Is `(Z,*)` a group, where `a*b = a+b-10` and `Z` is the set of integers?

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To demonstrate that the set `Z` (the integers) is a group with respect to the binary operation (`*`) where `a*b = a + b-10` we need to show that the four axioms hold, ie i)* closure, *ii)* associativity *iii)* identity *and iv) *invertibility*.

i) (`Z`, `*`) is closed if the operation on any elements of `Z` results in another element of `Z` .

Since the operation is simply a combination of addition and subtraction, then the group is closed.

ii) (`Z`, `*`) is associative if the sub-operations within the operation can be performed in a different order.

The sub-operations in this operation are addition and subtraction so that we can perform `(a+b)` then add `-10` *or *`a` plus `b-10` . Since addition/subtraction is associative then the group is associative.

iii) (`Z`, `*`) has an identity element if there exists an element `b` in `Z` such that `a` is unchanged under the binary operation (`*`). In the case of addition/subtraction (which the operation of interest consists of) the identity element is `b=0`.

iv) (`Z`, `*`) is invertible if the binary operation `f(a|b) = (*)` has an inverse. If we take `(*)` such that (`Z`, `*`) is a group if (`a+b-10 in Z` when `a,b in Z`) then the inverse operation `f'(a|b) = (*')` is given by

`a - b + 10`

We can see this if we let `z = a + b - 10`. Performing`f'`on `z|b`,

`f'(z|b) = z - b + 10 = a+b-10 - b + 10 = a`

Thus `f'(f(a)) = a` and `f'` is the inverse of `f`. Note that `f(f'(a)) = a` also.

**(Z, `*` ****) is a group with respect to the binary operation a + b -10, where Z is the set of integers (If Z is the integers, that is also a group, but if Z is the naturals it is not since a + b may be <10 ) **

**Sources:**

` ``Z={ZZ,*}` where `a*b=a+b-10` . Is `Z` a group?

For Z to be a group it must obey the four group axioms: closure, associativity, identity, and inverses.

(1) closure -- since we are adding/subtracting integers the set is closed under `*`

(2) associativity -- we need to show that `(a*b)*c=a*(b*c)` for any `a,b,c in Z`

`(a*b)*c=(a+b-10)*c=(a+b-10)+c-10=a+b+c-20`

`a*(b*c)=a*(b+c-10)=a+(b+c-10)-10=a+b+c-20`

Since this is true for arbitrary a,b,c the operation is associative.

(Note that subtraction is not associative -- in general (a-b)-c is not a-(b-c). But this operation is associative.)

(3)The identity element e is an element such that `a*e=e*a=a` . Let the identity element be 10.

`a*10=a+10-10=a` and `10*a=10+a-10=a`

(4) inverses -- we need to show that for every `a in Z` there exists an inverse `a'` such that `a*a'=a'*a=10`

Then `a*a'=10==>a*a'=a+a'-10=10`

So `a'=20-a`

`a*a'=a+(20-a)-10=20-10=10`

`a'*a=(20-a)+a-10=20-10=10`

Thus every element has an inverse.

**Z is a group -- it is closed under `*` , `*` is an associative operation, there is an identity element (10), and every element has an inverse (if a is the element, then 20-a is the inverse.)**

` `

If Z is the Reals that is also a group.

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