0

I've got the following problem:

For $M:={0,1,2,3}$ we look at the Monoid $F:=(M^M, \circ)$.

Now I need to state, how many elements F contains. Since every element in a Monoid need to have a neutral element, I get this 6 elements. Is that correct?

Furthermore, I want to state how many elements of F are invertible. Since I can give the inverse of every of these 6 elements, all 6 are invertible. Is this true as well?

Bowueewa
  • 25
  • 1
  • 3

1 Answers1

0

Your answers are incorrect. The monoid $M^M$ has $4^4 = 256$ elements. It consists of all functions from the set $\{0, 1, 2, 3\}$ to itself. The inversible elements are the bijections. Therefore there are $4! = 24$ inversible elements in this monoid.

J.-E. Pin
  • 40,163