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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.

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    $\begingroup$ 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
    Commented Sep 11, 2014 at 20:34
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    $\begingroup$ $xRy\iff xy^{-1}\iff (x^{-1}y)^{-1}$ is nonsense. You mean $xy^{-1}\in H,\dots$. $\endgroup$ Commented Sep 11, 2014 at 20:35
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    $\begingroup$ 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. $\endgroup$ Commented Sep 11, 2014 at 20:36
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    $\begingroup$ Note that (generally) $(ab)^{-1} = b^{-1}a^{-1} \neq a^{-1}b^{-1}$ $\endgroup$ Commented Sep 11, 2014 at 20:36
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    $\begingroup$ Yes, if $G$ is non commutative, it is not true that $xy^{-1}= (x^{-1}y)^{-1}$. $\endgroup$ Commented Sep 11, 2014 at 20:37

1 Answer 1

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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.

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  • $\begingroup$ Thank you that what I was worried about $\endgroup$ Commented 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
    Commented Sep 11, 2014 at 21:14

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