Index of G/H when H is not normal I have just begun my course of Abstract Algebra. I know what is the index and normality, my question is: 
$$ [G:H]=\frac{|G|}{|H|}$$
if H is normal this is true, but in general I don't know, but is kind of intuitive because if I make "parts" of G dividing it in $|H|$ parts it should survive only $\frac{|G|}{|H|}$ things, but I don't know how it works. Please help me to understand this concept. 
Thanks
 A: It's true, whether $H$ is normal in $G$ or not.  The critical fact here may be expressed as follows:
Fact:  Two cosets $aH$ and $bH$ are either disjoint or identical.
For if there is $k \in G$ such that $k \in aH \cap bH$, then we have $h_1, h_2 \in H$ with $ah_1 = k = bh_2$; then $a = bh_2h_1^{-1}$, whence $a \in bH$; but then for any $h \in H$ we have $ah = bh_2h_1^{-1}h \in bH$, hence $aH \subset bH$; the parallel argument with $b = ah_1h_2^{-1}$ shows $bH \subset aH$; thus $aH = bH$ as claimed. 
By virtue of this Fact, we see that $G$ may be partitioned into a disjoint union of distinct cosets of $H$:
$G = \bigcup_{g \in R} gH, \tag 1$
where the index set $R$ contains precisely one element from each coset of $H$.  Clearly,
$\bigcup_{g \in R} gH \subset G, \tag 2$
and since every $g \in G$ is an element of some coset (i.e., $g \in gH$), we also have
$G \subset \bigcup_{g \in R} gH; \tag 3$
thus (1) binds.
The above discussion pertains to any subset $H$ of any group $G$.  Now if $G$, and hence $H$, are finite, since each coset of $H$ contains the same number of elements $\vert H \vert$, and since the number of cosets is $[G:H]$, we have
$[G:H] \vert H \vert = \vert G \vert, \tag 4$
or
$[G:H] = \dfrac{\vert G \vert}{\vert H \vert}. \tag 5$
This holds whether $H$ is a normal subgroup of $G$, or not.
