Subgroups of an infinite product

Question

$G := \prod\limits_{n=1}^{\infty} (\mathbb{Z}/3^{n} \rtimes \mathbb{Z}/2 )$

$g_m = ((m,0),(m,0), (m,0),\dots) \in G$

$H = \prod\limits_{n=1}^{\infty} (\{0\},\mathbb{Z}/2 ) \subset G$

Questions:

• What are the subgroups $K_m:= \langle g_m,H \rangle$ ?

• Does $K_r \subsetneq K_s$ iff $s<r$ and $s\mid r$?

• What is the index $[K_s:K_r]?$

Notation: $\mathbb{Z}/k$ is the cyclic group $\mathbb{Z}/k\mathbb{Z}$, and $\mathbb{Z}/k \rtimes \mathbb{Z}/2$ is the dihedral group $D_k$.

Context and application: An inclusions of finite groups $(H \subset G)$ produces the subfactor $(R \rtimes H \subset R \rtimes G)$ of index $[G:H]$ (with $R$ the hyperfinite ${\rm II}_{1}$ factor). Every subfactor do not come from groups (because a subfactor can have a non-integer index) but those coming from groups are very useful because it is much more easy to compute with (there are also integral index subfactors not coming from groups). There is a Galois correspondence: an intermediate subfactor $P$ of $(R \rtimes H \subset R \rtimes G)$ [i.e. $R \rtimes H \subset P \subset R \rtimes G$ with $P$ a factor] is given by an intermediate subgroup $K$ of $(H \subset G)$ [i.e. $H \le K \le G $] in the sense that $P = R \rtimes K$. Then the subfactor $(R \rtimes H \subset R \rtimes G)$ is maximal iff the inclusion $(H \subset G)$ is maximal. There is a famous series of maximal subfactors due to Vaughan Jones and called the Temperley-Lieb-Jones subfactors which are completely determined by the Dynkin diagram $A_n$ ($n \ge 3$) and of index $4 cos^{2}(\pi/n)$. For $n=3$ (resp. $5$) the index is the integer $2$ (res. $3$) and for all the other value of $n$ the index is not an integer. For $n=3$ or $5$ the subfactor is given by groups, i.e. $(R \subset R \rtimes \mathbb{Z}/2)$ or $(R \rtimes \mathbb{Z}/2 \subset R \rtimes (\mathbb{Z}/3 \rtimes \mathbb{Z}/2))$.
This MO post asks about a process producing a series of subfactors from a given maximal subfactor. If we start with the maximal subfactor $(R \subset R \rtimes \mathbb{Z}/2)$ then we get every subfactor $(R \subset R \rtimes \mathbb{Z}/n)$. What is happening by starting from $(R \rtimes \mathbb{Z}/2 \subset R \rtimes (\mathbb{Z}/3 \rtimes \mathbb{Z}/2))$ can be reformulated into the questions above. Now we got the answer below.

Answer 1

According to the composition on a semi-direct product, we have that:

$((m,1),(m,1),\dots,(m,1),(m,0),(m,1),\dots)^{2} = ((0,0),(,0,0),\dots,(0,0),(2m,0),(0,0),\dots)$

But for $2m \in \mathbb{Z}/3^{n}$:

• if $m \wedge3 = 1$ then $\langle 2m \rangle = \mathbb{Z}/3^{n}$
• if $v_3(m) = r$ with $r<n$, then $\langle 2m \rangle = \mathbb{Z}/3^{n-r}$
• if $v_3(m) = r$ with $r \ge n$, then $\langle 2m \rangle = 0$

Let $r = v_3(m)$, then $K_m = \prod \limits_{n=1}^{r} (\{0 \}, \mathbb{Z}/2) \times \prod \limits_{n=r+1}^{\infty} (\mathbb{Z}/3^{n-r}, \mathbb{Z}/2)$
Next $K_r \subsetneq K_s$ iff $v_{3}(s) < v_{3}(r) $ and $[K_s,K_r] = \infty$