I recently had as an assignment, to find cycles $\sigma,\tau\in S_{\mathbb{N}}$ (i.e. permutations over the naturals) such that $ord(\sigma)=ord(\tau)=2$ and $\tau\circ\sigma$ has order infinity. This is just a background; my question here is not to find the answer to this question, but rather a notational question (see bottom).

By intuition I was able to guess $\sigma=(0,1)\circ(2,3)\circ(4,5)\circ\ldots$ , $\tau=(1,2)\circ(3,4)\circ\ldots$, which, each being the composition of an arbitrary number of disjoint transpositions, have order 2.

To show that their composition is of order infinity (calling it $\rho$:

Writing tau backwards (because composition commutes over disjoint cycles) gives
$\rho=\ldots\circ(3,4)\circ[(1,2)\circ(0,1)\circ(2,3)]\circ(4,5)\circ\ldots$ (note brackets)

Now noting that given a cycle of order $n$, which can be written as $X=(X_1,X_2,\ldots,X_n)$, and a transposition which is disjoint to X, say $A=(a,b)$, and taking a transposition containing one element of X and one element of A, e.g. $J=(X_1,a)$, and composing them all, the resulting cycle $J\circ X\circ A=(a,b,X_1,X_2,\ldots,X_n)$ which is of order $n+2$.

The bracketed expression above, $(1,2)\circ(0,1)\circ(2,3)$, is exactly such a collection of cycles, $(0,1)$ being disjoint from $(2,3)$ and $(1,2)$ containing one element of each, and thus $(1,2)\circ(0,1)\circ(2,3)$ is of order 4.

Calling $\sigma_n=(2n-2,2n-1)$, we can say that $\sigma=\sigma_1\circ\sigma_2\circ\sigma_3\circ\ldots$
$\tau_n=(2n-1,2n)$, we can call $\tau=\tau_1\circ\tau_2\circ\ldots$
Now call $\rho_1=\sigma_1=(0,1)$, and recursively $\rho_{n+1}=\tau_{n}\circ\rho_n\circ\sigma_{n+1}$

So $\sigma_1=(0,1)$, order 2.
$sigma_2=(1,2)\circ(0,1)\circ(2,3)$, order $ord((0,1))+2=4$.
In general, $ord(\sigma_n)=ord(\sigma_{n-1})+2=2n$.

In my proof for the assignment, I concluded with "Because $\rho$ is the result of infinitely many of these iterations which increase the order by 2 each time, we can conclude that the order of rho is infinity." I feel that the answer is here, but I have not expressed it in an adequate manner. How could I more formally express the notion that my sequence, $\rho_n$, does indeed "approach" or "converge" to the actual (infinite) cycle $\rho$ ?

  • $\begingroup$ Have you examined what iterations of $\rho$ do to some specific element, such as $1$? If you can show where it is after $n$ iterations, and it's not back at $1$ for any $n$, then the order of $\rho$ is infinite. $\endgroup$ – G Tony Jacobs Mar 14 '14 at 17:13
  • $\begingroup$ @GTonyJacobs Yes, I could; however I'm not so much interested in answering the original question using a different mehod, as correcting the notation/explanation of the method I constructed. $\endgroup$ – JustAskin Mar 14 '14 at 17:18

It's not really a good argument, because permutation groups have no well-behaved notion of convergence in general.

Hmm, scratch that. On further thought it seems perfectly well-behaved to say that $\sigma_1, \sigma_2, \sigma_3,\ldots$ converges iff for every $x$ there is an $N$ such that $\sigma_N(x)=\sigma_{N+1}(x)=\sigma_{N+2}(x)=\cdots$.

With this definition limits distribute over composition, but I don't think this is a standard notion, so you may have to include definitions and lemmas in your argument.

It's more of a problem that you seem to assume that limits commute with taking the order of a permutation. Consider the sequence $(\sigma_n)_n$ where $\sigma_n$ is a cyclic permutation of the elements $n$ through $2n-1$. Then $\sigma_n$ has order $n$, but the limit of the sequence is the identity permutation, which has order $1$! So the limit of a sequence of permutations of high order doesn't necessarily have high order itself.

It's better to look at $\tau\sigma$ as a finished whole and consider what it does to individual elements: Every even number $n$ is mapped to $n+2$, and every odd number $n$ is mapped to $n-2$, except that $1$ maps to $0$.

Since no amount of iterations of $\tau\sigma$ can make, say, 0, return to its original position, the permutation has infinite order.

  • $\begingroup$ Hm. What if I constructed some argument such as "For any natural number $m$, there exists a $\rho_n$ which has order greater than $m$." Then showed that the order of the real $\rho$ is always greater than the order of $\rho_n$ for any $n$. That would establish that the order of $\rho$ is greater than all natural numbers, without having to use "convergence" of $\rho_n$ to $\rho$ at all, just comparing orders. $\endgroup$ – JustAskin Mar 14 '14 at 17:24
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    $\begingroup$ @Justaskin: I was a bit too quick in rejecting your approach (see edit), but I don't think that particular suggestion will work. (See further edit). $\endgroup$ – Henning Makholm Mar 14 '14 at 17:32

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