Let G be a group and H be a subgroup of G. Define R by $xRy \iff xy^{-1} \in H$. Show R is an equivalence relation.
$\textbf{Definition:}$ R is a relation on X. R is an equivalence relation of X if R satifies the following for $\forall x,y,z \in X$:
- xRx (relexive)
- If xRy, then yRx (symmetric)
- If xRy and yRz, then xRz (transitive)
$\textbf{Proof:}$
-
Reflexive
$$xRx \iff xx^{-1}=1$$
Since $1 \in H$. Hence R is relexive. -
Symmetric
$$xRy \iff xy^{-1} \iff (x^{-1}y)^{-1}$$
So $x^{-1}y \in H$ so does it's inverse. Hence R is symmetric.
- Transitive
Let $x=(a,b),y=(c,d),z=(e,f) \in A$. Assume the following:
\begin{equation*}
\begin{aligned}
xRy & \iff xy^{-1} \\
yRz & \iff yz^{-1} \\
\end{aligned}
\end{equation*}
Using this facts, if we multiply xRy and yRz, we have $xz^{-1} \in H$. Hence xRz.Thus R is transitive.
Since all the properties are satified, R is an equivalence class.
Best Answer
The ideas are good, but badly exposed.
Reflexivity Let $x\in G$; we want to show that $x\mathrel{R}x$, that is, $xx^{-1}\in H$. This is true, because $xx^{-1}=1\in H$.
Simmetry Let $x,y\in G$ and suppose $x\mathrel{R}y$. We want to show that $y\mathrel{R}x$, that is, $yx^{-1}\in H$. By hypothesis, $xy^{-1}\in H$, so, by the properties of a subgroup, $(xy^{-1})^{-1}\in H$; but, by general rule, $(xy^{-1})^{-1}=yx^{-1}$, so we are done.
Transitivity Let $x,y,z\in G$ and suppose $x\mathrel{R}y$ and $y\mathrel{R}z$. We want to show that $x\mathrel{R}z$, that is $xz^{-1}\in H$. We know that $xy^{-1}\in H$ and $yz^{-1}\in H$, so $(xy^{-1})(yz^{-1})\in H$. But $(xy^{-1})(yz^{-1})=xz^{-1}$, so we are done.