Middle Cancellation in Groups

Question

For a, b, c, d, x elements of a group G. If ab = cd does that mean that axb = cxd? What if ab = cd only in this one instance, does the equality still hold?

In any case, the answer is no in general. Let $b = a^{-1}$ and $d = c^{-1}$. Then $ab = cd = 1$, but $axb$ is conjugating $x$ by $a$, and $cxd$ is conjugating $x$ by $c$; no reason these should be equal. — , Oct 29 '17 at 06:48

score 10 · Answer 1 · answered Oct 29 '17 at 07:02

10

This happens if and only if the group is abelian:

$y \cdot y^{-1} = 1\cdot 1 \implies yxy^{-1} = x \implies yx = xy$

answered Oct 29 '17 at 07:02

Mike

1,312

So would $bxb = cxc^2$ imply that $b^2 = c^3$? – Tom Jones Oct 29 '17 at 15:47
1

@TomJones For abelian groups, Yes. – M. Winter Oct 29 '17 at 18:04

score 0 · Answer 2 · answered Oct 29 '17 at 21:38

Suppose they were equal. Let $y=axb=cxd$. Then $$axa^{-1}=yb^{-1}a^{-1}=y(ab)^{-1}=y(cd)^{-1}=yd^{-1}c^{-1}=cxc^{-1}.$$ Therefore \begin{align*} xa^{-1}c&=a^{-1}cx& \end{align*} so $a^{-1}c$ is in the centre $Z(G)$. Similarly, $bd^{-1}$ is in $Z(G)$.

So it's not necessarily true that $G$ is abelian (i.e. that $Z(G)=G$), but does say something about what is in $Z(G)$.

Middle Cancellation in Groups

2 Answers2