Is every direct product of (finite) cyclic groups abelian?

Question

Long story short I'm halfway through a proof and have hit a step where I show that a group order $p^2$ is abelian. I did this by splitting into the $C_{p^2}$ case and the $C_p\times C_p$ case, the first being trivial as all cyclic groups are abelian. The second case I proved by using the definition of the direct product and showing it directly from the definition of the two $C_p$ groups.

However this feels odd. Something feels off - especially added to the fact that my proof seemingly holds for any two cyclic groups, implying the direct product of any two cyclic groups is abelian, and yet I can't find anything online talking about this. Is it just so trivial nobody mentions it? The only thing I consistently see brought up is that "every cyclic group is an abelian group" (trivial), and "every finitely generated abelian group is a direct product of cyclic groups", which is the converse of what I'm trying to show.

Answer 1

Yes.

If $G\cong \langle a\mid a^m\rangle, H\cong\langle b\mid b^n\rangle$, then

$$G\times H\cong\langle a,b\mid a^m, b^n, ab=ba\rangle.$$

See here for details.

Answer 2

If you look at the way the direct product is defined, in particular the group operation, it's pretty easy to see that the answer is yes. For we have $(a,b)*(c,d)=(ac,bd)=(ca,db)=(c,d)*(a,b)$.

Thus every direct product of abelian groups is abelian.