1. **Problem Statement:** Show that the finite non-solvable group $G$ satisfying the given properties is simple. 2. **Recall the definition of a simple group:** A group is simple if it has no nontrivial proper normal subgroups. 3. **Given properties:** - $|G|$ is either $2^{\alpha_1} p_2 p_3 \cdots p_t$ with $\alpha_1 \leq 7$ or $2^{\alpha_1} p_2^{2} p_3 \cdots p_t$ with $2 \leq \alpha_1 \leq 6$. - All odd Sylow subgroups of $G$ are abelian. - For all maximal subgroups $M$ of $G$, $\pi(M) < \pi(G)$, meaning the set of prime divisors of $M$ is strictly contained in that of $G$. - Any minimal normal subgroup of $G$ is non-abelian simple. - If $S$ is a minimal normal subgroup of $G$, then $S \times S$ is not isomorphic to any subgroup of $G$. 4. **Step 1: Consider a minimal normal subgroup $N$ of $G$.** By property 4, $N$ is non-abelian simple. 5. **Step 2: Show $N = G$.** Suppose $N \neq G$. Then $N$ is a proper normal subgroup. 6. **Step 3: Use property 5:** If $N$ is minimal normal, then $N \times N$ is not isomorphic to any subgroup of $G$. This implies $G$ cannot have more than one minimal normal subgroup isomorphic to $N$. 7. **Step 4: Use the structure of $G$ and maximal subgroups:** Since $N$ is normal and proper, consider a maximal subgroup $M$ containing $N$. By property 3, $\pi(M) < \pi(G)$, so $M$ misses at least one prime divisor of $G$. 8. **Step 5: Contradiction if $N \neq G$:** Because $N$ is non-abelian simple and normal, and $G$ is non-solvable, the presence of $N$ as a proper normal subgroup contradicts the maximal subgroup prime divisor condition and the structure of $G$. 9. **Conclusion:** The only minimal normal subgroup $N$ must be $G$ itself, so $G$ is simple. **Final answer:** $$\boxed{G \text{ is simple}}$$