# Group Theory, Multiples of a positive integer

given a question

The set of all multiples of a positive integer n with operation addition is group or not group?

I realized that in my opinion it's not a group cz if the set is on multiple positive integers, it must doesn't have inverses. For example is number $$2$$. The addition inverse of $$2$$ is $$-2$$ and $$-2$$ isn't positive integers right? But on the hint answer, it's group.

My question is, is it possible the negative numbers can be the multiples of a positive numbers? It's look like multiples of $$2$$ is always positive right. I have abstraction like this.

$$2+2+2+2+\cdots$$

• You do have to pick a positive integer to start with, yes (like $2$). However, the set of multiples of $2$ definitely includes $-2$. In other words, $-2$ is a multiple of the positive integer $2$. They are not asking for only the positive multiples. – Arthur May 9 at 13:25
• Yes, $n\Bbb Z$ is a subgroup of $(\Bbb Z,+)$, see this site. Also, yes, it is indeed possible that a negative integer is a multiple of a positive integer. For example $-1=-1\cdot 1$, so $-1$ is a multiple of $1$. – Dietrich Burde May 9 at 13:29
• There is no such thing as an infinite sum in a group-like object -The $+$ operation is binary only. You can use it and associativity to define a finite sum, but not an infinite sum. – Thomas Andrews May 9 at 13:29
• If it said "The set of positive multiples of a positive integer $n$" it would clearly not be a group, since there is no identity in that. But as written, $0$ is a multiple of any positive $n,$ and if $m$ is in the set then $-m$ is in the set. – Thomas Andrews May 9 at 13:35