# Equivalence relation with permutation

$$\forall \sigma,\beta \in S_5 \quad \sigma\,R\, \beta \Leftrightarrow \quad \sigma(1)=\beta(1) \land \sigma(5)=\beta(5)$$

1) Prove that $R$ is an equivalence relation: The reflexivity,symmetry and transitivity of $R$ coming from the reflexivity, symmetry and transitivity of $\quad = \quad$ right?

2) Find the equivalence class of $[i_{S_5}]_R$ where $i_{S_5}$ is the identity permutation.

I have no clue how to solve the point 2). Can anyone could explain me the process? thanks

• (1) Yes, right. (2) Besides the identity, can you think of other permutations in $\;S_5\;$ that fix both $\;1,\,5\;$ ? For example, $\;(23), (234)\;$ , etc. – DonAntonio Sep 11 '16 at 13:13
• what is exactly $S_5$ ? in 1) it is "prove that R", no ? – user354674 Sep 11 '16 at 13:21
• @DonAntonio sorry I don't understand, what do you mean with fix both 1,5? – Alfonse Sep 11 '16 at 13:25
• @igael yes you're right! $S_5$ is the group of permutation of order 5 – Alfonse Sep 11 '16 at 13:26
• @DonAntonio: reading your answer , I understand the question ! TY – user354674 Sep 11 '16 at 14:01

## 1 Answer

$(1)$ Correct; it amounts to recognizing the relation is one of equality, which we know is reflexive, symmetric, and transitive.

$(2)$ $$\forall \sigma,\beta \in S_5: \quad \sigma \,R\,\beta \iff \left(\sigma(1)=\beta(1)) \land (\sigma(5)=\beta(5)\right)$$

Now, we are looking for all permutations $\sigma\in S_5$ such that:

• $id(1) = 1 = \sigma(1),$ and also

• $id(5) = 5 = \sigma(5)$

In other words, we are looking for all permutations that "fix" both $1$ and $5$: All permutations that permute 1 to 1, and 5 to 5.

$\left [id_{s_5}\right]$ includes $(2,3), (2,4), (3, 4), (2, 3, 4)$.

More formerly, the equivalence class $\left[id_{s_5}\right] = \left\{id_{s_5}, (2, 3), (2, 4), (3, 4), (2,3,4)\right\}$

• now it's totally clear! thanks – Alfonse Sep 11 '16 at 14:41