There is an equivalence relation between inclusion of finite groups coming from the world of subfactors:

Definition: $(H_{1} \subset G_{1}) \sim(H_{2} \subset G_{2})$ if $(R^{G_{1}} \subset R^{H_{1}})\cong(R^{G_{2}} \subset R^{H_{2}})$ as subfactors.

Here, $R$ is the hyperfinite $II_1$ factor (a particular von Neumann algebra), and the groups $G_1$ and $G_2$ act by outer automorphisms. The notation $R^G$ refers to the fixed-point algebra.

Theorem: Let $(H \subset G)$ be a subgroup and let $K$ be a normal subgroup of $G$, contained in $H$, then:
$(H \subset G) \sim (H/K \subset G/K)$. In particular, if $H$ is itself normal: $(H \subset G) \sim (\{1\} \subset G/K) $
Theorem : $(\{1\} \subset G_{1}) \sim(\{1\} \subset G_{2})$ iff $G_1 \simeq G_2$ as groups.

Remark : the relation $\sim$ remembers the groups, but not necessarily the subgroups:
Exemple (Kodiyalam-Sunder p47) : $(\langle (1234) \rangle \subset S_4) \sim (\langle (13),(24) \rangle \subset S_4)$

Is there a purely group-theoretic reformulation of the relation $\sim$ ?

Motivations: See here and here.

Some definitions: A subfactor is an inclusion of factors. A factor is a von Neumann algebra with a trivial center. The center is the intersection with the commutant. A von Neumann algebra is an algebra of bounded operators on an Hilbert space, closed by taking bicommutant and dual. Here, $R$ is the hyperfinite $II_{1}$ factor. $R^{G}$ is the subfactor of $R$ containing all the elements of $R$ invariant under the natural action of the finite group $G$. In its thesis, Vaughan Jones shows that, for all finite group $G$, this action exists and is unique (up to outer conjugacy, see here p8), and the subfactor $R^{G} \subset R$ completely characterizes the group $G$. See the book Introduction to subfactors (1997) by Jones-Sunder.

  • $\begingroup$ What is a "subfactor"? It might help if you defined the notation $R^G$. $\endgroup$ Jul 7 '13 at 17:55
  • $\begingroup$ Ok @ViditNanda, I have edited some definitions and an hyperlink to The book. $\endgroup$ Jul 7 '13 at 19:04
  • $\begingroup$ Sebastien: In order to help people who don't know what factors are understand your question, you might want to include some purely group-theoretic consequences of your notion of equivalence. Alternatively, you might want to include some examples of pairs H1⊂G1 and H2⊂G2 that are equivalent in your sense. $\endgroup$ Jul 7 '13 at 19:35
  • $\begingroup$ Ok André, thank you for your comment, I will edit. $\endgroup$ Jul 7 '13 at 19:38

For finite groups, the answer was given by Izumi in his paper "Characterization of isomorphic group-subgroup subfactors" (MR1920326). There he looks at the crossed product subfactor, but you can always take duals.

Edit after @Andre's comment:

The actual condition between the two pairs of subgroups is quite technical, and it would basically require reproducing an entire page of a 10 page article. Here is a link to the article: http://imrn.oxfordjournals.org/content/2002/34/1791.short

See also this video (27:30) of a talk of M. Izumi on this subject, at the Sunder Fest 2012.

  • 1
    $\begingroup$ Dave: It would be even more useful if you could state the answer here, instead of just giving a pointer to the literature. $\endgroup$ Jul 8 '13 at 21:15
  • $\begingroup$ I don't read its proof about the classification of subfactors at index 5 in this paper. Do you know if it is published somewhere ? My original motivation is about index 6 (see here), in particular, the list of maximal subgroups at index 6. Is it known ? $\endgroup$ Jul 9 '13 at 9:50
  • $\begingroup$ @Sébastien : according to my answer at mathoverflow.net/a/136176/6451 and the wikipedia article en.wikipedia.org/wiki/Primitive_permutation_group , the list seems to be known, of length $4$. $\endgroup$
    – BS.
    Jul 9 '13 at 10:49
  • 1
    $\begingroup$ Yes @BS., thanks to your answer here, the classification of group-subgroup maximal subfactors at index $n$ reduced to the knowledge of primitive permutation groups (already classified for all n<4096). There is (up to equiv.) 5, 4, 7, 7, 11 such objects for n= 5, 6, 7, 8, 9. $\endgroup$ Jul 9 '13 at 15:16
  • $\begingroup$ The last question to answer is whether or not some of these objects are subfactor-equivalent ? $\endgroup$ Jul 9 '13 at 15:43

This is somewhere between a comment and an answer. But it is too long for a comment, so I put it here.

To me the natural thing to look at is the action $G\curvearrowright G/H$. $|G/H|$ captures the index, which you surely want to do, and the action should in some sense capture the position of $H$ inside $G$. The issue then would be to try to come up with an appropriate notion of equivalence of these two objects. Here are 3 possibilities:

So let $H\subset G$ and $\Lambda\subset \Gamma$ be two inclusions.

1.) We can say $H\subset G \simeq_1 \Lambda\subset\Gamma$ if there is an isomorphism of groups $\Phi:G\rightarrow\Gamma$ such that $\Phi(H)=\Lambda$. This is the strongest notion of equivalence and would certainly imply that the two actions on the cosets spaces are the "same". The issue is that it requires the groups involved to be isomorphic, which you certainly don't want.

