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Happy Peaceful New Year !

In this question, I recalled that if $H$ is a proper subgroup of a finite group $G$, such that $$({\bf A1})\qquad(g\not\in H)\Longrightarrow(g^{-1}Hg\cap H=(1)),$$ then $$N:=(1)\bigcup\left(G\setminus\bigcup_g g^{-1}Hg\right)$$ is a normal subgroup. The normality is obvious, but the fact that $N$ is a group is proved by character theory (one doesn't know a direct proof). The terminology is that $G$ is a Frobenius group, $H$ is the Frobenius complement of $G$ and $N$ is the Frobenius kernel.

What happens when we allow $G$ to be infinite ? I suppose that the status of this question is well-known. I suspect that there be some counter-example, a pair $(G,H)$ for which $N$ is not a subgroup. Perhaps the Frobenius Theorem above adapts with some extra assumption; an additional hypothesis could have a topological flavour, in the spirit of functional analysis, where we use topology in order to extend linear algebra in the infinite dimensional context.

Edit. After the negative answers, essentially based on the use of free groups or free product, I came to the following strengthen context. Let $K$ be the subgroup generated by the union of the conjugate subgroups $g^{-1}Hg$ (notice that $K$ is a normal subgroup). Then $G$ is the union of $K$ and $N$, where $K\ne(1)$ and $N\ne G$. If the same conclusion as in Frobenius Theorem holds true ($N$ a subgroup), we have easily that $K=G$. This leads me to add the following assumption:

Assume in addition that $$({\bf A2})\qquad\bigcup_g g^{-1}Hg\quad\hbox{generates}\quad G.$$

Then can we say that $N$ is a subgroup ?

Let me discuss a little the case where $H=\langle a\rangle$ is a cyclic subgroup of $\mathbb{L}_2$, the free group with generators $a,b$. Then $K$ is a proper subgroup, the kernel of the morphism $\phi:\mathbb{L}_2\rightarrow\mathbb{Z}$ defined by $\phi(a)=0$ and $\phi(b)=1$. This explains why $N$ cannot be a subgroup of $\mathbb{L}_2$. In a general configuration, suppose that $H$ is a malnormal subgroup of $G_0$, but that $K$ is proper. We might replace $G$ by $G_1=K$, and $H$ is still malnormal. However, the new $K$, which I denote $K(H,G_1)$ is smaller than $K$, because there are less conjugate subgroups $g^{-1}Hg$ (the constraint is now $g\in G_1$). Whence the necessity to define $G_2=K(H,G_1)$. This is the beginning of an induction. This induction is not necessatily finite or denumerable, it can be transfinite, but its length is bounded by the cardinal of $G/H$. Eventually, we reach a subgroup $G^\dagger$ in which $H$ is malnormal and satisfies (A2). It is unclear to me whether $G^\dagger$ is bigger than $H$ or not.

What is $G^\dagger$ in the case described above, where $G=\mathbb{L}_2$ and $H=\langle a\rangle$ ? In particular, can $G^\dagger$ be equal to $H$ ?

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    $\begingroup$ If $G$ is commutative the hypothesis implies $H=\{0\} $ or $G$, hence the conclusion is trivially satisfied. $\endgroup$ – abx Jan 4 '16 at 7:47
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    $\begingroup$ Might this arxiv.org/abs/1104.3065 be relevant? If I recall correctly it has an example of a malnormal H where the Frobenius complement is not a subgroup (these examples predate the paper) $\endgroup$ – Yemon Choi Jan 4 '16 at 8:05
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    $\begingroup$ Take the free product of two finite groups $A*B$ . Certainly $gAg^{-1}\cap A$ is reduced to $1$. But $abab^{-1}$ is of infinite order and is therefore not conjugate in $A$. $\endgroup$ – Thomas Jan 4 '16 at 9:13
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    $\begingroup$ @Thomas. A sentence such as "$c$ is not conjugate in $A$" is meaningless. Only a pair of elements can be conjugate. $\endgroup$ – Denis Serre Jan 4 '16 at 13:07
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    $\begingroup$ I think there are simpler examples, such as any of the subgroups $\langle x \rangle$, $\langle y \rangle$ or $\langle xy \rangle$ in the Hurwitz gropup $\langle x,y \mid x^2=y^3=(xy)^7=1 \rangle$. $\endgroup$ – Derek Holt Jan 12 '16 at 16:55
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A subgroup satisfying the condition $g \not\in H \Rightarrow H \cap g^{-1}Hg=1$ is called malnormal. This class of subgroups has been much studied.

For many types of infinite groups, such as word-hyperbolic groups, there are lots of examples. As a simple example, if $G$ is free and $a$ is an element in a free generating set, then $H = \langle a \rangle$ is malnormal in $G$.

