If $G,H$ are groups and $H$ is not trivial then $G \times H \ncong G$ My instinct tells me this should be true, and I hope there aren't any weird counterexamples. If $G,H$ are groups and $H$ is not trivial, then I think $G \times H \ncong G$ (i.e., $G \times H$ is not isomorphic to $G$). Surely this is right? Is there an elementary group theory based argument that I'm not seeing?
 A: This is false.  For instance, let $H$ be any nontrivial group and let $G=H^\mathbb{N}$, an infinite product of copies of $H$.  Then $G\times H\cong G$, since adding one more copy of $H$ to the product doesn't change the number of copies.
A: Let $H$ be any non-trivial group, and $G=\prod_{n=1}^\infty H$.
Then $G\times H\cong G$.
A: As others have pointed out, the claim is false.  However, if $G$ can be expressed as a finite direct product of groups each of which is itself indecomposable (i.e., cannot itself be expressed as a non-trivial direct product), then the claim is true by the Krull-Schmidt Theorem (which isn't the most obvious result in the world), so I would hardly call this elementary even in that case.  
A: For sake of completeness, I want to add something. 
All counterexamples mentioned here were infinitely generated; but what if both $G$ and $H$ are finitely generated? You'd say that it's completely impossible for f. g. group to be isomorphic to its own square, but it turns out that almost every reasonable nontrivial statement about all f. g. groups is false — David Meyer constructed an example of finitely generated group $G$ such that $G \cong G \times G$, https://doi.org/10.1112/jlms/s2-26.2.265 5 years Later Gilbert Baumslag constructed an example of finitely presented group isomorphic to its square. (G. Baumslag, C. F. Miller III, Some odd finitely presented groups).
