Cosets of S3 and Permutations Quoting:
"Let H be the subgroup of S3 defined by the permutations
{(1); (123); (132)}. The left cosets of H are
(1)H = (123)H = (132)H = {(1); (123); (132)}
(12)H = (13)H = (23)H = {(12); (13); (23)} "
I am a bit stock here, I am not understanding the meaning of the first permutation (1).
I believe I understand the rest such as (12) means that 1 is mapped onto 2, 2 is mapped onto 1, and 3 onto 3.
Second, I am not understanding why (1)H, (123)H, and (132)H are equal.
I believe that {(1); (123); (132)} and {(12); (13); (23)} are the two cosets.
 A: $(1)H$ is shorthand for: The set of elements that you get when you multiply $(1)h$ for each of the elements $h \in H$.
Similarly, $(123)H$ is shorthand for: The set of elements that you get when you multiply $(123)h$ for each of the elements $h \in H$.
Finally, $(132)H$ is shorthand for: The set of elements that you get when you multiply $(132)h$ for each of the elements $h \in H$.
Try listing out the set resulting from each of these three scenarios, and you will see that it is the same every time.
A: The permutation $(1)$ simply means the identity, which sends:
$1 \mapsto 1\\2 \mapsto 2\\3 \mapsto 3.$
We could also denote it by $(1)(2)(3)$ (three disjoint $1$-cycles), which might be a bit less confusing. It's common practice to omit $1$-cycles from permutations, but in the case of the identity map, that leaves us "nothing to write down" (which of course is what the identity map does-namely, "nothing").
If you apply the permutation $(1\ 2)$ to each element of $H$, which is what is meant by:
$(1\ 2)H$, you obtain (applying the pemutation on the right first, as is often done with composition):
$(1\ 2)(1) = (1\ 2)$
$(1\ 2)(1\ 2\ 3) = (2\ 3)$
$(1\ 2)(1\ 3\ 2) = (1\ 3)$, which is what you surmised.
It's not that hard to see that $\{(1), (1\ 2\ 3), (1\ 3\ 2)\}$ form a cyclic subgroup of order $3$, which we might also denote as:
$H = \{1,a,a^2\}$ (since $(1\ 2\ 3)(1\ 2\ 3) = (1\ 3\ 2)$).
then $(1)H  = 1H = H$ (this is obvious), while:
$(1\ 2\ 3)H = aH = \{a,a^2,a^3 = 1\}$, which is the same SET (the order of elements doesn't matter in a set) as $H$.
Similarly, $(1\ 3\ 2)H = a^2H = \{a^2,a^3 = 1,a^4 = a^3a = a\} = H$.
It turns out in this case (and you might well speculate as to what happens in general) that for $\sigma \in S_3,$ we have:
$\sigma H = H \iff \sigma \in H$
