# Show Zm with binary operation is a group

1. Feb 17, 2008

### karnten07

1. The problem statement, all variables and given/known data
Let m$$\in$$N. We defined a binary operation on Zm:= {$$_{}0$$,$$_{}1$$..., $$_{}m-1$$}. Show that Zm together with this binary operation is a group. (Hint. For associativity you need to distinguish several cases)

2. Relevant equations

3. The attempt at a solution

I don't know what the lines above the numbers mean, does it mean they can be positive an negative? (there are meant to be lines above the numbers in Zm)

Example of Z3:

+ 0 1 2
0 0 1 2
1 1 2 0
2 2 0 1

For the cases i need to use to show associativity, i think they are m=1, m>1 and m>2 because my lecturer said that, although im not sure i remember exactly what he said so i could be wrong. I can show m=1 as it is just a group table of all zeros. I think the case where m>1 it is meant to be 2>/=m>1 so it becomes Z2. But how do i show z>2 as it has elements that go up to m-1?

2. Feb 17, 2008

### HallsofIvy

Staff Emeritus
What binary operation? You told us what Zm is (which I suspect many of us already knew) but you haven't told us what the "binary operation defined on Zm" is! (s it "addition modulo m"?

From what you are saying, then, the operation is addition modulo m. I can't be certain what the "lines above the numbers" mean, especially since I don't see any lines above in numbers in what you wrote! Do you have $\overline{0}, \overline{1}, \overline{2},\cdot\cdot\cdot, \overline{m-1}$? Zm, technically, is the set whose elements are "equivalence classes"- that is, sets of integers such that difference is a multiple of m. Those, I suspect are what your "lines above the numbers mean"- $\overline{1}$ is the set containing {1, m+1, -m+1, m+2, -m+2, ...}.

In that case, the usual binary operation is this: to add "A+ B", choose an integer from set A and a number from set B. Add them. If the result is in set C, then A+ B= C. Of course, you have to prove that if you were to choose different integers from the same sets you would get the same resutl but that is relatively easy. Once you have done that, since each set contains exactly one non-negative member less than m and it is simplest to "represent" the set by that number. To prove "associativity" you need to show that (a+ b)+ c= a+ (b+ c) for any numbers in the set. One thing you might do is this- assuming that a, b, c are all in {0, 1, 2, ..., m-1} consider the cases when a+b< m, a+b> m, b+ c< m, b+ c> m. If a+b< m then their sum "modulo m" is just a+ b itself. If a+ b> m, their sum "modulo m" is a+ b- m since a+ b- m will be less than m and (a+b)- (a+b- m)= m.

After you have proven associativity you will need to prove that this set has an "additive identity": there is an e in the set such thata+ e= a for all a in the set. Try the obvious! You will also need to prove that every member has an "additive inverse": for every a in the set, there is b in the set so that a+ b= e.

While each of the sets making up Zm contain both positive and negative numbers, in general, a negative of a number in the set is NOT in the set. n- (-n)= 2n is not, in general, divisible by m.

You will want to notice that, for any number a, (a+ m)- a= m is divisible by m. If a is in a particular set in Zm, so is a+ m. That should tell you what the additive identity is. Also notice that if a is in such a set, then a+ (m-a)= m. That tells you what the additive identity of a is.

You might want to look at Z4 as an example to see what I am talking about. What is the additive identity in Z4? What are the additive inverses of 1, 2 and 3 in Z3?

3. Feb 17, 2008

### karnten07

Yes you are right, it must be addition modulo m. I have just come to understand how it works now and think i will be able to do this question now. But you've been a great help once again, thanks

Last edited: Feb 17, 2008
4. Feb 17, 2008

### karnten07

For associativity can i use these cases:

a+b=<m, a+b=m and a+b>m. Do i have to do the other cases of b+c<m and b+c>m or should i make it so these conditions are met simultaneously in say 3 cases? Or should they all be separate or at least treated separately?