Proving $a \in Hb \implies Ha = Hb$ This seems like an elementary property of cosets, I saw if first in A Book of Abstract Algebra by Pinter:
Let $H$ be a subgroup of a group $G$
$$ a \in Hb \implies Ha = Hb $$
My guess is, since $Ha$ and $Hb$ are sets, to prove their equality first I have to prove that $Ha \subseteq Hb$ and $Hb \subseteq Ha$. Pinter's proof is as follows:
We are given that $a \in Hb$ which means that $a = h_1b$ for some $h_1 \in H$. We need to prove that $Ha = Hb$
Let $x \in Ha$, this means that $x = h_2a$ for some $h_2 \in H$. But $a = h_1b$ so $x = h_2a = (h_2h_1)b$ which is clearly in $Hb$. This proves that every $x \in Ha$ is in $Hb$ and analogously we may show that every $y \in Hb$ is in $Ha$, and therefore $Ha = Hb$
My problem is with this "analogous" proof of $Hb \subseteq Ha$, I don't see at all how to start to develop it (maybe because I don't really get cosets yet). I reached for Contemporary Abstract Algebra by Gallian and the same part of the proof is left out. This seems so trivial everyones skips it yet I can't find out how to do it
 A: I think you should see this property as a corollary to the fact that $H\backslash G=\{\,Hg\mid g\in G\,\}$ is a  partition of the set $G$. Because it is a (defining) property of partitions that whenever two parts have an element in common, they are identical (or by contrapositive: if two parts differ, their intersection is empty). For, with obviously $a\in Ha$, the condition $a\in Hb$ means that the parts $Ha $ and $Hb$ have an element in common, namely $a$; the cited property then immediately gives you the conclusion.
So why is $H\backslash G$ a partition of the set$~G$? Well the three defining properties are easy to establish. (1) The already mentioned $a\in Ha$ ensures that all parts are nonempty , and (2) since it holds for all $a\in G$ , the union of all$~Ha$ clearly (contains, and therefore) is equal to $G$. The crucial condition (3) , which is really all that is needed here, says that if $\def\thru{\cap}Ha\thru Hb$ is non-empty, then $Ha=Hb$. Let $c\in Ha\thru Hb$, so there exist $h_1,h_2\in H$ such that $c=h_1a=h_2b$. Now $a=h_1^{-1}h_2 b$ and $b=h_2^{-1}h_1 a$ so any $ha\in Ha$ can be written $ha=hh_1^{-1}h_2 b\in Hb$ and any $hb\in Hb$ can be written $hb=hh_2^{-1}h_1 a\in Ha$ , proving both inclusions of $Ha=Hb$.
This was actually a bit of extra work, but knowing that $H\backslash G$, or indeed any set of $H$-orbits for an action of $G$ (which is proved similarly) is a partition, is a fundamental fact that is useful for many other occasions.
