Given nonabelian simple groups $G_1,\ldots,G_n (n\ge 2)$. If $H$ is a subgroup of $G=G_1\times \cdots \times G_n$ such that the projection of $H$ to each $G_i \times G_j$ is surjective. Prove that $H=G$.
I am trying the following approach: for any $a,b\in G_1$, there exists $h_a = (a,1,*,\cdots)$ and $h_b = (b,*,1,\cdots)$ in $H$, because of the surjectivity onto $G_1\times G_2$ and $G_1\times G_3$. Then the commutator $[h_a,h_b]\in H$ has the form $([a,b],1,1,
\cdots)$. Since $G_1$ is nonabelian simple, for any $g\in G_1$, we can write $g=[a,b]$ for some $a,b\in G_1$. Repeating this will show that $(g,1,\cdots,1)\in H$ for any $g\in G$.
This question is from a competition and it looks like this approach is a false proof because I (very likely) didn't get credits from this question. Please help me point out where it goes wrong.
Edit: Thanks for the comments. I looked up my answers and found that I didn't mention one can write $g=[a,b]$ explicitly, but went straight with a conclusion that for any $g\in G_1$ there is an element in $H$ in form of $(g,1,1,*,\cdots)$, using the fact that $G_1$ is nonabelian simple. So I guess the submitted version isn't quite wrong, but I didn't really understand everything, and I deserve a punishment for being too sloppy.
Best Answer
This approach could work but there is a serious mistake.
You have assumed that every element of the commutator subgroup of $G_i$ is a commutator. This has been proved for finite simple groups (with a very long and difficult proof) but not for infinite simple groups. This assumption is completely unnecessary in your proof - you just need the fact that $[G_i,G_i]$ is generated by the commutators.
Also "Repeating this will show that ..." is a bit vague and unclear. You should set it up as a formal induction.