You ensure that at least $n!-k+1$ of the products of the $g_i$ are equal -- but can you also prove that the remaining $k-1$ are pairwise distinct? -- Your relations might force some of them to be equal as well, or at least I don't see why they don't.

@Someone: I think the sheer size of the wreath product renders this approach useless -- if the groups $N$ and $H$ have both order $p^n$, their standard wreath product has order $p^{n(p^n+1)}$.

@Marc: The OP asked for a proof of the implication "three squares theorem => four squares theorem" rather than for a proof of the three squares theorem.

It's quite plausible that there are heuristics suggesting the existence of counterexamples. -- But the abc conjecture doesn't tell very much directly: what we have is $a+b=c$ where $a,b,c$ are coprime and the squarefree part $q(abc)$ of $abc$ equals the product of the first $n$ primes. Now the abc conjecture asserts that for any $\epsilon > 0$ there is a constant $C_{\epsilon}$ such that $c < C_{\epsilon} \cdot q(abc)^{1+\epsilon}$. Since not all of the first $n$ primes divide $c$ (in general maybe only something like half of them), there remains still some room to vary exponents of powers.

I think this doesn't make a difference -- if $G$ works for $k_1$ and $H$ works for $k_2$, then $G \times H$ works for both.

@Alain: The notation ${\rm L}_n(q)$ for ${\rm PSL}(n,q)$ is often used by people working on finite simple groups, in the spirit that series of simple groups deserve simple (i.e. one-letter) names.