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**1. The problem statement, all variables and given/known data**

Show that there is no simple group of order ##10^4##.

**2. Relevant equations**

**3. The attempt at a solution**

By way of contradiction, suppose ##G## is simple and ##|G| = 10000 = 5^42^4##. Sylow theory gives ##|\operatorname{Syl}_2(G)| = 1## or ##16##. If ##|\operatorname{Syl}_2(G)| = 1##, then there is a Sylow 2-subgroup that is normal, and so we would have a contradiction. So suppose that ##|\operatorname{Syl}_2(G)| = 16##. Consider the action of ##G## on ##\operatorname{Syl}_2(G)## by conjugation and let $$\phi : G \to S_{16}$$ be the associated permutation representation. The map ##\phi## is nontrivial since the action is transitive by the second part of Sylow theory, which says that all Sylow p-subgroups are conjugate of each other. This show that the kernel of ##\phi## is not all of ##G##. Also, note that ##10^4## does not divide ##16!##, since ##16! = 2^{15}×3^6×5^3×7^2×11×13##, and this prime factorization does not contain ##5^4##. Hence ##\phi## is not injective, and so the kernel is not trivial. Hence ##\ker(\phi)## is a proper nontrivial normal subgroup of ##G##, which contradicts out assumption that ##G## is simple.