Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(H_1 \subset G_1)$ and $(H_2 \subset G_2)$ be core-free maximal inclusions of finite groups.

Their product, the inclusion $(H_1 \times H_2 \subset G_1 \times G_2)$, admits four obvious intermediate subgroups : $H_1 \times H_2$, $G_1 \times G_2$, $H_1 \times G_2$ and $G_1 \times H_2$.

Question : If $(H_1 \times H_2 \subset G_1 \times G_2)$ admits a non-obvious intermediate subgroup, what are the properties of $(H_1 \subset G_1)$ and $(H_2 \subset G_2)$ ?

Here are examples of "prospective" properties, ordered from the weakest to the strongest :

  • $[G_1 : H_1]=[G_2 : H_2]$ ?
  • $(H_1 \subset G_1) \sim (H_2 \subset G_2)$ ?
  • $(H_1 \subset G_1) \sim (H_2 \subset G_2) \sim (\{e\} \subset \mathbb{Z}_p)$ ?
share|improve this question

2 Answers 2

up vote 2 down vote accepted

If I understand correctly, each $H_i$ is maximal in $G_i$ and core-free. In particular, unless $G_i$ has prime order $p$, $H_i$ is not normal in $G_i$.

Let $X$ be a subgroup of $G_1 \times G_2$. For $i=1,2$ let $X^i$ be the projection of $X$ onto $G_i$ and let $X_i$ be the intersection of $X$ with $G_i$. Then $X_i$ is a normal subgroup of $X^i$, and $X^1/X_1 \cong X^2/X_2$. If $X$ contains $H_1 \times H_2$ as in your situation, then each $X_i$ contains $H_i$. As $H_i$ is maximal in $G_i$, $X_i \in \{H_i,G_i\}$. If I have understood your assumptions correctly, unless $G_i$ has prime order $p$, it follows that $X_i=X^i$ (in which case also $X_{3-i}=X^{3-i}$ and $X=X^1 \times X^2$). So, unless each $G_i$ has prime order $p$, the four groups you list are the only intermediate subgroups.

share|improve this answer
    
Thank you for your answer. You seems to have correctly understood my assumptions. For my part, if I understand your conclusion correctly, you have proved that $G_1 \simeq G_2 \simeq \mathbb{Z}_p$, it's wonderful! Is it a usual result? Now I need to read your proof very closely because I'm not an expert. –  Sébastien Palcoux Feb 17 at 11:18
    
Perhaps your argument is generalizable by induction for proving the following result : let $(H_i \subset G_i)$ be core-free maximal inclusions of groups ($i \in I$), then $(\prod_{i∈I} H_i \subset \prod_{i∈I} G_i)$ admits a non-obvious intermediate subgroup (i.e. other than the $2^{|I|}$ obvious intermediates) iff $\exists i_1,i_2 \in I$, $ i_1 \neq i_2$ and $p$ prime such that $G_{i_1} \simeq G_{i_2} \simeq \mathbb{Z}_p $. –  Sébastien Palcoux Feb 17 at 12:00
    
Perhaps there is a way to use your argument here for answering this new post : Is $\mathcal{L}(H_1 \times H_2 \subset G_1 \times G_2)$ modular if $\mathcal{L}(H_i \subset G_i)$ distributive? –  Sébastien Palcoux Feb 18 at 12:02

Here is a development of the argument of John Shareshian :

Theorem: Let $(H_i \subset G_i)$ be core-free maximal inclusions of groups, then $(H_1 \times H_2 \subset G_1 \times G_2)$ admits a non-obvious intermediate subgroup iff $G_1 \simeq G_2 \simeq \mathbb{Z}_p$.

Proof: $H_i$ is core-free in $G_i$ (i.e. if $K \subset H_i$ is a normal subgroup of $G_i$ then $K=\{ e \}$), so if $H_i$ is a normal subgroup, then $H_i=\{ e \}$, and $G_i \simeq \mathbb{Z}_p$ ($p$ prime) by maximality assumption.
So $G_i \not\simeq \mathbb{Z}_p$ (for a $p$ prime) iff $H_i$ is not normal in $G_i$.

