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

## 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.