I am looking at a problem and really confused.
Let $X$ be a set and $R$ a subset of $ X×X $. We write $x1 ∼ x2$ if and only if $(x1, x2) ∈ R$
Suppose now that $R$ defines an equivalence relation and P1, P2, . . . , Pk are the disjoint nonempty subsets of $X$ giving the corresponding partition of $X$, that is, $X = ∪(k,i=1)Pi$ and if $x ∈ Pi$ then $y ∈ Pi$ if and only if $x ∼ y$. Show that $R = ∪(k,i=1)Pi × Pi$
I have done this by showing that R is contained within $∪(k,i=1)Pi × Pi$ and then proof by contradiction that all elements of $∪(k,i=1)Pi × Pi$ lie in R.
Then it says:
Let G be a finite group and H a subgroup of G. For $g1, g2 ∈ G$, define $g1 ∼ g2$ if and only if $(g^−1)1g2 ∈ H$.show that the number of elements in the set $R$, as above, is the product of the number of elements in $G$ and the number of elements in $H$.
I am having a hard time believing this, let alone proving it. thank you. Also, sorry I am new to this site and so my formatting is horrendous.
Show that for each element $g$ in $G$, there are exactly $|H|$ elements in $G$ which would satisfy the (equivalence) relation