Prove that a simple group of order 168 is isomorphic to a subgroup of A_8

Asked Dec 05 '17 at 17:02

Active Dec 05 '17 at 17:02

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Let $G$ be a simple group of order $168=2^3 \cdot 3 \cdot 7$. By simplicity we get that $n_7=8$. Let $G$ act on the set of $7$-Sylow P subgroups (of which there are 8).

This induces an injective homomorphism $\psi:G \to S_8$, and by the first isomorphism theorem $G/\ker(\psi) = G \cong \psi(G)$ How do I show that $\psi(x)$ is necessarily even to complete this proof?

asked Dec 05 '17 at 17:02

How do you know the homomorphism is injective? – Arthur Dec 05 '17 at 17:04
Otherwise $\psi(G) \cap A_8$ would be a normal subgroup of index $2$ in $\psi(G)$, contradicting its simplicity. – Derek Holt Dec 05 '17 at 17:05

Prove that a simple group of order 168 is isomorphic to a subgroup of A_8

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