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.
Best Answer
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.