\}$. This needs $h=b\lor h=ab$. In the first case $ha=ba\ne ab$, so this fails. Therefore $h=ab$ and $ha=aba=b$. Similarly, $bab=a$. This implies $aa=abab=bb$. We conclude $$a=bab=bbabb=aaaaa,$$ hence $a^4=1$ and similarly $b^4=1$.

Now take $A=\{1,a,b\}$ and $g=b$. Then there is $h\in G$ such that $\{h,ha,hb\}=\{b,ab,b^2\}$.

• $h=b$: Then $hb=b^2$ implies $ba=ha=ab$, contradiction
• $h=ab$: Then $ha=aba=b$ implies $hb=b^2$, i.e. $a=1$ and of course $ab=ba$, contradiciton.
• $h=b^2=a^2$: Then $ha=a^3=a^{-1}\ne b$, hence $ha=ab$, i.e. $a^2=b$ and of course $ab=ba$, contradiction
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