In a monoid, does $x \cdot y=e$ imply $y \cdot x=e$?

Question

A monoid is a set $S$ together with a binary operation $\cdot:S \times S \rightarrow S$ such that:

The binary operation $\cdot$ is associative, that is, $(a\cdot b) \cdot c=a\cdot (b \cdot c)$ for all $a,b,c \in S$.
There is an identity element $e \in S$, that is, there exists $e \in S$ such that $e \cdot a=a$ and $a \cdot e=a$ for all $a \in S$.

Question: Suppose, $x,y \in S$ such that $x \cdot y=e$. Does $y \cdot x=e$?

This question was motivated by the question here, where the author attempts to prove a special case of the above in the context of matrix multiplication. It was subsequently proved, but the proofs require the properties of the matrix.

I attempted to use Prover9 to prove the statement. Here's the input:

formulas(assumptions).

% associativity
(x * y) * z = x * (y * z).

% identity element a
x * a = x.
a * x = x.

end_of_list.

formulas(goals).

x * y = a -> y * x = a.

end_of_list.

and it returned sos_empty, which, I guess, implies that no proof of the above statement is possible from the axioms of monoids alone. I ran Mace4 on the same input, and found no counter-examples for monoids of sizes $1,2,\ldots,82$.

A comment by Martin Brandenburg here regarding K-algebras might also apply here. For example, the property might be true for finite monoids, but not all infinite monoids. A counter-example would (obviously) need to be non-commutative.

There are counterexamples for matrix algebras, so a fortiori for monoids. — Martin Brandenburg, Oct 19 '12 at 08:10

Chris Eagle · Accepted Answer · 2012-10-18T07:28:53.670

22

Let $M$ be the monoid of all functions from $\mathbb N$ to $\mathbb N$, with function composition as the operation. Let $y(n)=n+1$ for all $n$, while $x(n)=n-1$ for $n\ge 1$, $x(0)=0$. Then $xy$ is the identity function, while $yx$ is not.

As you conjecture, the statement is true for finite monoids. Suppose $xy=e$. If we have a $z$ such that $zx=e$, then $zxy=ey$, so $ze=ey$, so $z=y$. Thus it's enough to show $x$ has a left inverse $z$. To do this, consider the function $f$ from $M$ to $M$ given by $f(a)=ax$. If $ax=bx$, then $axy=bxy$, hence $a=b$. Thus $f$ is injective, and so (since $M$ is finite) $f$ must be surjective, so in particular $e$ is in its image, and we're done.

In short: every monoid $M$ is isomorphic to a set of functions from $M$ to $M$ under composition, and $fg=\mathrm{id} \rightarrow gf=\mathrm{id}$ only holds on finite sets.

edited Oct 18 '12 at 07:28

answered Oct 18 '12 at 07:13

Chris Eagle

33,306

In functional analysis the left- and right-shift operator. – vesszabo Oct 18 '12 at 07:15
Great! Nice proof. Thanks for that (and thanks to the other answerers). – Douglas S. Stones Oct 18 '12 at 07:25
In fact, $x$ has two right inverses (another way to show it is a counterexample using Isaac Solomon's remark). We already have $y$ as constructed in the answer, but also there is $z$ defined by $z(0)=0$ and $z(n)=n+1$ for all other $n\in\Bbb{N}$, and $xz=\text{id}$. – yearning4pi Oct 18 '12 at 07:39
This is a nice example because it's also useful for constructing a ring example. If you have a countable dimensional vector space $V$, and you define $f(b_i)=b_{i+1}$ and $g(b_i)=b_{i-1}$ for $i>0$, and $g(b_0)=0$, then the linear transformations of $V$ determined by $f$ and $g$ are such that $gf=1$, but $fg\neq 1$. So even in a monoid as rigid as the monoid of a ring, $ab=1$ need not imply $ba=1$. – rschwieb Oct 18 '12 at 12:32

score 9 · Answer 2 · answered Oct 18 '12 at 07:08

9

If a left inverse and a right inverse exist in a monoid, they are equal. However, the existence of a left inverse need not imply the existence of a right inverse, and vice versa.

