Simple non-abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup of finite index

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Statement

Suppose G is a simple non-abelian group and H is a proper subgroup of G that is a subgroup of finite index in G. Then, G is isomorphic to a subgroup of the alternating group on the left coset space G/H.

Facts used

  1. Simple non-abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup
  2. Normality satisfies transfer condition: The intersection of a normal subgroup and any subgroup is normal in the latter subgroup.
  3. Second isomorphism theorem

Proof

Given: A simple non-Abelian group G, a proper subgroup H of finite index in G.

To prove: G is isomorphic to a subgroup of the alternating group on G/H.

Proof:

  1. By fact (1), G is isomorphic to a subgroup, say K, of \operatorname{Sym}(G/H).
  2. By definition, the alternating group, \operatorname{Alt}(G/H), is normal in \operatorname{Sym}(G/H). Thus, by fact (2), \operatorname{Alt}(G/H) \cap K is normal in K.
  3. Since K is simple, \operatorname{Alt}(G/H) \cap K = K or \operatorname{Alt}(G/H) intersects K trivially. We consider both cases:
    1. \operatorname{Alt}(G/H) \cap K = K: In this case, K \le \operatorname{Alt}(G/H), so G is isomorphic to a subgroup, K, of \operatorname{Alt}(G/H).
    2. \operatorname{Alt}(G/H) \cap K is trivial: In this case, the second isomorphism theorem (fact (3)) yields that K \operatorname{Alt}(G/H)/\operatorname{Alt}(G/H) \cong K. The left side is a subgroup of \operatorname{Sym}(G/H)/\operatorname{Alt}(G/H), which is a group of order two. But a group of order two has no simple non-Abelian subgroups, so this case is not possible.