If I take union of the left cosets and right cosets of a subgroup, should I end up getting the original Group? I am really having hard time understanding the concept of cosets.
Does the cosets exhaust all elements in the subgroup?
In addition to that, can I consider cosets as subsets of the subgroup?
 A: For a fixed subgroup $H$ of a group $G$, each left coset is a subset of $G$. The set of left cosets is a partition of $G$ - every element belongs to exactly one left coset.
I think the best way to understand this concept (in fact, any new concept) is to compute some examples. Perhaps start by writing down the three the left cosets of a two element subgroup of the symmetries of a triangle.
Do the same for as many examples of subgroups you have looked at.
A: If $H$ is a subgroup of $G$, then you can define an equivalence relation $\sim$ on the set $G$ by
$$x \sim y \iff y^{-1}x \in H.$$
Here $x,y \in G$. So, what you learned from set theory, this yields a partition, and the equivalence classes are in this case the left cosets. Try for yourself what the equivalence should be for the right cosets. Anyhow, the union of all equivalence classes make up the whole set and are disjoint. Note that the only groups among the cosets is $H$ itself, the others do not contain the trivial element and are just subsets. It looks pretty much like tiling $G$ with "copies" of the subgroup $H$ and you need $|G:H|$ (the index) such "tiles".
A: i made some quick illustrations and put it into a small video.
hope it helps :)
https://streamable.com/hgw430
most of the arguments boil down to the fact that multiplying a group by a particular element is a one-to-one map.
