# Common quotients of direct products

In a paper I read recently, the authors use the fact that if two groups G and H have no nontrivial common quotient, then neither do GxG and HxH. It's unclear from the context whether this is true for all groups, or just groups of the type that are important for this paper, and they don't prove the claim.

I've been trying to prove the statement for general groups without any success. Is it true?

I've also tried restricting to the case I really care about, as follows. Let K, K', M, and M' be normal subgroups of U, with $U/K \simeq U/K'$ and $U/M \simeq U/M'$. Suppose that KK' = KM = KM' = K'M = K'M' = MM' = U. Is it also the case that $(K \cap K')(M \cap M') = U$? The reason this is connected to the above is that $U/(K \cap K')(M \cap M')$ is a common quotient of $U/(K \cap K') \simeq U/K \times U/K' \simeq U/K \times U/K$ and $U/(M \cap M') \simeq U/M \times U/M' \simeq U/M \times U/M$. So if U/K and U/M have no nontrivial common quotients, is the same true of $U/(K \cap K')$ and $U/(M \cap M')$?

Finally, if the result is not true in general for either of these cases, what about if we restrict to finite groups G and H?

-

If $Q$ is a quotient of $G\times G$, then it has a normal subgroup $A$ (the image of $G\times 1$) such that both $A$ and $Q/A$ are quotients of $G$. If it is also a quotient of $H\times H$, then it has a normal subgroup $B$ such that both $B$ and $Q/B$ are quotients of $H$. Now assume that every common quotient of $G$ and $H$ is trivial. Then $Q/AB$ is trivial, so $AB=Q$. So $Q/A=AB/A=B/(A\cap B)$ is trivial, so $A=Q$. Likewise $B=Q$. So $Q$ is trivial.
Nice!  – Victor Protsak Jul 17 '10 at 19:38