$x\sim y\iff x^2\equiv y^2\pmod{\!6}$ is an equivalence relation.

Question

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 $\{3,6,9,…,3n \mid n\in \mathbb{Z}\}$.

Answer 1

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).

Answer 2

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.

Answer 3

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 $\rm\,S.\,$ It's 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.