Are there any conditions , such that 2 permutations in $S_4$ are commutative?

Question

It may a bit general question, but I'd like to ask anyway, and I'll try to detail as much as I can.

Well, I have the symmetric group, $S_4$. Lets take permutations $\sigma, \tau \in S_4 $.

So, $\sigma$ would look like this:

$$ \begin{pmatrix} 1 &2 &3 &4 \\ \sigma_{1} &\sigma_{2} &\sigma_{3} &\sigma_{4} \\ \end{pmatrix} $$

And $\tau$ would look like this: $$ \begin{pmatrix} 1 &2 &3 &4 \\ \tau_{1} &\tau_{2} &\tau_{3} &\tau_{4} \\ \end{pmatrix} $$

my question is:

Are there any terms for $\sigma$ and $\tau$, that make their action commutative?

I.e Are there any terms such that $\sigma \star \tau = \tau \star \sigma$ ?

I hope I could be understandable. thank-you.

Answer 1

There is a condition which classifies when two permutations commute. However, it is easier to state if we write the permutations using disjoint cyclic notation (see wikipedia).

So, write $\sigma=(a_{1, 1} \: a_{2, 1}\: \ldots \: a_{n_1, 1})\cdots (a_{1, i} \: a_{2, i}\: \ldots \: a_{n_i, i})$, then $\sigma$ and $\tau$ commute if and only if $(\tau(a_{1, 1}) \: \tau(a_{2, 1})\: \ldots \: \tau(a_{n_1, 1}))\cdots (\tau(a_{1, i}) \: \tau(a_{2, i})\: \ldots \: \tau(a_{n_i, i}))=\sigma$ (so apply $\tau$ to each of the "numbers" in $\sigma$).

For example, if $\sigma$ is a single cycle $(a_{1} \: a_{2}\: \ldots \: a_{n})$ then $\sigma$ and $\tau$ commute if and only if $(\tau(a_{1}) \: \tau(a_{2})\: \ldots \: \tau(a_{n}))$ is a cyclic shift of $(a_{1} \: a_{2}\: \ldots \: a_{n})$.

This follows from the fact that, in general, $$\tau^{-1}(a_{1, 1} \: a_{2, 1}\: \ldots \: a_{n_1, 1})\cdots (a_{1, i} \: a_{2, i}\: \ldots \: a_{n_i, i})\tau$$$$=(\tau(a_{1, 1}) \: \tau(a_{2, 1})\: \ldots \: \tau(a_{n_1, 1}))\cdots (\tau(a_{1, i}) \: \tau(a_{2, i})\: \ldots \: \tau(a_{n_i, i}))$$ which is quite well-known but I will leave proving it to the reader (it is in most advanced undergraduate texts). I will also leave interpreting this criterion to non-disjoint-cycle notation to the OP or any other interester person. I am hungry and have to run home for my dinner, so I have no time to do this myself!

Answer 2

Yup. If $\sigma=(1, 2)$ and $\tau= (3, 4)$, then $\sigma \tau = \tau \sigma$. If this isn't what you're looking for, then could you please be more specific?

Addendum: In light of recent (by a few seconds!) answers, perhaps I should say more than just a counterexample. The following exercises are useful to think about:

Exercise 1: Prove that if $\sigma,\tau\in S_4$ are disjoint permutations, then $\sigma\tau=\tau\sigma$.

Exercise 2: Prove that $(1,2,3,4)$ and $(1,3,2,4)$ commute although they aren't disjoint permutations. In fact, prove that $(1,3,2,4)$ is the square of $(1,2,3,4)$.

The following exercises constitute useful information but might be difficult to tackle depending on your knowledge of group theory. If you're familiar with the orbit-stabiliser formula for a group action and with the fact that two permutations in $S_n$ are conjugate if and only if they have the same cycle structure, then you have enough knowledge to have a crack at the following exercises. If you aren't familiar with these things, then you can still solve the following exercises by other clever techniques or even directly by the tedious approach of systematically "checking" elements.

Exercise 3 (Challenge):

(a) Prove that an element of $S_4$ commuting with a 4-cycle (e.g., such as $(1,2,3,4)$) must be a power of the 4-cycle in question.

(b) Prove the analogous assertion in (a) for an element of $S_4$ commuting with a 3-cycle.

(c) Prove that there are exactly four elements of $S_4$ commuting with a given 2-cycle (i.e., transposition!) in $S_4$. What are they?

Exercise 4: Is it true that for two non-disjoint commuting permutations in $S_4$, one must be a power of the other?

Exercise 5: A (2,2)-cycle in $S_4$ is an element of the form $(a,b)(c,d)$ where $\{a,b,c,d\}=\{1,2,3,4\}$. Prove that there are exactly eight elements in $S_4$ commuting with a given (2,2)-cycle. What are they, e.g., for the element $(1,2)(3,4)$ in $S_4$.

I hope this helps!

Answer 3

Let $\sigma$ interchange $1$ and $2$, and let $\tau$ interchange $3$ and $4$. They clearly commute, since they do not really interact. Or else one can check the fact that they commute slowly by seeing what $\sigma \tau$ and $\tau\sigma$ do to each of $1$. $2$, $3$, $4$.

One can find a much more boring example by letting $\sigma$ be the identity, and $\tau$ any permutation.

Answer 4

Yes, in $S_4$ there are such permutations. $\sigma\tau=\tau\sigma$ trivially holds when $\sigma=\tau$ or $\sigma=\tau^{-1}$ or $1\in\{\sigma,\tau\}$. But you can also consider disjoint cycles, such as $\sigma=(1\,2)$ and $\tau=(3\,4)$. But $\sigma=(1\,2)(3\,4)$ and $\tau=(1\,3)(2\,4)$ would also be fine.

Answer 5

One general condition is that the permutations are disjoint. Two permutations, $\tau$ and $\sigma$, are disjoint if, for all $i$ we have $\tau(i) \neq i \Rightarrow \sigma(i) = i$ and $\sigma(i) \neq i \Rightarrow \tau(i) = i$.

In plain language: $\tau$ only moves things that $\sigma$ fixes and vice versa. Or equivalently, the set of numbers not fixed by $\tau$ is disjoint from the set of numbers not fixed by $\sigma$.

Disjoint permutations always commute.

Answer 6

More generally, and perhaps somewhat more boring, you could always take one of the permutations to be trivial. Then it will commute with everything. In general, given any permutation (or any element of a group), the set of elements commuting with it is called its centralizer.