Finding all normal subgroups

Question

Stuck on finding the normal subgroups of the symmetry group of a cube. Trying to find 9 normal subgroups not too sure how to get the rest. Since I'm given 5 homomorphisms and I know the kernel of a homomorphism is a normal subgroup, I've identified.

1. ker($\phi$)
2. ker($det$)
3. ker($p$)
4. ker($q$)
5. ker($f$)

Now for 6, 7, 8 and 9. I know if H and K are normal subgroups, HK is. So,

6. $HK$ where $H = \operatorname{ker}(\phi)$ and $K = \operatorname{ker}(p)$

7. $HK$ where $H = \operatorname{ker}(p)$ and $K = \operatorname{ker}(\phi)$

8. $HK$ where $H = \operatorname{ker}(det)$ and $K = \operatorname{ker}(\phi)$

9. $HK$ where $H = \operatorname{ker}(det)$ and $K = \operatorname{ker}(p)$

Now while this is what I got, I'm skeptical if I really approached this right. Would appreciate any assistance. Especially because I have to organize this in a lattice after and if the subgroups are wrong, all the subsequent questions are wrong.

Answer 1

Just a few hints:

First of all, there are two trivially normal subgroups, namely $1$ and the group $S(C)$ itself.

Second, since (as the comment indicates) $f$ is an automorphism of $S(C)$, its kernel is trivial.

Third, it is a good idea to compute the orders of $\ker(\phi)$, $\ker(\det)$, $\ker(p)$ and $\ker(q)$.

Fourth: Show that $\ker(\phi)\le\ker(p)$ and that $\ker(\det)\cap\ker(p)\le\ker(q)$. What is the order of $\ker(\det)\cap\ker(p)$? Can you construct another nontrivial normal subgroup which involves $\ker(\det)\cap\ker(p)$?

And finally: What does $Z(S(C))$ look like?

Answer 2

As an addition to my other answer as well as the various comments, here's a drawing of the lattice of normal subgroups of the group $S(C))$:

Note that the three maximal normal subgroups intersect in a normal subgroup of order $12$ (which is, by the way, isomorphic to an $A_4$), and the intersection of that subgroup with $\ker(\phi)$ is an elementary abelian group of order $4$.