# Prove whether the given satifies the four group axioms. If not determine whether it is a semigroup or monoid. 1. <R, *> where a*b = `a^(2)+ b` 2. <z, *> where a*b = a + b + ab ` `

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### 1 Answer

The four group axioms are closure, associativity of the binary operation, an identity element, and inverses.

(1) `{RR,*},a*b=a^2+b`

(a) The set is closed under this operation. (Squaring a real number and adding a real number to the result will always give a real number.)

(b) Associativity: Let `a,b,c in RR`

`(a*b)*c=(a^2+b)*c=(a^2+b)^2+c=a^4+2a^2b+b^2+c`

`a*(b*c)=a^2+(b*c)=a^2+b^2+c`

The operation is not associative.

**The set with the given operation is not a group. Further, it is neither a semigroup or a monoid as both requires an associative binary operation.**

(2) `{ZZ,*},a*b=a+b+ab`

(a) The set is closed under this operation -- an integer plus an integer plus the product of integers is an integer.

(b) Associativity -- let `a,b,c in ZZ`

`(a*b)*c=(a+b+ab)*c=(a+b+ab)+c+(a+b+ab)c`

`=a+b+ab+c+ac+bc+abc`

`=a+b+c+ab+ac+bc+abc`

`a*(b*c)=a*(b+c+bc)=a+(b+c+bc)+a(b+c+bc)`

`=a+b+c+bc+ab+ac+abc`

`=a+b+c+ab+ac+bc+abc`

So the operation is associative

(c) The identity is 0

`a*0=a+0+a0=a`

`0*a=0+a+0a=a`

for any `a in ZZ`

(d) Inverse: Suppose b is the inverse of a with `a,b in ZZ` .Then

`a*b=0`

`==>a+b+ab=0`

`==>b+ab=-a`

`==>b(1+a)=-a`

`==>b=(-a)/(1+a)` which is not an integer for every `a in ZZ` so inverses do not exist.

**This set with the given operation is not a group.**

**Since the binary operation is associative it is a semigroup. Since there is an identity element it is a monoid.**