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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), 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 $M = (K_1 \cap K_2) \rtimes (\mathbb{Z}/2)^2$?

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

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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 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 of order $2n$, because any finite quotient of the group $\mathbb{Z}/2 * \mathbb{Z}/2$ admits the presentation $\langle a,b \mid a^2, b^2, (ab)^n \rangle$.

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