If $b$ is another element in the free basis, then neither $b$ nor $b^{-1}a$ lies in a conjugate of $H$, but their product does, so there is no equivalent of the Frobenius kernel.

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  • $\begingroup$ To answer the question it is preferable to take $A*B$, where $A$ finite ; see my former comments $\endgroup$ – Thomas Jan 5 '16 at 8:25
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    $\begingroup$ @Thomas I agree that you can do that, but I don't see why it is preferable in particular. The example with the free group is more elementary. $\endgroup$ – Derek Holt Jan 5 '16 at 9:22
  • $\begingroup$ I think the question asked that $H$ is finite, but the question was not so clear. The case of $Z_2*Z_2$ gives an amenable example $\endgroup$ – Thomas Jan 5 '16 at 12:32
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First answer: Let $G$ be a finite index torsion-free (neat) subgroup of $SL_3({\mathbb Z})$: it is known that there exist such neat subgroups (almost any congruence subgroup will do; for details, one may look at Raghunathan's book). Suppose $K/{\mathbb Q}$ is a totally real cubic extension. The group $O^*$ of units of $K$ acts by multiplication of the ring $O\simeq {\mathbb Z}^3$ of integers in $K$. Hence we get an embedding of $O^*$ in $GL_3({\mathbb Z})$. Denote by $H$ the intersection of $O^*$ with the subgroup $G$ of $GL_3({\mathbb Z})$.

One can show that if $g\in G\setminus H$, then $H\cap gHg^{-1}$ is trivial (by using that non-trivial elements of $H$ have distinct eigenvalues and also that $G$ is neat; the latter is needed to ensure that if $g\in G$ normalises $H$, then it centralises $H$ and hence lies in $H$). Your set $N$ is infinite, since there are unipotent elements in $G$ which never lie in a conjugate of $H$.

If $N$ were a subgroup of $G$, then by a result of Bass-Milnor and Serre (the "other" Serre) $N$ would have finite index and hence would intersect $H$ in a finite index subgroup, contradicting the definition of $N$.

Second Example: Even the revised question has a negative answer. Take $G=SL_2({\mathbb Q})$ and $H$ to be the normaliser of the diagonals $T$ in $G$. Then the only elements of $G$ which normalise $H$ already lie in $H$. The group generated by conjugates of $H$ is $G$ (since $G$ is simple modulo centre).

However, $N$ cannot be a subgroup, because it it were, being a union of conjugacy classes, it would be a normal subgroup and hence would be all of $G$; but then it would contain $H$.

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    $\begingroup$ That must be the first time where an answer to a question of mine involves a result by the ""other Serre". Incidentally, the "other Serre" is my uncle. However, I don't understand the argument that $N$ is infinite. Not only that, but it also hurts my knowledge of the finite case, where the order of $N$ is the index of $H$. $\endgroup$ – Denis Serre Jan 4 '16 at 15:07
  • $\begingroup$ @ denis. OOps! Pls convey my apologies to your uncle. $\endgroup$ – Venkataramana Jan 4 '16 at 15:14
  • $\begingroup$ @denis: I have added a line about the infinitude of $N$. There are unipotent elements in $G$, and they can never conjugate into $H$. Why does it say anything about the finite case? $\endgroup$ – Venkataramana Jan 4 '16 at 15:18
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I think this question has been asked in some form on MO before, though I can't remember where or when. Anyway an extreme example for sufficiently large primes $p$ is a Tarski Monster $G$, by which I mean a simple infinite group in which every proper non-identity subgroup has order $p$. Then when $H$ is a proper subgroup of $G$, we certainly have $H \cap g^{-1}Hg = 1$ for every $g \in G \backslash H$ whenever $H$ is a proper non-identity subgroup of $G$.

There is a dichotomy however : if $p$ is sufficiently large, there are Tarski monsters in which all proper non-identity subgroups are conjugate. In such a Tarski monster $G$, we have $\{1_{G}\} = \{1_{G}\} \cup \left( G\backslash \bigcup_{g \in G} g^{-1}Hg \right)$, so the analogue of the Frobenius kernel is the identity subgroup.

On the other hand, if we have a (simple) Tarski monster $G$ in which not all non-identity proper subgroups of $G$ are conjugate ( I believe there are such groups, but I haven't double checked), then for every proper non-identity subgroup $H$ of $G$, we see that the non-identity normal subset $\{1_{G} \} \cup \left( G \backslash \bigcup_{g \in G} g^{-1}Hg \right)$ is neither $G$ nor $\{1_{G} \}$, so can't be a subgroup ( if it were a subgroup it would be normal).

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