Showing $\mathbb{Z_n}$ is abelian So I'm reviewing my notes and I just realized that I can't think of how to show that a particular integer mod group is abelian. I know how to do it with symmetric but not with integers themselves.
For example, lets say I was asked to show $\mathbb{Z_5}$ is abelian.
I know for symmetric groups, lets say $S_5$ i can pick two elements in $S_5$, for example (123),(23) and if (123)(23)=(23)(123) then I know it is abelian but how would I go about the integers?
 A: The symmetric group $S_n$ is not abelian for $n\geq 3$ since $(12)(123)=(23)$ and $(123)(12)=(13)$.
On the other hand any cyclic group is abelian. This is because in reality every element $m$ can be thought of as $\underbrace{1+1+1\dots+1}_{\text{m times}}$.
And so
$m+n=(\underbrace{1+1+1\dots+1}_{\text{m times}})+(\underbrace{1+1+1\dots+1}_{\text{n times}})=(\underbrace{1+1+1\dots+1}_{\text{n times}})+(\underbrace{1+1+1\dots+1}_{\text{m times}})=n+m$
Where the equality in the middle is just the associativity rule.
A: Note that symmetric groups are not Abelian unless $n < 3$.  See my answer here for a proof.
As for how to see that $\Bbb{Z}_n$ is Abelian, note that the group $\Bbb{Z}$ is Abelian.  Therefore, any quotient group of $\Bbb{Z}$ by a subgroup is also Abelian.  Since $\Bbb{Z}_n \cong \Bbb{Z} / n \Bbb{Z}$, we are done.
A: What you really want to show, for any cyclic group $G = \langle a\rangle$, is that:
$a^ka^m = a^ma^k$ for any integers $k,m$.
This takes care of $\Bbb Z$ and $\Bbb Z_n$ (for any $n$) "all at once".
Can you show how to leverage the "rules of exponents" and commutativity of addition in the integers, here?
