Neutral element in monoid.

Question

Let $(G, *) $ be a monoid . Let $g \in G$ and $ g = g * g $. Can I assume, that $g$ must be neutral element? Why?

score 15 · Answer 1 · answered Mar 01 '15 at 19:15

15

No, you cannot. $\mathbb{N}^{\geq 0}$ is a monoid under multiplication, but $g=0$ is not the identity element even though it satisfies your condition.

answered Mar 01 '15 at 19:15

Michael Burr

32,867

score 7 · Answer 2 · answered Mar 01 '15 at 23:41

7

In fact there are monoids all of whose elements have this property (which is called being idempotent). For example, you can take the underlying set of any poset with finite joins or finite meets (semilattices).

As an example of that example, the totally ordered set $\{ 1, 2, \dots n \}$ has joins and meets given by taking max and taking min respectively. Both of these define a monoid structure where every element is idempotent.

answered Mar 01 '15 at 23:41

Qiaochu Yuan

419,620

And another example of such, again based upon matrices, is the set of all diagonal matrices with only zeroes and ones on the diagonal! Cheers! – Robert Lewis Mar 03 '15 at 17:33
@Robert: that is, of course, a special case of the above: take the poset to be the poset of subsets of a finite set! – Qiaochu Yuan Mar 03 '15 at 17:35
Thanks for confirming my hunch in this matter! – Robert Lewis Mar 03 '15 at 17:43

score 5 · Answer 3 · answered Mar 01 '15 at 19:14

5

No. Take $G=\mathbb Z$ and $g=0$.

Far less drastic examples can be found in matrices.

answered Mar 01 '15 at 19:14

lhf

216,483

score 3 · Answer 4 · answered Mar 03 '15 at 16:57

3

Minimal counterexample. The monoid $\{0, 1\}$ with the usual multiplication of integers. Then $0 * 0 = 0$ and $1 * 1$ but $0 \not= 1$.

answered Mar 03 '15 at 16:57

J.-E. Pin

40,163

score 2 · Accepted Answer · edited Mar 02 '15 at 01:16

No, and here's a more complex example: let $\Bbb N$ be the natural numbers, i.e. $\Bbb N = \{ 0, 1, 2, 3, \ldots \}$; note I am allowing $0 \in \Bbb N$. For any $n \in \Bbb N$, $n \ge 2$, let $M_n(\Bbb N)$ be the set of $n \times n$ matrices with entries taken from $\Bbb N$. It is clear that $M_n(\Bbb N)$, with the usual operation of matrix multiplication, forms a (non-commutative) monoid with identity $I$. However, any matrix $G$ formed by replacing at least one diagonal entry of $I$ with $0$ also satisfies $G^2 = GG = G$, but it is easy to construct examples showing no such $G$ satisfies $GA = AG = A$ for all $A \in M_n(\Bbb N)$; e.g. take $A = I - G \ne 0$ for some such $G$. Then $AG = GA = G(I - G) = G - G^2 = 0 \ne A$. Such $G$, though idempotent, are not the identity element of $M_n(\Bbb N)$.

MattAllegro · Answer 6 · 2015-04-06T07:49:11.967

As pointed out already by five answers before mine, no. It is easy to construct the Cayley table of a finite counterexample. Let $G=\{1,2,3\}$, then $$\begin{array}{c|ccc} \ast & 1 & 2 & 3\\\hline 1 & \color{red}{1} & 2 & 3\\ 2 & 2 & \color{red}{2} & 3\\ 3 & 3 & 3 & \color{red}{3}\\ \end{array}$$ defines a commutative (idempotent) monoid $(G,\ast)$ where $2\ast 2=2$ and $3\ast 3=3$ but the neutral element is $1$.

Neutral element in monoid.

6 Answers6