I'm trying to find all fully invariant subgroups of the group $S_4$.
Meaning subgroups $H \leq S_4$, such that for all $\phi \in \operatorname{End}(S_4)$ the following holds: $\phi (H) \subseteq H$.
Of course the trivial subgroups $\{ e \} ,S_4$ are fully invariant, so I'm just interested in non-trivial subgroups.
I know that only characteristic subgroups (invariant under automorphisms) can be fully invariant. As $A_4, V_4$ are the only normal subgroups (invariant under inner automorphisms) these are the only possible subgroups that could be characteristic. That they are characteristic can be seen using the fact that being a characteristic subgroup is a transitive property and that automorphisms preserve order of elements. So $A_4$ must be characteristic in $S_4$, as it is generated by all elements of order $3$. By the same argument $V_4$ is characteristic in $A_4$, so in $S_4$.
Now I have to find out whether $A_4$ and $V_4$ are fully invariant. I'm missing arguments there, because I don't know many properties of endomorphisms on symmetric groups. Maybe someone has an idea?
Thanks!
Best Answer
Here is the solution(just the comments above in one answer):
We know, both $A_4$ and $V_4$ are characteristic subgroups.
First I show, that $A_4$ is fully invariant. We know that the order of the image divides the order of the preimage. As $A_4$ is generated by all elements of order 3, the image of these generators must have order 3 or 1. In both cases it is in $A_4$.
For showing that $V_4$ is fully-invariant, we look at an endomorphism which is not an automorphism. It follows that the kernel is non-trivial, as it is normal it must be $S_4$, $A_4$, $V_4$. In all cases $V_4$ is contained in the kernel, though mapped to $e \in V_4$.