Cancelling $a$ in $ab=ca$

Question

In Pinter's A Book of Abstract Algebra, the cancellation law is explained like this:

$ab=ac$

$a^{-1}ab=a^{-1}ac$

$eb=ec$

But then it goes on to say that

In general, we cannot cancel $a$ in the equation $ab=ca$. (Why not?)

At first I thought that it had to do with commutativity, but that can't be it because multiplication by inverses and by the identity is generally commutative right?

So if we apply $a^{-1}$ on from the left on the LHS, and from the right on the RHS, we get

$a^{-1}ab=caa^{-1}$

Then by the law of associativity

$(a^{-1}a)b=c(aa^{-1})$

Since an operation with inverses is commutative,

$eb=ce$

And since operations involving the identity are commutative,

$b=c$

So what's wrong with that?

Answer 1

Here's your mistake:

So if we apply $a^{-1}$ on from the left on the LHS, and from the right on the RHS, we get

$a^{-1}ab=caa^{-1}$

Once you multiply on the left (or the right) of a word (i.e., a string of elements) in an equation, you have to do the same on the other side of the equation, like so:

$$a^{-1}(ab)=\color{red}{a^{-1}}(ca).$$

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For a counterexample to the cancellation candidate, consider the group given by the presentation $$\langle a,b,c\mid ab=ca \rangle;$$ that is, via Tietze transformations, the group given by $$\langle a,b,c\mid c=aba^{-1}\rangle;$$ i.e., the presentation $$\langle a,b\mid\varnothing \rangle,$$ which yields the free product $\Bbb Z\ast\Bbb Z$.

Answer 2

In general, we cannot cancel a in the equation $ab=ca$. (Why not?)

Think of the group $GL_n(K)$ of matrices. We have $AB=CA$ which means $C=ABA^{-1}$, so both matrices $B$ and $C$ are conjugated. Why should it be allowed to cancel $A$? Certainly conjugated matrices need not be equal. So the Jordan normal form makes sense.

If you have $ab=ac$ then you have to multiply with $a^{-1}$ from the left (or right) consistently, i.e., $a^{-1}(ab)=a^{-1}ca$, i.e., $b=a^{-1}ca$.

Answer 3

Let's take as an example the smallest group where this fails : $\mathfrak S_3$. Its elements are permutations of $\{1,2,3\}$ and multiplication is composition.

A permutation is typically represented as a product of cycles : a cycle $(m_1 ... m_n)$ represents a permutation that sends $m_1$ to $m_2$, $m_2$ to $m_3$,... $m_{n-1}$ to $m_n$, and $m_n$ to $m_1$.

Then if you compute $(1 2)(1 3)= (2 3)(12)$.

The reason your "proof" is not one is that you multiply on the right on one side, and on the left on the other side : why would you be allowed to do that ?

You are allowed to do that if the multiplication is commutative (or abelian), that is $ab = ba$ for all $a,b$.

In fact, the cancellativity property you ask about is equivalent (if you allow it for all $a,b,c$) to the commutativity of the operation (assuming it's already associative) : indeed if you have it, then $a(ba) = (ab)a$ gives you, by "cancellativity" $ba= ab$.