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?

  • 2
    $\begingroup$ $n$ 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 $\mathbb{Z}$? $\endgroup$ – carmichael561 Sep 15 '16 at 5:14
  • 2
    $\begingroup$ "real integers"? That's a new one. :) $\endgroup$ – 6005 Sep 15 '16 at 5:16
  • $\begingroup$ @6005 As in $\Bbb Z[i]\cap \Bbb R$, perhaps ... $\endgroup$ – Hagen von Eitzen Sep 15 '16 at 5:17
  • 2
    $\begingroup$ @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. $\endgroup$ – 6005 Sep 15 '16 at 5:20
  • $\begingroup$ 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. $\endgroup$ – fleablood Sep 15 '16 at 7:02

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


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, $$x+y = nk_1+nk_2 = n(\underbrace{k_1+k_2}_{\in \mathbb{Z}}) \in n\mathbb{Z}$$


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 $\mathbb{Z}$.


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


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 $n\mathbb{Z}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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