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I know how to regularly prove that multiplication modulo n is commutative on some set, but I don't know why is here underline that it has to be set which includes zero (only specific thing I know about this set that there is no inverse for zero value - so it cannot be group).

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    Hint: What happens if you calculate $2 \cdot 3 \mod 6$? Nevertheless, the condition seems to be too weak and somewhat arbitrary - you want the set to be closed unter multiplication.2012-04-04

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As Johannes points out in the comments, for arithmetic modulo $n$, one does not always have the property that "non-zero times non-zero equals non-zero". Non-zero elements which multiply together to be zero are called zero divisors. It turns out that in $\mathbb{Z}_n$ (the ring of equivalent classes of integers mod $n$), every element is either zero, a zero divisor, or a unit (an element with a multiplicative inverse). So if you exclude zero but include a zero divisor, your set will not be closed under multiplication.

In any ring (with $1$) one can prove that "unit times unit equals a unit" so the set of units of a ring form a group under multiplication. In $\mathbb{Z}_n$ the group of units is exactly the equivalence classes represented by the integers between $1$ and $n$ which are relatively prime to $n$.

Example: $\mathbb{Z}_6 = \{0,1,2,3,4,5\}$. $\mathbb{Z}_6$ itself is closed under multiplication (since any integer is equivalent to one of $0,1,\dots,5$ mod $6$). The set $\{1,2,3,4,5\}$ is not closed since for example: $2 \cdot 3 =0$.

Notice that $\{1,5\}$ is closed under multiplication since these are the units of $\mathbb{Z}_6$. On the other hand, $2,3,4$ are zero divisors because $2\cdot 3=0$ and $4 \cdot 3=0$.

Now if $n$ is prime, then $1,2,\dots,n-1$ are all relatively prime to $n$. Thus the group of units for $\mathbb{Z}_n$ (if $n$ is prime) is $\mathbb{Z}_n-\{0\}=\{1,2,\dots,n-1\}$. So if $n$ is prime (and only if $n$ is prime), we have that the non-zero elements are closed under multiplication.