No group of order 10,000 is simple

[Mr Davis 97]

Homework Statement

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

Homework Equations

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.

 

[member 587159]

Perhaps a late answer, but it looks ok to me.