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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.