# Two questions about equivalence relations

Question 1: Let $x,y \in S$ such that $x\sim y$ if $x^2 =y^2\pmod6$. Show that $\sim$ is an equivalence relation.

This is what I tried:
Reflexive: $x^2\pmod6 = x^2$ implying $x\sim x$
Symmetry: suppose $x\sim y$, then $x^2 =y^2\pmod6$
Operating by inverse of $x^2$ both sides we have $e=(x^2)^{-1} y^2\pmod6$
Then I have $y^2=x^2\pmod6$, hence $y\sim x$
I would like to get some corrections up to here because am not sure with this

Question 2. Find the orbits of the given permutation: $$v:\mathbb{Z}\to\mathbb{Z}\quad\text{ defined by }\quad v(n)=n+3$$

Here is my solution which I would also want somebody to correct me!
Let the relation $\sim$ be defined on $\mathbb{Z}$ by $a\sim b$ if $b=a+3$
I consider that the orbits of $v$ are the classes of $\mathbb{Z}$ with respect to $\sim$ given by $\{b\mid b=a+3\}$. Thus I get only three orbits which are $\{1,4,7,10,…,3n-2\mid n\in \mathbb{Z}\}$, $\{2,5,8,…,3n-1\mid n\in \mathbb{Z}\}$, and $\{3,6,9,…,3n \mid n\in \mathbb{Z}\}$.

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What is $S$? Does x2 mean $x^2$? –  lhf Dec 17 '11 at 16:36
And please try to use latex! –  sxd Dec 17 '11 at 16:37
Despite the fact that you correctly write $n \in \mathbb{Z}$, your explanation makes it sound as if the only numbers you are considering are the positive integers. –  André Nicolas Dec 17 '11 at 17:06

Your argument for symmetry of the first relation is flawed, because you do not know that there is such a thing as the inverse of $x$ modulo $6$ (for example, what if $x^2 \equiv 0 \pmod{6}$? ) But you don't have to: if $x^2\equiv y^2\pmod{6}$, then $6|x^2-y^2$, hence $6|y^2-x^2$, so $y^2\equiv x^2\pmod{6}$.

And you are still missing transitivity: if $x^2\equiv y^2\pmod{6}$ and $y^2\equiv z^2\pmod{6}$, why must $x^2\equiv z^2\pmod{6}$? (Hint. Just use the definitions).

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;thanks sir, do yo mean a divides b by b|a instead of b/a? –  neema Dec 17 '11 at 21:08
what about question 2, am i correct? –  neema Dec 17 '11 at 21:13
"$a|b$" means "$a$ divides $b$" means "there exists an integer $k$ such that $ak=b$". This is different from $b/a$ which denotes the rational (possibly integer) number you get by actually dividing $b$ by $a$. $a|b$ is a relation between $a$ and $b$; $b/a$ is a number resulting from performing an operation on $a$ and $b$. –  Arturo Magidin Dec 18 '11 at 0:16
Question 2 is badly written, because the sets you describe look as if they were finite and start at the least positive integers. It is also false to claim that $\sim$ as you define it is an equivalence relation: it's not reflexive, it's not symmetric, and it's not transitive. So while it is true that there are three orbits, your argument is very confused. –  Arturo Magidin Dec 18 '11 at 0:18
:okay..i want to know more, is it possible for the permutation of a given set to have orbits, given that the relation involving a and b is not an equivalent relation? –  neema Dec 18 '11 at 14:18

HINT $\$ Show more generally that if $\rm\ f\::\:S\to T\:$ and $\rm\:\: \equiv\:\:$ is an equivalence relation on $\rm\:T,\:$ then $\rm\ \{\: (x,y)\ :\ f(x)\:\:\equiv\:\: f(y)\:\}\$ is an equivalence relation on $\rm\:S\:.\:$ This is called the (equivalence) kernel of $\rm\:f\:,\:$ if we view the codomain as the quotient set $\rm\ T/\equiv\:,\:$ which above is $\rm\ \mathbb Z/6\:,\:$ with $\rm\ f(x) = x^2\:.$

See also the more general notions of difference kernels and equalizers, and see this prior question where the fiber viewpoint is mentioned.

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Hint for #1. Any relation defined via equality of the values of a function is an equivalence relation. More precisely:

If $f:X\to Y$ and $\sim$ is defined on $X$ by $x\sim y$ iff $f(x)= f(y)$, then $\sim$ is an equivalence relation.

Conversely, every equivalence relation arises in this way.

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Note that to apply this directly requires viewing $\:\mathbb Z/6\:$ as a quotient set, because it uses an equality $\rm\:f(x) = f(y)\:$ vs. a congruence $\rm\:f(x)\equiv f(y)\:.\:$ But even if the OP has not yet learned about quotient sets, they may still comprehend this viewpoint by using congruence instead of equality - see my answer. –  Bill Dubuque Dec 17 '11 at 18:51
@Bill, yes, nicely put, thanks. –  lhf Dec 17 '11 at 19:28