I can't understand this proof. If $G$ is a group and $H,KI know this question it's proved in prove that $[G:K]$ is finite and $[G:K]=[G:H][H:K]$ but I'd like to understand the demonstration which I was given to
This bellow is the demonstration that I have got from my monitor but I didn't get some steps.
$G=\bigcup_{a \in G}aH$ and $H=\bigcup_{b \in K}bK$
1-Why has he written $G$ and $H$ instead $G:H$ and $H:K$?
Then he mixed $G$ and $H$ and he got:
$G=\bigcup_{i \in A, j \in A}a_ib_jK$
2- Why is it $G$ and not $G:K$ ?
3-and if it's $G:K$ so why $G:H \cup H:K=G:K$?
He tried to prove that the last union (the last $G$) has $(G:H)(H:K)$ distinct elements as written bellow:
-If $i \neq i'$ so $a_ib_jK \neq a_{i'} b_j K$ if they both were equal so $a_ib_j \sim a_{i'} b_j$ (4- where does this equivalence come from?) that's there exists $k \in K$ such that:
$a_{i'}b_j=a_i b_j K$
5-Is this correct because the definition of right equivalence?
The he has written:
$a_{i'}=a_i(b_jKb_j^{-1})$
Then he argued that $b_jKb_j^{-1} \in H$, (6-why?, 7-why does this help?)
he argued that $a_{i'}=a_i(b_jKb_j^{-1})$ is equivalent to:
$a_i H=a_{i'}H$ so there's a contradiction. (8-why contradiction?)
Why does this equivalence hold? why is it not $a_{i'}=a_iH$?
Then he made another step:
-If $j \neq j'$ so we have $a_i b_j K \neq a_i b_{j'} K$ if they were equal so $a_ib_j \sim a_i b_{j'}$ (9- why does he use equivalence and not equality here?) that's there is some $k \in K$ such that $a_ib_{j'}=a_ib_jK$ so:
$b_{j'}=b_jK$
then the same problem again... he argued this is equivalent to:
$b_{j'}K=b_{j}K$
and the proof is complete.
PS:
even if this demonstration is correct, is it poorly clean?
 A: If $x$ is an element of $G$, and $H$ a subgroup of $G$, then $xH$ is a subset of $G$.  By definition, $xH = \{ xh : h \in H\}$.  Sets of the forms $xH$ are called left cosets of $H$ in $G$.
$[G : H]$ is the index of $H$ in $G$.  It is by definition the number of distinct subsets of $G$ of the form $xH$.
On the other hand, $G/H$ is the set of left cosets of $H$ in $G$.  So, $G/H$ is not a subset of $G$, but rather its elements are themselves subsets of $G$.  Thus $[G : H]$ is the cardinality of the set $G/H$.
1, 2, 3: The notation
$$G = \bigcup\limits_{a \in G} aH$$
is correct.
This is to say, $G$ is the union of its left cosets.
The statement
$$H = \bigcup\limits_{b \in H} bK$$
is similarly correct.  The statement
$$G = \bigcup\limits_{i,j} a_i b_j K$$
is also correct and not a mix up.  This is because
$$a_i  \Bigg( \bigcup\limits_j b_j K \Bigg) = \bigcup\limits_j a_i b_j K$$
and therefore
$$G = \bigcup\limits_i a_i H = \bigcup\limits_i a_i \Bigg( \bigcup\limits_j b_j K \Bigg) = \bigcup\limits_i \bigcup\limits_j a_i b_j K = \bigcup\limits_{i,j} a_i b_j K.$$
