I try to find a monoid with three elements that it doesn't be a group but I can't. Dose this monoid exist? How can I find this? If I can't,how can I prove this monoid dosen't exist?
monoid with three elements
0
$\begingroup$
abstract-algebra
group-theory
-
0How does the monoid's having only three elements entail existence of inverses? – 2017-02-10
-
0Yes it has only three elements – 2017-02-10
1 Answers
4
Try $\{-1,0,1\}$ with multiplication, or $\{0,1,x\}$ with $x^2=0$.
In fact there is a non-group monoid of order $n$ for all $n>2$: one idea is to adjoint a $0$ element to any multiplicative cyclic group of order $n-1$, or another is to adjoint $0$ and $1$ to any set and stipulate the product of any two nonidentity elements is $0$.
There is also a nonabelian monoid of any order $n>2$, given by adjoining $1$ to any set and stipulating that $xy=x$ for all $x,y$ in the set.
-
0What if the monoid is assumed to be nonabelian? Do you know which $n$ works? – 2017-02-10
-
0@Batominovski Any $n>2$ still. Updated answer. (I'll leave it as an exercise to check associativity.) – 2017-02-10