# Mathematical structures problem

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?

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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

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.

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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 $r_a+r_b$?

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