first my definition of a coset that I got is: Definition. Let $H$ be a subgroup of a group $G$. A left coset of $H$ in $G$ is a subset of $G$ that is of the form $xH$, where $x\in G$ and $xH = \{y\in G | y = xh \text{ for some }h\in H\}.$
And here is the theorem that is confusing me.
Let $H$ be a subgroup of a group $G$. Then the left cosets of $H$ in $G$ have the following properties:
- $x\in xH$ for all $x\in G$;
- if $x$ and $y$ are elements of $G$, and if $y = xa$ for some $a\in H$, then $xH = yH$;
- if $x$ and $y$ are elements of $G$, and if $xH \cap yH$ is non-empty then $xH = yH$.
$\textbf{Proof.}$ Suppose $x\in G$. Then $x = x1$, but $1$ is in $G$ and $H$ since $H$ is a subgroup of $G$. Thus $x\in xH$, which proves (1).
Suppose $x$ and $y$ are elements of $G$ and we can choose some $a\in H$ such that $y = xa$. Now suppose $z\in yH$. It follows that $z = yh$ for some $h\in H$. But $y = xa$, so we know that $z = xah$. Since $H$ is a subgroup and $a\in H$ and $h\in H$ it follows that $ah \in H$, hence $z\in xH$. Since $z$ was an arbitrary element of $yH$ we can conclude that $yH\subseteq xH$. Now suppose $z\in xH$. It follows that $z = xh$ for some $h\in H$. But $x =ya^{-1}$, so $z = ya^{-1}h$. Since $H$ is a subgroup and $a\in H$ it follows that $a^{-1}$ is in $H$. Similarly, since $a^{-1}\in H$ and $h\in H$ it follows that $a^{-1}h \in H$, thus $z\in yH$. Since $z$ was an arbitrary element of $xH$ we can conclude that $xH \subseteq yH$. Therefore we can conclude that $xH = yH$, which proves (2).
Now suppose that $x$ and $y$ are elements of $G$ and $xH \cap yH$ is non-empty. Since $xH \cap yH$ is non-empty we can choose some $z\in xH \cap yH$. Since $z\in xH\cap yH$ it follows that $z\in xH$ and $z\in yH$. Thus $z = xh_x$ with $h_x \in H$ and $z = yh_y$ with $h_y \in H$. Thus, since $z = z$ in all groups, $yh_y = xh_x$. Hence, $y =xh_xh_y^{-1}$. Since $H$ is a subgroup $x\in H$, $h_x\in H$ and $h_y\in H$, we can conclude that $xh_xh_y^{-1} \in H$. Therefore by (2) we know that $xH = yH$, which proves (3). $\Box$
Now my confusion stems from (1) and (3).
Suppose $G = \{x_1, x_2, ..., x_n\}$ under $\star$ Then each element $x_i$ of $\{x_1, x_2, ..., x_n\}$ is a member of $x_iH$ by (1).
By (3) we know that no element $x_i$ can be a member of two sets $x_iH$ and $x_jH$.
But since each $x_iH$ is a subset of $\{x_1, x_2, x_3, ..., x_n\}$, so it would seem to me that each coset would have to contain only one element.
Where am I wrong in my thinking here? Thank you very much