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Assume there is an element $a\in M$, where $M$ is a non-commutative monoid. If there exists $a b\in M$ such that $a * b = n$, where $n$ is the neutral, does it follow that $b * a = n$?

I want to say no, because the axioms which define a group shouldn't be redundant ($a * b = b * a = n$ is the 4th axiom). But the following logic seems to prove that the inverse is automatically commutative:

$b * n = b$

$b * (a * b) = b$

$(b*a) * b = b$

Thus for the last line to be true, $b*a$ has to evaluate to $n$.

OSCAR
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ConfusedEngineer
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    Note, however that the group axioms as usually stated are redundant (this is generally stated as a remark or given as an exercise in basic algebra textbooks), in the sense that it is equivalent to state that there is a left identity, and that every element has a left inverse. – fkraiem Jan 28 '14 at 22:38

3 Answers3

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The answer is yes if $M$ is finite, no otherwise. For this you only have to consider for each $a\in M$, the function $$\begin{align*} T_a:M&\rightarrow M\\ x&\mapsto a*x \end{align*}\text{.}$$ Then, if there is a $b$ such that $$b*a=1\text{,}$$ this means that $T_a$ is injective since $$\text{id}_M=T_b\circ T_a=T_{b*a}\text{.}$$ In this case, if $M$ is finite, this implies that $T_a$ is bijective and $T_b$ its inverse. Thus $$b*a=1$$ due to $$T_{b*a}=T_b\circ T_a\text{.}$$

However, when $M$ is infinite, this implication is false, i.e., there are functions that are injective and no surjective. For a concrete example with a monoid consider the monoid $M$ of functions $\mathbb{N}_0\rightarrow \mathbb{N}_0$ generated by the functions given by $$ f:n\mapsto\begin{cases} 0 & \text{ if }n=0\\ n-1& \text{ if }n\neq 0 \end{cases}\text{ and } g:n\mapsto n+1\text{.} $$ Then $f\circ g=\text{id}$ but not in the other way. (Note here how we are using the injective but not surjective function trick.)

Josué Tonelli-Cueto
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The answer is no. The bicyclic monoid is a counter-example. However, in a finite monoid, the answer is yes.

J.-E. Pin
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In your question, you called $M$ a non-commutative monoid. It should follow just from that, that the operations included in $M$ are not commutative. Thus, $a*b$ is not necessarily equal to $b*a$.

Normally, when someone twists my arm for an example of an associative but non-commutative binary operation, I answer "matrix multiplication."

jasotastic
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  • I downvoted your answer because it is misleading. Typically "matrix multiplication" is a good place to look. However, see here. – Karl Kroningfeld Jan 28 '14 at 22:45
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    Not helpful for the OP's question since that concerns a special case. Even for multiplication of square matrices, $AB$ does not equal $BA$ in general, but $AB=I$ does imply $BA=I$. – David Jan 28 '14 at 22:47
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    Fair enough. I suppose saying "non-square matrix multiplication" might have enhanced what I was trying to convey. – jasotastic Jan 28 '14 at 22:52
  • @jasotastic Then there are other problems, such as not being a monoid... Again, multiplication of square matrices is typically a good example--just not in this instance. – Karl Kroningfeld Jan 28 '14 at 22:57