Left cosets of $H$ in $G$. Symmetric group of order $3$ I’m having a hard time understanding cosets. I’m working through Gallians contemporary abstract and I don’t  even understand example one!
Let $G=S_3$ and $H=\{(1),(1\,3)\}$. Then it shows me the left cosets. I know $G$ is of order $6$ and $H$ is of order $2$, so there are definitely three cosets by Lagrange. 
The coset def is for ANY $a \in G$, $Ha=\{ha\mid h \in H\}$
The preliminary step is:
$(1)H,\;(12)H,\;(13)H,\;(23)H$
If $S_3$ is $\{ 123,231,321,132,312,213\}$ why are our elements $a \in G$ above $12,13,23$. I think I get the identity one!
Further we have $1H=H$ and $13H=H$. I presume that's bc $1$ and $13$ are elements of $H$.
I totally do not get why $12H$ and $23H$ are considered and how they result in $132H$ and $123H$ respectively. I attempted the cyclic multiplication and for $12H$, I got $132H$ but with different steps and for $23H$, I got $321H$ which Ii suppose is the same as $123H$ per Gallian? But if those are identical why do we have two symmmetric members of $S_3$, $123$ and $321$, that it seems to me are not identical?
I’m spending dozens of hours trying to understand my first abstract course, which I suppose is normal, but nonetheless incredibly frustrating. 
Thanks for your help
Marco
 A: "The cosets are displayed as (1)H,(12)H,(13)H,(23)H".
This is definitely wrong. But you are right, there are 3 cosets, since the group has order 6 and the subgroup has order 2.
The cosets are 
$H = (1)H = (13)H$, 
$(123)H =\{(123)(12)\}$ and 
$(132)H = \{(132),(23)\}$.
A: First of all the elements of $S_3$ are permutations, so $(123)$ or $(12)$ means $1\to 2 \to 3 \to 1$ and $1 \to 2 \to 1$ respectively. What I understand from the question is that you have some difficulty in figuring out how to multiply the permutations, so first I will try to explain that and then we will get to cosets.
The group in consideration is $S_3 = \{ (1), (12),(23),(13),(123),(132)\}$ i.e. permutation on three elements namely $1,2$ and $3$.
Pick two elements from $G =S_3$ say $g_1 = (23), g_2 =(132)$, then $g_1 \cdot g_2 = (23)(132).$ So we need to figure out what this new pwermutation is? Since we know that $G$ is a group, this new element must also be in $G$ i.e. a permutation. So begin acting it on some element say $1$, then 
$g_1\cdot g_2 [1] = (23)(132) [1] = (23)[2] = 3$, notice we first apply $(132)$ and then $(23)$. So $g_1 \cdot g_2$ sends $1$ to $3$. Similarly you can check that $2$ is fixed and $3$ goes to $1$. So $g_1 \cdot g_2$ is nothing but $(13) \in G = S_3$.
Now assuming you understood how to multiply permutations, we proceed with cosets.
As you mentioned, the cosets of a subgroup $H$ are either $Hg$ or $gH$ depending on which type of cosets you are looking for (right or left).
$\textbf{Remark}$: If $H$ is not a normal subgroup, these are different.
In our case we have $H = \{(1), (13)\}$, so to find all (left) cosets, we need to find $gH = \{gh\ |\ h \in H\}\ \forall\ g \in G$
These are:
$1)\ g =(1), gH = H$ 
$2)\ g =(12), gH = \{(12)(1), (12)(13)\} = \{(12),(132)\}$
$3)\ g =(23), gH = \{(23)(1), (23)(13)\} = \{(23),(123)\}$
$4)\ g =(13), gH = \{(13)(1), (13)(13)\} = \{(13), (1)\} = H$
$5)\ g =(123), gH = \{(123)(1), (123)(13)\} = \{(123), (12)\}$
$6)\ g =(132), gH = \{(132)(1), (132)(13)\} = \{(132),(23)\}$
Hence there are precisely $3$ distinct cosets. 