2.) We can forget about the acting group per se and only consider the action on the coset space (in particular look at the orbit equivalence relation). We define this equivalence relation by saying $g_1H\simeq g_2H\Leftrightarrow \exists g\in G$ such that $gg_1H=g_2H$. We do a similar thing on $\Gamma/\Lambda$. Then we say $H\subset G \simeq_2 \Lambda\subset\Gamma$ if there exists a bijection $\Phi:G/H\rightarrow \Gamma/\Lambda$ that takes equivalence classes to equivalence classes.

3.) The last one is exactly what you suggest for subfactors.

Just a quick final few comments. $1\Rightarrow 2$ but I'm not sure if $2\Rightarrow 3$. The motivation for these comes from three notions of equivalence you can give to a measure preserving action of a discrete group. In this case (for the analogous three equivalences) we have $1\Rightarrow 2\Rightarrow3$ and none of the implication are reversible, in general. Much of Popa's deformation/rigidity program is in trying to reverse these arrows in certain cases. In the ergodic theory case, $1\Rightarrow 2$ is obvious and $2\not\Rightarrow 1$ is not to hard. $2\Rightarrow 3$ isn't to hard either but $3\not\Rightarrow 2$ was quite difficult.

  • $\begingroup$ Thank you Owen for this detailed overview. Do you have some references for these results ? The answer of Dave gives a reference with an explicit group-theoretic characterization of the third possibility for finite groups. Do you suggest (through your answer) there is no such characterization in general ? $\endgroup$ Jul 8 '13 at 20:52
  • $\begingroup$ For the ergodic theory motivation, check out Popa's ICM talk math.ucla.edu/~popa/ICMpopafinal.pdf. There is much more there than I have suggested here. As far as what is true in general I have not given it thought. Though I should say that I left out one additional notion of equivalence. Namely isomorphism of the inclusion $L(H)\subset L(G)$ of group von Neumann algebras. Note that for countably infinite discrete groups the nicest case would be when both groups are icc, then you get $II_1$ factors. In the finite index case I don't know any references for the standard invariant. $\endgroup$ Jul 8 '13 at 21:18
  • $\begingroup$ Do you know if $3\not\Rightarrow 2$ is also true for finite groups ? $\endgroup$ Jul 9 '13 at 14:44
  • $\begingroup$ Owen, it seems you need to improve your 2.) because (if I'm not mistaken) the action of $G$ on $G/H$ is transitive, so that, $G/H$ itself is the unique equivalent class. Maybe what you had in mind about 2.) is : let $X = G/H$ and $Y = \Gamma / \Lambda $, let $\pi: G \to S_{X}$ and $\mu : \Gamma \to S_{Y}$ be the natural representations, then $H\subset G \simeq_2 \Lambda\subset\Gamma$ if $\pi (G)$ and $\mu(\Gamma)$ are permutation isomorphic (see here p 156). $\endgroup$ Jul 10 '13 at 8:51
  • $\begingroup$ @Sebastien: Yes I was being careless, your notion is perhaps the correct one. Maybe just looking at the subgroup action $H\curvearrowright G/H$ is better. $\endgroup$ Jul 10 '13 at 21:22

We give here an easy and purely group-theoretic sufficient condition (but not necessary) :

Definition : Let $\sim_2$ be the equivalence relation on inclusions of finite groups, defined by :
$(A \subset B) \sim_2 (C \subset D)$ if $(A/A_B \subset B/A_B) \simeq (C/C_D \subset D/C_D)$ with $A_B$ the normal core.

Remark : $\sim$ does not imply $\sim_2$ thanks to the Kodiyalam-Sunder example (see above and here).

Theorem : $\sim_2$ implies $\sim$.
Proof: here p47 + Izumi's thm2.3 p3 here, with $N= \{ 1\}$ and $\omega = 1$ (there is certainly a direct proof).

Corollary: Let $G$ be a finite group, $H$ a subgroup, $\phi \in Aut(G) $ then : $(H \subset G ) \sim (\phi(H) \subset G)$.

Warning : the dual version of the corollary means that for every $\phi \in Aut(G)$ and subgroup $H \subset G$, it exists $\Phi \in Aut(R \rtimes G)$, such that $\Phi(R \rtimes H) = R \rtimes \phi(H)$. Let $\sigma$ be a (faithful) outer action of $G$ on $R$ ($\sigma_g(x) . u_g=u_g . x$) and $\phi \neq id$, then $\Phi$ is not defined by $\Phi(x) = x$ if $x \in R$ and $\Phi(u_g) = u_{\phi(g)}$, because else $\Phi(u_g . x) = \Phi(u_g).x $ and $ \Phi(\sigma_g(x) . u_g) = \sigma_g(x) . \Phi(u_g) $. So $ \sigma_g(x) . u_{\phi(g)}= u_{\phi(g)}.x$, then $\forall g \in G$, $ \sigma_g = \sigma_{\phi(g)}$, so $\phi = id$ because $\sigma$ faithful, contradiction.

Remark: We have the following useful properties (see M Izumi here, 23:30 and 27:50):
If $(A \subset B) \sim (C \subset D)$ then $Rep(A/B_A) \simeq Rep(C/D_C)$.
If moreover the inclusions are maximal then $(A \subset B) \sim_2 (C \subset D)$.

Corollary: $\sim_2$ $\Leftrightarrow$ $\sim$ if the inclusions are maximal.
Problem: Extension to all the "natural" inclusions (see the definition in the optional part of this post).
The Kodiyalam-Sunder counter-examples are not "natural" inclusions, because they are single chain but not homogeneous single chain.


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