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