Prove there is $\sigma\in S_3$ such that $H_{\sigma (i)} \cong\textrm{}K_i ,\space \forall i $

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In class they gave us a problem, After spending a long time trying to solve it, I turn to you =]

Let $H_1,H_2,H_3, K_1,K_2,K_3 \le G$ be simple groups, $G=\{{h_1}{h_2}{h_3}:h_1\in H_1,h_2\in H_2,h_3\in H_3\}=\{{k_1}{k_2}{k_3}:k_1\in K_1,k_2\in K_2,k_3\in K_3\}$

Prove there is $\sigma\in S_3$ such that $H_{\sigma (i)} \cong\textrm{}K_i ,\space \forall i $


Trying to solve this I realized, I don't know anything about the behavior of the Groups $H_i$,$K_j$,
I Can't prove that they are normal in G
I Can't see if they intersect in {e} only
I Can't Claim $H_1*H_2$ is a group, for example
And I cannot find any composition series(Jordan-Holder) of G since I cannot find any normal subgroup of G.

Any Ideas?