2
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ No. For instance, $x\mathrel{R}x$ if and only if $xx^{-1}\in H$. The fact that $xx^{-1}=1$ is true by definition. Similar quirks are also in the other arguments. $\endgroup$
    – egreg
    Sep 11, 2014 at 20:34
  • 2
    $\begingroup$ $xRy\iff xy^{-1}\iff (x^{-1}y)^{-1}$ is nonsense. You mean $xy^{-1}\in H,\dots$. $\endgroup$ Sep 11, 2014 at 20:35
  • 1
    $\begingroup$ San with transitive. You can only write "$P\iff Q$" when $P$ and $Q$ are statements. $xy^{-1}$ is not a statement, it is an expression. $xy^{-1}\in H$ is a statement. $\endgroup$ Sep 11, 2014 at 20:36
  • 1
    $\begingroup$ Note that (generally) $(ab)^{-1} = b^{-1}a^{-1} \neq a^{-1}b^{-1}$ $\endgroup$ Sep 11, 2014 at 20:36
  • 1
    $\begingroup$ Yes, if $G$ is non commutative, it is not true that $xy^{-1}= (x^{-1}y)^{-1}$. $\endgroup$ Sep 11, 2014 at 20:37

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thank you that what I was worried about $\endgroup$ Sep 11, 2014 at 21:13
  • $\begingroup$ @UsernameUnknown I find it more useful and instructing to write proofs verbosely, not as a sequence of symbols. As others commented, be careful that $(xy)^{-1}=y^{-1}x^{-1}$, not $x^{-1}y^{-1}$. $\endgroup$
    – egreg
    Sep 11, 2014 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.