Let $G$ be a group of order $360$ and suppose $M$ is a maximal subgroup of $G$ which is isomorphic to $A_5$. Prove that $G$ is isomorphic to $A_6$.

Let $G$ be a group of order $360$ and suppose $M$ is a maximal subgroup of $G$ which is isomorphic to $A_5$. Prove that $G$ is isomorphic to $A_6$.

Help me some hints.

Thanks a lot.

• I think we should prove that $G$ is simple so $G$ is isomorphic to $A_6$ – chuyenvien94 Jan 2 '14 at 2:30
• And see page $58,59$ of Finite Group Theory I. Martin Isaacs – chuyenvien94 Jan 2 '14 at 2:34

Since $|G| = 360$ and $|M| = |A_5| = 60$, the natural action of $G$ on cosets of $M$ induces a homomorphism $\phi : G \to S_6$. We will show that $\phi$ is an embedding whose image is $A_6$.
Now, $\ker \phi$ is the intersection of all conjugates of $M$ in $G$, i.e. $\ker \phi = \bigcap_{g \in G} gMg^{-1}$. Since this is a normal subgroup of $G$ contained in $M$, and $M$ is simple, we must have $\ker \phi = \{e\}$ (since $\ker \phi \neq M$, else $M \unlhd G$ - why is this impossible?). Thus $\phi : G \to S_6$ is an embedding.
To show that $\phi(G) = A_6$, notice that $\phi(G) \subseteq S_6$ is a subgroup of index 2. Then $\phi(G) \unlhd S_6 \implies \phi(G) \cap A_6 \unlhd A_6 \implies \phi(G) = A_6$ (note $\phi(G) \cap A_6 \neq \{e\}$, since at least half of the elements in $\phi(G)$ are even permutations).
Edit: Just to be clear on why $M \unlhd G$ is impossible, notice that a normal maximal subgroup must be of prime index, as the quotient by such a subgroup cannot have nontrivial proper subgroups.