In the definition of a) Cosets and b) Classes, there always seems to be some demands for the order, since we first pick up some element to generate left coset $aH$ from a subgroup $H$, while we first pick up some $a$ to form class $C_a$; ... then we demand next element not in the previous cosets (classes) we already have. Say, $z\notin aH, bH, cH...$, ($z\notin C_a, C_b, C_c...$).
So, the set of cosets (classes) we have seem to depend on the order of elements we choose. I'm not sure if we would end up having exactly the same classes if we first choose b.
And my questions are, A: is the definition of cosets for a subgroup $H$ really order-dependent? B:Is the definition of classes really order-dependent?
Thank you in advance!
Cheers, Collin
P.S.: I'm a physics student, not major in mathematics, so please involve as few math concepts as possible. I would really appreciate that.
(Below are the definitions from our professor, those who familiar with them surely don't need them):
a) Definition for Left Cosets:
Suppose $H=\{e, h_1, h_2, h_3... h_k\}$ is a subgroup of $G$. Now we try to form $H$'s left cosets:
$a\notin H$: $$aH:=\{a, ah_1,..., ah_k\}.$$ $b\notin H, b\notin aH,$
$$bH:=\{b,bh_1,..., bh_k\}.$$ $c\notin H, c\notin aH, c\notin bH$
$$cH:=\{c,ch_1,..., ch_k\}.$$ ... Until we run out of all the remaining elements.
b)Definition of Classes:
Classes basically are sets, and constructed as follows, pick a group element a and construct the set $\{gag^{-1}: \text{for all g}\in G\}.$ We call it $C_a$. Having constructed $C_a$, pick a group element b that does not lie in $C_a$ and construct the set $gbg^{-1}$ for all g in G. This is the class $C_b$.
No, there is no order dependence, and the one-at-a-time sequential construction you have apparently been taught is definitely not the standard way to define the concepts.
A much more usual definition of, say, conjugacy classes would be
In general it will happen that most of the conjugacy classes are produced more than one time by this definition -- but the class will be the same each time it is produced, so that doesn't matter.
You should be able to prove that if $\{gag^{-1}\mid g\in G\}$ and $\{gbg^{-1}\mid g\in G\}$ have any element in common, then they are the same set. So the conjugacy classes actually partition the group.
It's the same way with cosets.
If you want to write down the conjugacy classes concretely, you can save some time by doing the step-by-step construction you describe. This saves you some time by not computing a class you have already found once -- but the outcome of it is still exactly all the subsets of the group that can be produced as $\{gbg^{-1}\mid g\in G\}$ for any $b$.