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For a permutation p: X → X, let $p^{k}$ denote the permutation arising by a k-fold composition of p, i.e. $p^{1} = p \ and \ p^{k} = p◦p^{k−1}$. Define a relation ≈ on the set X as follows: i ≈ j if and only if there exists a k ≥ 1 such that $p^{k}(i) = j$. Prove that ≈ is an equivalence relation on X, and that its classes are the cycles of p.

I am new to this, what does it mean by its classes? does it mean its elements in the set X? I do see the connection between the k-fold composition and the permutation of p. I think I just do not understand the language. Please give me some hints.

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An equivalence relation is a relation that is reflexive ($i \approx i \;\forall i$), symmetric $(i \approx j \iff j \approx i)$, and transitive $(i \approx j$ and $j \approx k \implies i \simeq k$). Because of those axioms, an equivalence relation induces a partition of the set $X$ into subsets of elements which are all equivalent to each other. That is, if we look at the sets $Y_i = \{j \in X : j \approx i\}$ for various $i \in X$, then each pair $Y_i$ and $Y_j$ are either the same set or disjoint from each other. The subsets $Y_i$, more commonly denoted $[i]$, are called the "equivalence classes" for the relation $\approx$. $[i]$ is called "the equivalence class of $i$".

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    Thanks, do you have a book or site recommended so I can study these things in a systematic way? the course I am taking has stuff from every where so it makes it so difficult for me to follow. – 2017-02-02
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    Try this chapter from "Book of Proof" by Richard Hammack. The whole book is on his website free. http://www.people.vcu.edu/~rhammack/BookOfProof/Relations.pdf – 2017-02-02
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    @ Nathaniel Mayer Would you show me how to prove the question as well? An example would start me off! – 2017-02-02
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    Ok, here's an example. Let $X = \{1,2,3,4,5\}$, and $p = (125)(34)$ in cycle notation. (If you're not familiar with cycle notation, that means $p(1) = 2$, $p(2) = 5$, $p(5) = 1$, $p(3) = 4$, and $p(4) = 3$.) If I iterate $p$, I can get 1 to map to 1, 2, or 5, but never to 3 or 4. The equivalence class of 1 in the relation defined in the problem is $[1] = \{1,2,5\}$. – 2017-02-02
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    @ Nathaniel Mayer Thanks! I understand your explanation, it is a 2 cycle permutation when it is mapped to the identity permutation. However, I still don't understand the notation $p^{k}$ and what does it mean by k-fold composition. – 2017-02-02
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    $p$ is a function that sends elements of $X$ to other elements of $X$. $p^k$ means applying $p$ repeatedly, $k$ times. So in my earlier example, $p^2(1) = p(p(1)) = p(2) = 5.$ $p^3(1) = p(p^2(1)) = p(5) = 1$. – 2017-02-02
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52910/discussion-between-tmm-and-nathaniel-mayer). – 2017-02-02