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Let $\Omega$ denote the set of integer tuples $(a,b)$ where $a + b$ is odd. We define two permutations of $\Omega$ as:

  1. $ \sigma : (a,b) \mapsto (a+1,b-1)$ and
  2. $ \delta: (a,b) \mapsto (b,a)$

Let $G$ be the group generated $\sigma$ and $\delta$. We need to construct a strictly descending infinite sequence of non-commutative subgroups such that: $G > G_1 > G_2 > ...$

A couple of questions:

  • Is there a rule I have to apply to find such a sequence? Like substracting an element from the set of tuples generated by the two permutations? Is the "descending" refering to the number of elements of the set that the group applies to?
  • Non-commutative subgroups (non-abelian?)? I see that the main group operation is the addition here (which is commutative on integers) and I really don't understand what are these exactly.
  • How is it possible for such a sequence never to end? The trivial subgroup will never be reached?

Any help?

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    non-commutative probably does mean non-abelian. Descending simply refers to the fact that $G_{i+1}$\sigma^n$? Does it commute with $\delta$? – 2017-02-09
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    What I undestand is that by applying the $\sigma$ permutation $b-a$ times in a tuple then we get the same result if we applied the $\delta$ permutation just once to that tuple: that is the reversing the tuple elements. – 2017-02-09

2 Answers 2

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To avoid a lengthy discussion, I will provide a full answer:

You can easily check that $\sigma^n(a,b)=(a+n,b-n)$ for any positive integer $n$. Note that this is the case for all pairs $(a,b)$. In particular, $\sigma^n\delta(0,0)=(n,-n)$ and $\delta\sigma^n(0,0)=(-n,n)$ so $\sigma^n\delta\ne\delta\sigma^n$.

I describe one possible descending sequence. Let $G_i=\langle\delta,\sigma^{2^i}\rangle$. From the above observation $G_i$ is non-abelian. As $\sigma^{2^{i+1}}=(\sigma^{2^i})^2$ we have $G_{i+1}\le G_i$. You can check that each permuation in $G_{i+1}$ is of the form $(a,b)\mapsto (a+2^{i+1}k,b+2^{i+1}k)$ or $(a,b)\mapsto (b+2^{i+1}k,a+2^{i+1}k)$ for some integer $k$. In particular, $\sigma^{2^i}\notin G_{i+1}$ so $G_{i+1}

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    Thanks for the answer! Just some clarifications: how $k$ is entered in the form of the permutations and if $G_{i+1} \lt G_i$ actually means that $G_{i+1}$ "produces" less tuples. – 2017-02-09
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    You might think of it as the orbits of $G_{I+1}$ are strictly contained in the orbits of $G_i$. – 2017-02-10
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    In the case where the group $G$ was the one that had the first degree functions - $G=\{ ax+b|a \neq 0, a,b \in \mathbb R \}$ - and the group operation was the composition of functions, how we would be able to construct an infinite chain of non-abelian subgroups as in the problem above? I was thinking of putting $G_i=$, where the composition would be applied $2^i$ times and so I get the proper subgroups, but I don't get the non-abelian property (I am confused!). Can you suggest anything? – 2017-02-22
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    Intuitively, you need some freedom in both $a$ and $b$ to ensure the non-abelian property. I can see two relatively simple ways to get the chain, the first is $G_i=\langle -x,x+2^i\rangle$ and the second is $G_i=\langle 2^{2^i}x,x+1\rangle$. – 2017-02-22
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$\newcommand{\N}{\mathbb{N}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$This is an addendum, which is too long for a comment, to the excellent answer by Robert Chamberlain (+1).

Note that

  • $\delta^{-1} \sigma \delta = \sigma^{-1}$ (direct calculation)
  • $\delta$ has order $2$, and
  • $\sigma$ has infinite order (as shown by R.C.).

Then $G$ is the infinite dihedral group.

The subgroups of $T = \Span{\sigma}$ are of the form $T_{n} = \Span{\sigma^{n}}$ for $n \in \N$. Each such $T_{n}$ is immediately seen to be characteristic in $G$, so that for $n > 0$ the subgroup $U_{n} = \Span{T_{n}, \delta}$ is non-abelian.

Now it is only a matter of choosing a strictly increasing sequence of positive integers $n_{i}$ such that $$ n_{0} = 1, n_{i} \mid n_{i+1}, $$ and then $$ G = U_{n_{0}} > U_{n_{1}} > \dots $$ will be a sequence as requested. R.C.'s example is the case $n_{i} = 2^{i}$.