answered Oct 18 '12 at 07:08

Elchanan Solomon

30,807

Do you have a proof of this? – Oct 18 '12 at 07:13
1

Great! Prover9 can easily prove the first statement if I add the goal x * y = a & z * x = a -> y = z.. Could you also provide an example of the second claim please? [Stet] – Douglas S. Stones Oct 18 '12 at 07:13

Makoto Kato · Answer 3 · 2012-10-18T07:25:15.733

4

Let $S$ be the set of maps $\mathbb{C} \rightarrow \mathbb{C}$. Let $f \in S$ be the map defined by $f(x) = x^2$. Since $f$ is surjective, there exists $g \in S$ such that $f\circ g = 1$. Since $f$ is not bijective, $g\circ f \neq 1$.

edited Oct 18 '12 at 07:25

answered Oct 18 '12 at 07:14

Makoto Kato

42,602

score 0 · Answer 4 · answered Apr 03 '23 at 15:39

There are monoids in which $a\cdot b=e$ does not imply that $b\cdot a=e$.

To prove this statement, note that a ring is a monoid under its multiplication operation, and so it suffices to find a ring where $xy=1$ does not imply $yx=1$.

Definition. A ring is said to be Dedekind-finite if $ab=1$ implies $ba=1$.

(Why Dedekind-finite rings are called "Dedekind-finite" is discussed in this post. As suggested by the name, all finite rings are Dedekind-finite. Indeed, this is a special case of the theorem mentioned by Chris Eagle in his answer.)

Theorem. There are rings which are not Dedekind-finite.

Proof. An example is given in this answer by Henning Makholm:

Take the ring of linear transformations on the space of infinite real sequences, and let $y$ be the shift-right operator, and $x$ be the shift-left operator. [Then, $xy=1$ but $yx\neq 1$, where ring multiplication is given by composition and $1$ denotes the identity operator.]

The next section of my post fleshes out Henning's answer in more detail for those who are less fluent with the linear algebra that is needed to understand it.

Let $V$ be the real vector space of sequences $(a_n)$ in $\mathbb R$. Vector addition and scalar multiplication are defined pointwise, so that $(a_n)+(b_n)=(a_n+b_n)$ and $c(a_n)=(ca_n)$.

Henning then asks to consider the set $L(V)$ consisting of linear maps from $V$ to $V$. Each member of $L(V)$ is a function satisfying certain properties which takes in a sequence and spits out another sequence.

We may endow $L(V)$ with a ring structure by defining the ring addition and multiplication in the following way:

Given two linear maps $f,g\in L(V)$, we define $f+g$ as the function given by $$\bigl(f+g\bigr)\bigl((a_n)\bigr)=f\bigl((a_n)\bigr)+g\bigl((a_n)\bigr)$$
The product $fg$ is defined as the composite function $f\circ g$, i.e. the function given by $$ \bigl(f\circ g\bigr)\bigl((a_n)\bigr)=f\bigl(g\bigl((a_n)\bigr)\bigr) $$

The multiplicative identity $1$ of $L(V)$ is the identity function $\DeclareMathOperator{id}{id}\id:L(V)\to L(V)$ such that $\id\bigl((a_n)\bigr)=(a_n)$ for every sequence $(a_n)$.

The left-shift operator is the map $x:L(V)\times L(V)\to L(V)$ given by $$ x(a_0,a_1,a_2,\dots)=(a_1,a_2,a_3,\dots) $$ and similarly the right-shift operator $y$ is given by $$ y(a_0,a_1,a_2,\dots)=(0,a_0,a_1,a_2,\dots) $$ Notice that $0$ is appended to the start of the sequence.

Exercise. Verify that $x$ and $y$ are indeed linear maps (so that they belong to $L(V)$), and that $xy=1$ but $yx\neq 1$.

In a monoid, does $x \cdot y=e$ imply $y \cdot x=e$?

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