Prove Theorem 6.5.5. Complete the proof of Theorem 6.5.6. Let m element Z^+, m >

Question

Prove Theorem 6.5.5. Complete the proof of Theorem 6.5.6. Let m element Z^+, m > 1. (a) Show that (Z_m, +m) is a group by verifying the group axioms in Table 6.2: [a]_m +m ([b]_m +_m [c]_m) = ([a]_m +_m [b]_m) +_m [c]_m. (ii) Show that for all [a]_m element Z_m, [a]_m +m [0]_m = [a]_m = [0]_m +_m [a]_m. (iii) Show that for all [a]_m element Z_m there exists [b]_m element Z_m such that [a]_m +_m [b]_m = [0]_m = [b]_m +_m (b) In fact, +_m is commutative, making (Z_m, +_m) an abelian^1 group: show that for all [a]_m, [b]_m element Z_m, [a]_m +_m [b]_m = [b]_m +_m [a]_m. Prove that if n element Z^+ is odd, then the sum of all the elements in Z_n is [0]_n. Given m element Z^+, m > 1, let Z*_m = Z_m - {[0]_m}. (a) Find the multiplication. table for (Z*_4, 4) and (Z*_5, 5). (b) Is (Z*_4, 4) a group? Is (Z*_5, 5) a group? Why are we considering Z*_m, rather than Z_m, under multiplication modulo m?

Explanation / Answer

Let m > 1 be a positive integer.Then the set <Zm ,+m> is a group as we show above.

The set Zm = {0,1,2,3,...,m-1} , the remainder set on division by m . a +m b means the remainder when the sum

a +b is divided by m.

ie., 2+4 modulo 5 means that the remainder obtained when 6 is divided by 5 = 1.

Let a, b,c are any threee elements of Zm.

Then [a]m + ([b]m + [c]m) = [a]m + [b+c]m = [a+b+c]m. (If the sum b+c is divisivble by m, then [b+c]m = 0, Then [a]m + [0]m = [a]m ).

Also, ([a]m + [b]m) +[c]m = [a+b]m + [c]m = [a+b+c]m .

Associativity holds.

Now [0]mis clearly an element of Zm ., which acts as the identity so that [a]m +[0]m = [a+0]m = [a]m and [0]m +[a]m = [0+a]m = [a]m

Also for all [a]m there exists an inverse modulo m , [b]m such that [a]m +[b]m= [a+b]m = identity = [0]m, since b = a-1 .Note that the remainder on division by m , lies in {0,1,2,...m-1}.This b can be obtained by simply doing m - a. For eg, in Z4 , 3-1 = 4-3 = 1 , 2-1 = 4-2 = 2 etc...

Modulo addition is similar to addition operation and hence is commutative as well , since a+b and b+a are both the same and hence leaves the same remainder on division by m ie., [a]m + [b]m = [a+b]m = [b]m + [a]m = [b+a]m

Thus to conclude <Zm , +m > is am abelian group

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