With respect to addition, show that the set of even integers, is a group but tha

Question

With respect to addition, show that the set of even integers, is a group but that the set of odd integers, is a group but that the set of odd integers, 2Z +1, is not a group. With respect to complex multiplication, show plus mines {1, i} is a group. With respect to multiplication, show (5^a | a sum Q) is a group. With respect to addition, show {a sum Z | a identical to is a group but that {a sum Z | a identical to 1 mod (3) is not. With respect to multiplication in R, show (0, 1] is not a group. Show that U_9 = Z_9 {[0], [3], [6]} and that {[1], [4], [7]} subset or equal to U_9 is a group with respect to multiplication.

Explanation / Answer

(A) set of even integer 2Z

Closure property

let x and y be two even integers then x+y is also even

Identity 0 is even as 0=2.0

0 belongs to 2Z

Inverse--- if x in 2Z then - x is also even and hence belongs to 2Z and x+(-x)=0

Hence 2Z forms a group

Set of odd integers is not a group with addition

As if we take two odd integers then their sum is even not odd hence it is not closed under addition .

Hence not a group with addition.

(B) {+1,-1,i,-i}

Closure----

+1. -1. i. -i

+1 +1 -1 i. -i

-1 -1 1. -i. i

i i -i. -1 +1

-i. i. i. +1. -1

By table it clear that set is closed under complex multiplication .

And identity is 1

Inverse of 1=1

Inverse of i= -i

Inverse of -1 =-1

Inverse of -i is i

Hence it forms group under complex multiplication .

C) {5^a:a in Q} forms a group

Let two elements from set 5^a and 5^b

5^a.5^b=5^(a+b) and a+b is also in Q as a,b in Q

5^a+b belongs to the set

5^0=1 belongs to the set as 0 in Q

And for every 5^a there is 5^-a such that 5^a.5^-a=5^(a-a) =5^0=1

Hence 5^-a is the inverse of 5^a in the set as -a belongs to Q as a in Q.

And associative rule also satisfied here

Hence it forms a group under multiplication

D) {a in Z ,a congruent to 0 MOD 3}

Let a,b in the set a cong to 0 mod3 ----> 3/a

b cong to 0mod 3 -----> 3/b

Which implies 3/(a+b)

a+b cong to 0 mod 3

Set is closed under addition

And 3/0 always means 0 is cong to 0 MOD 3

Identity element in the set is 0

And do every a in the set the is -a in the set as if 3/a then 3/(-a)

Such that a+(-a)=0

Means Every element have inverse in the set

Assosictiv also satisfied since in congruenc e associative rule satisfied .

Hence it forms a group .

a in Z ,a cong to 1 MOD 3

It is not closed under add as if a,b in the set

3/(a-1) and 3/(b-1) it doesn't implies that 3/(a+b-1)

Hence not closed under addition , and not form a group.

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