Subgroup $Q$ of $M\times M'$ is not a direct product of subgroups of $M$ & $M'$, do $M$ and $M'$ share no simple subgroups?

Question

Let $M$ and $M'$ be groups.
Let $M\times M'$ be a direct product.
If a subgroup $Q$ of $M\times M'$ is not a direct product of subgroups of $M$ and $M'$,
in other words, $Q\neq \{(m,m') \mid m\in P\le M \text{ and } m'\in P'\le M'\})$,
can we say that $M$ and $M'$ share no simple subgroups?
In other words, for every simple subgroup $S$ of $M$ and simple subgroup $S'$ of $M'$, $S$ is not isomorphic to $S'$.

Answer 1

It has been carefully explained in the comments that the answer to your question is no, we cannot say that share no simple subgroups. Let me try again.

Suppose that $M=\{1,x,x^2,\ldots,x^{p-1}\}$ and $M' = \{ 1,y,y^2,\ldots,y^{p-1}\}$ are both cyclic groups of order $p$. So $M$ and $M'$ are isomorphic simple groups, and hence they "share simple subgroups". Do you agree, or do you mean something different by "share simple subgroups"?

Anyway $M \times M'$ has the subgroup $Q=\{ (x^i,y^i) \mid 0 \le i \le p-1 \}$ which is not a direct product of subgroups of $M$ and $M'$.

I think you have got the question wrong. What is true is that if all subgroups of $M \times M'$ are of the form $N \times N'$ with $N \le M$ and $N' \le M'$ (i.e. if there are NO subgroups $Q$ of the type you describe), then $M$ and $N$ share no simple subgroups.