Non-trivial groups pairwise non-isomorphic with isomorphic direct products Are there four non-trivial groups $G_1,G_2,H_1,H_2$ pairwise non-isomorphic but such that $G_1 \times G_2$ is isomorphic to $H_1 \times H_2$?
My idea was to consider the abelian groups and something related to their decomposition as direct product of cyclic groups, but it doesn't work.
 A: Let $C_n$ denote the cyclic group with $n$ elements.
Relevant fact: If $a$ and $b$ are relatively prime, then $C_a \times C_b = C_{ab}$.
Idea of proof of fact (or look in any group theory book): It suffices to show that $C_{ab}$ is cyclic - i.e. is generated by a single element. The naive guess is $(1,1)$. Observe that $a*(1,1) = (0,a)$, and since $a$ is a unit modulo $b$ (this is equivalent to them being relatively prime), there is some c so that $ca*(1,1) = (0,1)$. Similarly you can get $(1,0)$ as a mutliple of $(1,1)$.
Do you see how to use this fact?
There are other ways also, such as what is being suggested in the comments.
A: It would be more fun if we were to insist that $G_{1}\simeq G_{2}$ and $H_{1}\simeq H_2$ and then give a counterexample.  
Tony Corner's example of an abelian group $A\not\simeq A\oplus A$ but with $A\simeq A\oplus A\oplus A$  will give an example. Take $B=A\oplus A$ and then
$$
A\oplus A\simeq A\oplus (A\oplus A\oplus A)\simeq (A\oplus A)\oplus (A\oplus A)\simeq B\oplus B
$$ 
(For the group $A$ see MR0169905 Corner, A.L.S., On a conjecture of Pierce concerning direct decomposition of Abelian groups. 1964 Proc. Colloq. Abelian Groups (Tihany, 1963) pp.43--48 Akademiai Kiado, Budapest.)
