I don't understand yet:in order to find orbits of a given permutation of a set $A$, is it necessary that the relation $\sim$ involving elements $a$ and $b$ to be an equivalence relation? Nice day.
Finding orbits of a given permutation
0
$\begingroup$
abstract-algebra
group-theory
-
0if you mean the relation $a\sim b$ iff $\exists g\in G : ga = b$ then yes this is an equivalence relation, and the orbits are precisely the equivalence classes. – 2011-12-20
-
0It is possible that neema means with orbit an equivalence class of an equivalence relation given by the cycles of the permutation. For example, Herstein does this in Topics in Algebra ($2$nd ed, pg. $77$). Fix some permutation $g \in S_A$. An orbit is an equivalence class of the equivalence relation $\sim$ on $A$, where $a \sim b$ if and only if $a = g^i(b)$ for some integer $i$. But I'm still confused by this question. – 2011-12-20