# Let $n \mathbb{Z} = \{nk \mid k \in \Bbb Z\}$. Prove that $n \Bbb Z$ is a group under addition.

Let $n \mathbb{Z} = \{nk \mid k \in \Bbb Z\}$. Prove that $n \Bbb Z$ is a group under addition.

$\Bbb Z$ is all real integers.

I know that for something to be a group it needs to be associative, have an identity, have closure, and have inverses. However, the notation is confusing to me because I am trying to prove addition but I'm multiplying $n$ and any integer? I know that addition is associative, that it has inverses as negative integers are included and that this is most likely closed. Is there an identity?

• $n$ is fixed for the purposes of the problem. So for instance if $n=2$ then you are trying to show that the set of even integers is a group under addition. What is the additive identity for $\mathbb{Z}$? – carmichael561 Sep 15 '16 at 5:14
• "real integers"? That's a new one. :) – 6005 Sep 15 '16 at 5:16
• @6005 As in $\Bbb Z[i]\cap \Bbb R$, perhaps ... – Hagen von Eitzen Sep 15 '16 at 5:17
• @HagenvonEitzen It would be very odd if someone new about the Guassian integers before knowing about the integers! Reminds me of a high school student who learned about modules but didn't know about vector spaces. When vector spaces came up, she asked what they were. Finally it was noted that a vector space is "just a module over a field", at which point she understood immediately. – 6005 Sep 15 '16 at 5:20
• The set nZ is {... -3n, -2n, -n, 0 ,n ,2n, 3n....}. So for example 4Z ={... -16, -12, -8, -4, 0, 4, 8, 12, 16, 20,...}. Can you show those set are groups under addition. The multiplication was how you create the set in the first place. It has nothing to do with group operators. – fleablood Sep 15 '16 at 7:02

## 1 Answer

Fix $n \in \mathbb{Z}$.

Closure

Consider any $x,y \in n\mathbb{Z}$. Then there exists $k_1, k_2 \in \mathbb{Z}$ such that $x = nk_1$ and $y = nk_2$. We see, $$x+y = nk_1+nk_2 = n(\underbrace{k_1+k_2}_{\in \mathbb{Z}}) \in n\mathbb{Z}$$

Associativity

Since $\mathbb{Z}$ is associative over addition and since $n\mathbb{Z} \subset \mathbb{Z}$ with the same addition operation, then associativity is inherited from $\mathbb{Z}$.

Identity

Notice that $0 = 0 \cdot n$ so $0 \in n\mathbb{Z}$. And for any $x \in n\mathbb{Z}$, we have $x+0 = x = 0+x$.

Inverses

Consider $x = nk_1 \in n\mathbb{Z}$ with $k_1 \in \mathbb{Z}$. Since $k_1 \in \mathbb{Z}$, then $-k_1 \in \mathbb{Z}$ and thus $y = n(-k_1) \in n\mathbb{Z}$. We see that $y$ is an invere of $x$: $$x+y = nk_1+n(-k_1) = nk_1-nk_1 = 0,$$ where $0$ is the identity element in $n\mathbb{Z}$.