I am doing some research of my own and I have a brief question and I don't recall studying this particular thing. Is it generally true that if a set coupled with a binary operation is closed under inverses and has an identity then $ab=cd$ implies $ba=dc$ for all $a,b,c,d$ in the set? If not, would this statement hold true for groups?
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Is the binary operation associative ? – Amr Jun 28 '13 at 12:23
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Not necessarily. That's why I make the distinction later by mentioning groups. – Jebruho Jun 28 '13 at 12:24
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3Consider for example $GL(2,\mathbb R)$, with $A$ identitiy matrix. If $B=CD$, then, in general, it is not true that $B=DC$ as it implies that $C$ and $D$ commute. – Avitus Jun 28 '13 at 12:32
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1@Avitus You should make this an answer. I really like this response. – Jebruho Jun 28 '13 at 12:33
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1@Jebruho ok, but I have some troubles in writing matrices...I need to search for the LaTeX commands :-) Sorry about that – Avitus Jun 28 '13 at 12:38
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@Jebruho Notation improved. Nice question, btw. – Avitus Jun 28 '13 at 12:49
4 Answers
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As long as your operation has an identity, the property
$$ab = cd \Rightarrow ba=dc$$
is equivalent to commutativity: if the operation is commutative, the implication is obvious, for the other direction, put $c=1$ and $d=ab$ to get $ab = ba$.
sdcvvc
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Consider $GL(2,\mathbb R)$. It is a non empty set with binary associative composition, unit and inverses.
Let $A$ be the unit matrix, $B=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$, $C=\ \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$ and $D=\ \begin{bmatrix}-1 & 1\\1 & 0\end{bmatrix}$.
Then $AB=B=CD$; but $B=DC$ is not true.
Avitus
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Counterexample: Consider the group $S_3$. $$(13)(12)=(123)e$$ However: $$(12)(13)\not=e(123)$$
Amr
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Another one may be arisen in $Q_8$, the quaternion group. Just take $a=x,y=b=c,d=x^{-1}$.
Mikasa
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