# Are these two inclusions of finite groups, equivalent?

Definition : Let $\sim$ be the equivalence relation on inclusions of finite groups, generated by :
$(H \subset G) \sim (\phi(H) \subset \phi(G))$, with $\phi: G \to L$ a group morphism, and $ker(\phi) \subset H$.

Question : $(\langle (1234) \rangle \subset S_4 ) \sim (\langle (12),(34) \rangle \subset S_4 )$ ?

Warning: Of course, there is no $\phi \in Aut(S_4)$ with $\phi(\langle (1234) \rangle) = \langle (12),(34) \rangle$, because $\langle (1234) \rangle \simeq \mathbb{Z}_4 \not\simeq \mathbb{Z}^2_2 \simeq \langle (12),(34) \rangle$, that why $\sim$ is the equivalence relation generated by the relation above, not just the relation alone.

Remark : the non-trivial normal subgroups of $S_4$ are $A_4$ and $\langle (12)(34), (13)(24) \rangle$ (see here).

Motivation : These inclusions are famous in the group-subgroup subfactor theory (see here p47).

• No. If $H \subset G$ is equivalent to $X \subset Y$ then the permutation action of $G$ on the cosets of $H$ is equivalent to the permutation action of $Y$ on the cosets of $X$. However, those two groups of degree 6 are not permutation isomorphic. – Jack Schmidt Feb 15 '14 at 22:16
• @JackSchmidt : I have developed your comment as an answer, I hope it's what you had in mind. – Sebastien Palcoux Feb 16 '14 at 15:41

I'm not an expert in finite groups theory, so I will try to develop the comment of Jack Schmidt :

Let $(H \subset G)$ and $\phi:G \to L$ with $ker(\phi) \subset H$.
Let $\Omega=G/H$ and $\Omega' = \phi(G) / \phi(H)$.

Let the map $\delta : \Omega \to \Omega'$ : $\delta(gH) = \phi(g)\phi(H)$
$\delta$ is well-defined because if $g_1^{-1}g_2 \in H$ then $\phi(g_1)^{-1}\phi(g_2) = \phi(g_1^{-1}g_2) \in \phi(H)$
$\delta$ is then obviously surjective, and it's injective because if $\phi(g_1)^{-1}\phi(g_2) \in \phi(H)$ then $g_1^{-1}g_2 \in H$, because for $h \in H$, $\phi^{-1} (\{\phi (h) \}) = h.ker(\phi) \subset H$.

Let $\pi: G \to S_{\Omega} : \pi(g_1)(g_2H) = g_1g_2H$, then $\pi(G) \simeq G/ker(\pi)$
$ker(\pi) = \{ g_1 \in G: g_1g_2H = g_2H, \forall g_2 \in G \} = \bigcap_{g_2 \in G} g_2^{-1}Hg_2 = core_{G}(H) \subset H$ is the normal core of $H$ in $G$ i.e the largest normal subgroup of $G$ contained in $H$.
Idem, let $\pi': \phi(G) \to S_{\Omega'}$, then $\pi'(\phi(G)) \simeq \phi(G)/ker(\pi')$

Now $\pi(G)$ and $\pi'(\phi(G))$ are permutation isomorphic because $\delta$ is a bijection and obviously $\delta(\pi(g_1)(g_2H)) = \pi'(\phi(g_1))(\phi(g_2)\phi(H))$; in particular $\pi(G) \simeq \pi'(\phi(G))$.

Conclusion, the equivalence relation generated by: $$(H \subset G) \sim (\phi(H) \subset \phi(G)) \text{ with } \phi: G \to L \text{ morphism and } ker(\phi) \subset H$$ implies this equivalence relation $\sim'$ as above (equivalence of permutation actions on the cosets).

Optional question : What's about the converse ?

If $(\langle (1234) \rangle \subset S_4 ) \sim (\langle (12),(34) \rangle \subset S_4 )$, then $(\langle (1234) \rangle \subset S_4 ) \sim' (\langle (12),(34) \rangle \subset S_4 )$ and so $\langle (1234) \rangle \simeq \langle (12),(34) \rangle$, because the normal cores are trival, contradiction.
The normal cores are trival because there is not non-trivial normal subgroup of $S_4$ contained in $\langle (1234) \rangle$ or $\langle (12),(34) \rangle$, thanks to the classification of the normal subgroups of $S_4$ given above.
• This seems fine. I think the equivalence relation looks like $(H \subset G) \to (H/H_G \subset G/H_G) \cong (X/X_Y \subset Y/X_Y) \leftarrow (X \subset Y)$ where $K_L$ is the largest normal subgroup of $L$ contained in $K$. In other words, every inclusion is equivalent to a core-free inclusion, and two inclusions are equivalent iff there core-free inclusions are permutation isomorphic. – Jack Schmidt Feb 16 '14 at 15:46