0

Assuming $(G,*)$ is an non-abelian group and $a,b\in G$.

I have these two equations: $a*x=b$ and $y*a=b$.

First I had to prove, that both of the equations are uniquely solvable ($x_1=x_2$ and $y_1=y_2$), which I did. But now I need to find an example, where $x\neq y$.

I thought, maybe I can just write it like this: $x=a^{-1}*b$ and $y=b*a^{-1}$, since $*$ is not commutative, $x\neq y$.

Would that be a proper example?

Arthur
  • 1,557
  • I am really not sure what you are asking and just because a group is non-abelian that doesn't mean $xy$ is never $yx$. It is just the case that this does not always occur. – Mr.Fry Feb 21 '15 at 07:41
  • What if I assumed that neither $b,x$ or $y$ are the neutral element of the group? – Arthur Feb 21 '15 at 07:45
  • @Arthur Non-abelian means that there is some $x,y$ in the group such that $xy\ne yx$. You don't get to pick $x,y$, and it is not necessarily true of all nonidentity elements. Hint: try using the $x,y$ that come out of this to construct your example. – Mario Carneiro Feb 21 '15 at 08:27
  • A general comment here for people doing homework: If you're asked for an example, it means to give an example rather than a hypothetical. In this case, that means picking a specific group and specific elements of that group. – John Brevik Feb 21 '15 at 17:52

1 Answers1

2

A simple example in $S_3$. We'll use the cycle notation:

$$(1\,2)(1\,2\,3)=(2\,3)=(1\,2\,3)(1\,3).$$

Bernard
  • 175,478