Let $X$ be a subgroup of $G_1 \times G_2$.

Let $X^1 = \{g_1 \in G_1 : \exists g_2 \in G_2 \text{ with } (g_1,g_2) \in X \}$ the projection of $X$ onto $G_1$.
Let $X_1 = \{h \in G_1 : (h,e) \in X \}$ the intersection of $X$ with $G_1 \times \{e \}$, projected onto $G_1$.

$X_1$ is a normal subgroup of $X^1$ because if $g_1 \in X^1$ and $h \in X_1$, then $g_1^{-1}hg_1 \in X_1$ because $(g_1^{-1}hg_1,e)=(g_1,g_2)^{-1}(h,e)(g_1,g_2) \in X$

By the same way, we define $X^2$ and a normal subgroup $X_2$.
Let $\phi : X^1/X_1 \to X^2/X_2$ with $\phi(g_1X_1)= g_2X_2$ such that $(g_1,g_2) \in X$
It is well-defined because if $(g_1^{-1}g'_1,g_2^{-1}g'_2) \in X_1 \times X_2$, then $(g'_1,g_2), (g_1,g'_2), (g'_1,g'_2) \in X$
It's obviously surjective, and injective because if $\phi(g_1X_1)= X_2$ then $(g_1,e) \in X$ and so $g_1 \in X_1$
Conclusion : $X^1/X_1 \simeq X^2/X_2$

Now if $X$ is an intermediate subgroup of $(H_1 \times H_2 \subset G_1 \times G_2)$ then $H_i \subset X_i$,
and the maximality assumption forces $X_i, X^{i} \in \{ H_i,G_i \}$.

Next if $G_1 \not\simeq \mathbb{Z}_p$ then $H_1$ is not normal in $G_1$, but $X_1$ is normal in $X^1$, so $X_1=X^1=H_1$ or $G_1$.
But, $X^1/X_1 \simeq X^2/X_2$, so $X_2=X^2=H_2$ or $G_2$ as well.
It follows that if $(g_1,g_2) \in X$ then $(g_1,e), (e,g_2) \in X$, so that $X = X_1 \times X_2$
Conclusion $X = H_1 \times H_2$, $G_1 \times H_2$, $H_1 \times G_2$ or $G_1 \times G_2$, i.e. there is not non-obvious intermediate.

Idem if $G_2 \not\simeq \mathbb{Z}_q$, ...

Finally , if $G_1 \simeq \mathbb{Z}_p$ and $G_2 \simeq \mathbb{Z}_q$, there is a non-obvious intermediate subgroup iff $p=q$. $\square$


Definitions: A lattice $(L, \wedge, \vee)$ is :
- Distributive if $a∨(b∧c) = (a∨b) ∧ (a∨c)$
- Modular if $a ≤ c \Rightarrow a ∨ (b ∧ c) = (a ∨ b) ∧ c$
$(\forall a,b,c \in L)$
Obviously, distributive $\Rightarrow$ modular.

Theorems : A finite group $G$ is
- Cyclic iff $\mathcal{L}(G)$ is distributive (Ore 1938)
- Abelian iff $\mathcal{L}(G \times G)$ is modular (Lukacs-Palfy 1986)
(see here thm2.3 p431 and thm6.5 p449)

Corollary: If $(H_i \subset G_i)$ are maximal inclusions of groups, then the lattice of intermediate subgroups $\mathcal{L}(H_1 \times H_2 \subset G_1 \times G_2)$ is modular.

Proof : If $(H_1 \times H_2 \subset G_1 \times G_2)$ admits no non-obvious intermediate subgroup, then the lattice is isomorphic to $\mathcal{L}(\mathbb{Z}_6)$ which is distributive (because $\mathbb{Z}_6$ cyclic), so modular.
Else, the lattice is $\mathcal{L}(\mathbb{Z}_p \times \mathbb{Z}_p)$ and is modular because $\mathbb{Z}_p$ is abelian. $\square$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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