Question

Show that the set of all positive even integers forms a semi-group under the usual addition and multiplication. Is it a moniod under each of the above operations?

Answer 1

Let $E=\{2x:x\in N\}$ be the set of even positive integers.

$\left(i\right)$Closure:Consider $(E,+)$

Let $2x,2y\in E$ then $2x+2y=2(x+y)\in E$

The closure property is satisfied.

$\left(ii\right)$Associativity:For $2x,2y,2z\in E$

$(2x+2y)+2z=2x+(2y+2z)$ since these are natural numbers and addition of natural numbers is associative.

$\therefore (E,+)$ is a semi-group.

$\left(iii\right)$Consider $(E,.)$

Closure :For $2x,2y,2z\in E$

$\left(2x\right)\left(2y\right)\left(2z\right)=2\left(4xyz\right)\in E$

(The product of even numbers is again an even number)

The closure property is satisfied.

$\left(iv\right)$Associativity:The associative property is inherited from the set of natural numbers(as with the associative property of addition over $E$)

Therefore $(E,.)$ is a semi-group.

Hence, $E$ is a semi-group under additions as well as multiplication.

Since $E\subset N$, then $\{E,+\}$ and $\{E,.\}$ are monoids with the identity elements $0$ and $1$ respectively.

Monoids are semigroups with identity. Hence from the above $E$ is a monoid.