Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would one go around this problem and what are the ways to reason about it?

Let $Z(n)$ denote the set $\{0, 1, 2, \dotsc ,(n - 1) \}$. For example, $Z(6) = \{0, 1, 2, 3, 4, 5\}$.

and let $a * b$ be the remainder when $a + b$ is divided by $n$. Is $Z(n)$ with this operation a group? If it is a group is it commutative?

share|improve this question
    
Can you state the operation more clearly? –  Metin Y. Mar 15 '13 at 22:40
    
@MetinY.: In this context operation is defined as "An operation, called the group operation, which will be denoted by *. The result of applying this operation to two group elements, a and b in that order will be a group element that will be denoted a * b.". That's the only thing given. –  Denys S. Mar 15 '13 at 22:43
    
This looks like the integers (mod n) under addition to me. –  JB King Mar 15 '13 at 22:49
    
By the way, what you are referring to by $Z(n)$ is usually denoted by $\Bbb Z/n\Bbb Z$ and it is called the set (ring, actually) of integers modulo $n$. –  A.P. Mar 15 '13 at 22:50
    
Do you know about quotient groups? –  Math Gems Mar 16 '13 at 0:37
add comment

2 Answers

up vote 1 down vote accepted

Yes, this is a group, and it is commutative. Here's how to consider it:

In general, we know that for any $c, n \in \mathbb{N}$, the remainder of $\frac{c}{n}$ will be greater than or equal to $0$ and less than $n$. Thus, given $a,b \in Z(n)$, we know that $a * b \in Z(n)$, so we know that the set is closed under the operation.

Since $0 \in Z(n)$, we have that $0 * a = a * 0 = a ~\forall a \in Z(n)$. This is because $a + 0 = 0 + a = a$. Thus, $0$ is the identity element in $Z(n)$.

Now we need to establish inverses. That is, given $a \in Z(n)$, can we find $b \in Z(n)$ such that $ a * b = b * a = 0$? How about $n - a$? That would give $0$ as its own inverse (which is required for identity elements), and it would mean that $2$, for example, is inversed by $(n - 2)$. To see that this makes sense, consider $2 * (n - 2) \rightarrow 2 + (n - 2) = n$ and the remainder when dividing $n$ by itself is of course $0$, our identity.

Further, since $a + b = b + a$, the operation will be commutative.

share|improve this answer
add comment

Hint: If $a,b\in \Bbb Z$ and $r_a,r_b\in Z(n)$ are the remainders, respectively, of $a$ and $b$ when divided by $n$, what can you say about $r_a+r_b$?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.