Middle cancellation in a group I can not find a such example, ( It's my first course in abstract algebra)
Give an example of a group with elements $a,b,c,d$ and $x$ such that $axb=cxd$ but $ab\neq cd$. (Hence "middle cancellation" is not valid in groups.)
 A: Let $e$ denote the neutral element of the group.
In a commutative group middle cancellation is evidently valid, so the group used as counterexample is not allowed to be commutative.
Find $a,b$ with $ab\neq ba$.
Then $x=a^{-1}b^{-1}$ does the job when $d=e$ and $c=ba$.
A: Notice that you can't find a counterexample in a commutative group. Let's try in a noncommutative one: $GL_2(\Bbb R)$, the group of regular $2\times2$ real matrices.
We'll choose randomly $a,b,c,x$, compute $d$ and check that it fails as expected.
$$a=\left(\begin{matrix}2 & 0 \\ 1 & 1\end{matrix}\right)$$
$$b=\left(\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right)$$
$$c=\left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right)$$
$$x=\left(\begin{matrix}2 & 1 \\ 1 & 0\end{matrix}\right)$$
Now define $d=x^{-1}c^{-1}axb$. You then have by definition of $d$, that $axb=cxd$. This requires a few matrix multiplications, but in dimension $2$ it's fairly easy (and getting the inverse of a $2\times2$ matrix is immediate).
$$d=\left(\begin{matrix}4 & 6 \\ -5 & -8\end{matrix}\right)$$
Then
$$ab=\left(\begin{matrix}2 & 2 \\ 1 & 2\end{matrix}\right)$$
$$cd=\left(\begin{matrix}-5 & -8 \\ 4 & 6\end{matrix}\right)$$
So $ab\neq cd$, and we found an example.
You may think it's just luck, but since the product is not commutative, it's rather common that this cancellation fails.
