Difference between left and right cosets From my teacher's notes:

"Suppose $H$ is a subgroup of $G$. For any $a\in G$, we define its associated left coset and right coset to be $$aH = \{ab | b \in H\}$$ $$Ha = \{ba | b\in H\}$$ respectively."



What is a coset? What are the differences between left and right?

Also from the notes:

"Suppose $H$ is a subgroup of $G$. The left cosets $\{aH | a\in G\}$ define a partition of $G$, so are the right cosets."
What does "define a partition of $G$" mean?

 A: Let $G$ be a group and $H$ be a subgroup of $G$. Let $a\in G$. The left coset of $H$ in $G$ with respect to $a$ is the set $$aH=\{ah:h\in H\}$$ while the right coset of $H$ in $G$ with respect to $a$ is the set $$Ha=\{ha:h\in H\}$$
For $a,b\in G$,  $ab=ba$ is not necessarily true, that is, $G$ is not necessarily abelian, so $Ha$ and $aH$ are different set.
For example, take $G=S_3$ and $H=\{1,(12)\}$. Try to verify that $H(13)\neq (13)H$. 
Let $\mathcal{A}=\{S_i:i\in I\}$ be a family of non-empty subsets of $G$. $\mathcal{A}$ partitions $G$ if
(i)For $i\neq j$, $S_i\cap S_j=\phi$
(ii) $\bigcup_{i\in I}S_i=G$
In other words, every element of $G$ belongs to exactly one set $S_i$ for some $i\in I$. 
So in this case, $\mathcal{A}=\{Ha:a\in G\}$ fulfils the conditions and form a partition of $G$.
A: Alan gives a good answer with the definitions of cosets and a partition. I hope to provide some intuition on cosets and partitions via examples. Given a set $S$, a partition of a set $S$ is a collection of disjoint subsets of $S$ whose union is all of $S$. Consider the set $S=\{1,2,3,4\}$. Then the subsets $\{1\}$, $\{2,3\}$, and $\{4\}$ are all disjoint with union $S$. 
The difference between left and right cosets depends on the structure of your group and which subgroups you choose to look at. For example, one of the comments above notes that in abelian groups, left and right cosets are always the same regardless of which subgroup you choose (try to prove this). 
Like other say, a good example to look at is $S_3$. It is important to notice that being nonabelian is not enough to say that the cosets will be different. Consider the subgroup of $S_3$ commonly referred to as the alternating group on $3$ letters: $A_3=\{1, (1\,2\,3),(1\,3\,2)\}$. Notice
$$ (1\,2)A_3=\{(1\,2),(1\,2)(1\,2\,3), (1\,2)(1\,3\,2)\}=\{(1\,2),(2\,3),(1\,3)\}.$$
Compute $A_3(1\,2)$ and see what happens! Hint: They should be the same. The subgroup $A_3$ is an example of what is called a normal subgroup of $S_3$. We can define normal subgroups in terms of left and right cosets by imposing the condition that they coincide, i.e., A subgroup $H$ of a group $G$ is said to be normal if for every $g\in G$,
$$gH=Hg.$$
If a subgroup is normal, then there are no differences between the cosets.
For an example of cosets paritioning a group, look at $A_3$ again. We have
$$A_3\cup (1\,2)A_3=S_3$$
and $A_3 \cap (1\,2)A_3=\emptyset$. No matter what elements you multiple $A_3$ on the left by, you will only get these two cosets (again, try it yourself). So the collection of left (or right) cosets of $A_3$ in $S_3$ partition $S_3$. Note that the subgroup need not be normal in the bigger group in order for the set of cosets to form a partition!
