Let $H\_1$ and $H\_2$ be two *distinct* index $2$ subgroups of a finite group $G$.  

We can deduce several properties about the intersection $H\_1 \cap H\_2$:


1. $H\_1$ and $H\_2$ are normal subgroups of $G$. Then $H\_1 \cap H\_2$ is also a normal subgroup of $G$.
2. $|G:H\_1 \cap H\_2| \le |G:H\_1| \cdot |G:H\_2|$ (see [here](https://math.stackexchange.com/a/170918/84284)), so $|G:H\_1 \cap H\_2| = 4$.
3. It follows that $G/(H\_1 \cap H\_2) \simeq (\mathbb{Z}/2)^2$.


Let $(N \subset M)$ be an irreducible finite index inclusion of hyperfinite ${\rm II}\_1$ factors.  

Let $K\_1$ and $K\_2$ be two *distinct* intermediate subfactors $N \subset K\_i \subset M$, such that $|M:K\_i| = 2$.


*Question*: What can be extended from the properties above? More precisely:


1. $K\_1$ and $K\_2$ are depth $2$ subfactors. Is $K\_1 \cap K\_2$ also of depth $2$?
2. Is it true that $|M:K\_1 \cap K\_2| \le |M:K\_1| \cdot |M:K\_2|$, and so that $|M:K\_1 \cap K\_2| = 4$?
3. Is it true that $M = (K\_1 \cap K\_2) \rtimes (\mathbb{Z}/2)^2$?




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**Edit:** It is false in general for 2. and 3. (see answer). So now only 1. is asked.



On the intersection of index 2 subfactors
It is false in general for 2. and 3. 


Let $R$ be the hyperfinite ${\rm II}\_1$ factor, and take the symmetric group $S\_3$ acting outerly on $R$.  

Now take the subfactor $(R^{S\_3} \subset R)$. Take the intermediate $K\_1= R^{\langle (1,2) \rangle}$ and $K\_2=R^{\langle (1,3) \rangle}$.  

Then $|R:K\_i| = 2$, but by Galois correspondence $R^{\langle (1,2) \rangle} \cap R^{\langle (1,3) \rangle} = R^{\langle (1,2),(1,3) \rangle} = R^{S\_3} $.  

It follows that $|R: R^{\langle (1,2) \rangle} \cap R^{\langle (1,3) \rangle}| = 6 > 4 $. So 2. and 3. are false in general. 


The point 1. is true for any finite group-subgroup subfactor $(R^G \subset R^H)$ because if $H\_1$ and $H\_2$ are two distinct intermediate subgroups $H \subset H\_i \subset G$, such that $H$ is a normal subgroup of $H\_i$, then $H$ is also a normal subgroup of $\langle H\_1, H\_2 \rangle$ 


The point 1. is true in general, see the last paragraph of [this answer](https://mathoverflow.net/a/249776/34538) of Zhengwei.  

In the same answer, Zhengwei shows what are the true statements for 2. and 3.  

2. $|M:K\_1 \cap K\_2| = 2n$  

3. $(K\_1 \cap K\_2) = M^{D\_{2n}}$ with $D\_{2n}$ the [dihedral group](https://fr.wikipedia.org/wiki/Groupe_di%C3%A9dral) of order $2n$, because any finite quotient of the group $\mathbb{Z}/2 \* \mathbb{Z}/2$ admits the presentation $\langle a,b \mid a^2, b^2, (ab)^n \rangle$.


