Why $\mathbb{Z}/n\mathbb{Z}$ isn't a subgroup of $\mathbb{Z}$ Could anyone explain to me why this isn't true? It's listed as an example in our textbook but no reason is given. I've checked the properties of a subgroup, and it seems to follow them. What am I missing?
 A: Usually $\mathbb{Z}/n\mathbb{Z}$ is thought to as the set of equivalence classes under congruence modulo $n$, so its elements are not in $\mathbb{Z}$, but are rather subsets of $\mathbb{Z}$.
Or you can identify $\mathbb{Z}/n\mathbb{Z}$ with a subset of $\mathbb{Z}$, precisely $\{0,1,2,\dots,n-1\}$, but then the operation is not the same as in the integers, because the sum between $(n-1)$ and $1$ is $0$ and not $n$. Thus the conditions for being a subgroup are not satisfied.
Note that it doesn't matter what subset of $\mathbb{Z}$ you identify $\mathbb{Z}/n\mathbb{Z}$ with; for instance $\{1,2,\dots,n\}$ would be as good, and $\{-n+1,-n+2,\dots,-1,0\}$ too. But in any case the operation wouldn't be the one inherited by $\mathbb{Z}$.
This is due to the fact that the group $\mathbb{Z}/n\mathbb{Z}$ has an element of order $n$, which $\mathbb{Z}$ hasn't, if $n>1$, so $\mathbb{Z}/n\mathbb{Z}$ can't even be isomorphic to a subgroup of $\mathbb{Z}$. The only cases in which this happens are precisely $n=1$ (with $\mathbb{Z}/1\mathbb{Z}$ isomorphic to $\{0\}$) and $n=0$ (with $\mathbb{Z}/0\mathbb{Z}$ isomorphic to $\mathbb{Z}$).
A: Even when "subgroup" is taken to mean "isomorphic to a subgroup that is a subset", this doesn't work out: For example, every element of $\mathbb Z$ has order $1$ or $\infty$, but $\mathbb Z/n\mathbb Z$ contains an element of order $n$, namely $1+n\mathbb Z$. Hence, there is no monomorphism from $\mathbb Z/n\mathbb Z$ to $\mathbb Z$, and therefore no isomorphism from $\mathbb Z/n\mathbb Z$ to a subgroup of $\mathbb Z$.
A: You could think of $\mathbb Z/n\mathbb Z$ as the set $\{0,1,2,\ldots,n-1\}$ with a funny addition, but then it is not a subgroup because the group operation on this subset does not match the operation on $\mathbb Z$. E.g. $n-1+1=0$ if you are thinking of this set as $\mathbb Z/n\mathbb Z$, but $n-1+1=n\neq 0$ if you are thinking of the usual group operation on integers.
A: It is related to $\mathbb Z$ in that it is a quotient group - and that is, indeed, what the notation $\mathbb Z/n\mathbb Z$ captures. In particular, any subgroup of $\mathbb Z$ can be written as $n \mathbb Z$ - to prove that, consider that if $G$ is a subgroup of $\mathbb Z$ and $a,b\in G$, then all multiples of $\gcd(a,b)$ are in $G$ as well, as they are linear sums of $a$ and $b$. So, all multiples of $\gcd(G)$ are in $G$ every element of $G$ is clearly a multiple of $\gcd(G)$, so $G=\gcd(G)\mathbb Z=n\mathbb Z$. (And obviously $\mathbb Z/n\mathbb Z$ is not isomorphic to $n\mathbb Z$ since there are either infinitely many elements or $1$ element of $n\mathbb Z$ whereas $\mathbb Z/n\mathbb Z$ has $n$ elements)
However, each of these subgroups has a number of cosets - for instance, for $n=2$, we get the even and odd numbers - $2\mathbb Z$ and $2\mathbb Z + 1$. As it happens, since the subgroups $n\mathbb Z$ are normal (that is, $a+(n\mathbb Z)=(n\mathbb Z)+a$ - so left and right cosets are the same), addition is well-defined on these cosets - meaning, for instance, if we sum two even numbers, we know we get another even number. More generally, knowing the parities of the summands, we know the parity of the sum. This allows us to do arithmetic on the cosets $2\mathbb Z$ and $2\mathbb Z + 1$. The group resulting is known as $\mathbb{Z}/2\mathbb{Z}$, since it is what happens when you identify every element of the subgroup $2\mathbb Z$ as the identity and perform arithmetic on cosets.
A: $\mathbb{Z}/n \mathbb{Z}$ isn't isomorphic to a subgroup of $\mathbb{Z}$ because $\mathbb{Z}$ is torsion free, for example.
