Is the set of integers modulo n ($\mathbb{Z}/n\mathbb{Z}$) a subgroup of $\mathbb{Z}$ with respect to addition?

I know various theorems for proving (or disproving) this question. For example:

Theorem: A finite subset of a group that is closed under the group operation is a subgroup of that group.

However, I'm not sure that the + operator in $\mathbb{Z}$ is the same as the + operator in $\mathbb{Z}/n\mathbb{Z}$.

Technically, the algorithms for computing sums in $\mathbb{Z}$ and in $\mathbb{Z}/n\mathbb{Z}$ are different, hence I believe that the two groups do not share the same operations, hence $\mathbb{Z}/n\mathbb{Z}$ should not be a subgroup of $\mathbb{Z}$.

Am I right?

  • 2
    $\begingroup$ No, all elements of $\mathbb Z$ have infinite order except for zero. All elements of $\mathbb Z/n\mathbb Z$ have finite order. $\endgroup$ Commented Mar 30, 2015 at 19:46
  • $\begingroup$ @GregoryGrant: clever! This is a consequence of Lagrange's Theorem I guess. If you turn your comment into an answer, I'm going to accept it. $\endgroup$
    – hey hey
    Commented Mar 30, 2015 at 19:47
  • 1
    $\begingroup$ An element of $\mathbb Z/n\mathbb Z$ isn't even an integer. It's a set of integers which are equivalent modulo $n$. So $\mathbb Z/n\mathbb Z$ isn't a subset of $\mathbb Z$ $\endgroup$
    – Kitegi
    Commented Mar 30, 2015 at 19:48
  • $\begingroup$ @heyhey Thank you. :-) $\endgroup$ Commented Mar 30, 2015 at 19:49
  • $\begingroup$ It doesn't really make sense to compare the two in this way; one is a set of integers, the other is a set of equivalence classes of integers. It doesn't make sense to compare $\Bbb Z_n$ with $\Bbb Z$ either (if you take $\Bbb Z_n$ to be $\{0,\ldots,n-1\}$ with addition being addition modulo $n$) since the algebraic structure isn't the same for $\Bbb Z$ and $\Bbb Z_n$, even though they are both now sets of integers. You could ask the better question: is $\Bbb Z/n\Bbb Z$ isomorphic to a subgroup of $\Bbb Z$ and the answer is, of course, no. $\endgroup$ Commented Mar 30, 2015 at 19:49

4 Answers 4


No, all elements of $\mathbb Z$ have infinite order except for zero. All elements of $\mathbb Z/n\mathbb Z$ have finite order.


To be a subgroup, the underlying set should be a subset in the first place. The elements of $\mathbb Z/n\mathbb Z$ are generally written in the form $a+n\mathbb Z$, which denotes the set $$\tag{*}\{\,\ldots, a-2n, a-n,a,a+n,a+2n,\ldots\,\}.$$ If $\mathbb Z/n\mathbb Z$ were a subset of $\mathbb Z$, then the element (*) must be an element of $\mathbb Z$ (and not the subset that it is).

However, someone may come up with the idea to pick any set of $n$ elements as underlying set of the group $\mathbb Z/n\mathbb Z$, and in doing so, why not an $n$-element subset of $\mathbb Z$? In effect, we are left with the question whether there exists an injective group homomorphims $\phi\colon\mathbb Z/n\mathbb Z\to\mathbb Z$. For in such a case, the image $f(\mathbb Z/n\mathbb Z)$ is isomorphic to $\mathbb Z/n\mathbb Z$ and is a subgroup of $\mathbb Z$. Then we could say that via that specific $f$ we might view the group $\mathbb Z/n\mathbb Z$ as a subgroup of $\mathbb Z$. But if $f\colon \mathbb Z/n\mathbb Z\to \mathbb Z$ is a group homomorphism, we see that for any $a\in\mathbb Z/n\mathbb Z$ we have that $\underbrace{a+a+\ldots+a}_n$ is the neutral element of $\mathbb Z/n\mathbb Z$, hence $f(a+a+\ldots+a)=0$. But also $f(a+a+\ldots+a)=f(a)+f(a)+\ldots+f(a)=nf(a)$. Hence (as long as $n\ne 0$) we conclude $f(a)=0$ for all $a$. This shows that $f$ is constant $0$ and cannot be injective - unless $n=0$ (in which case we cannot conclude that $f(a)=0$ for all $a$) or $n=1$ (in which case constant $0$ does not contradict injectivity).


The only subroups of $\mathbb Z$ are $n \mathbb Z$. That is if $H \subseteq \mathbb Z$ is a subgroup of $\mathbb Z$ then $H = n\mathbb Z$ for some $n \in \mathbb Z$.

Hint: Take $n = \min\{x \in H; x>0\}$.


The elements of the grp $\mathbb {Z}/n \mathbb {Z}$ are sets but the elements of $\mathbb {Z}$ are integers..so these two sets are not comparable..so not subgrp..


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