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Let $H$ be a subgroup of G, and define a relation $∼$ on G by the rules that $x∼y$ mean $x^{-1}y\in H $. Show that $∼$ is an equivalence relation and its equivalence classes are the left cosets of $H$.

My attempt

We know that a relation $R$ on a set $X$ is a set of ordered pairs of members of $X$ which satisfies the condition of the given relation.

To prove that the relation is an equivalence relation, we need to check if the relation $∼$ have all the properties

  • $∼$ is reflexive if $x∼x$ for all $x\in G$
  • $∼$ is symmetric if $x∼y \Rightarrow y∼x$ for all $x,y\in G$
  • $∼$ is transitive if $x∼y$ & $y∼z$ then $x∼z$ for all $x,y,z\in G$

So, we can see that the relation is reflexive since $$x^{-1}x=1 \forall x\in H$$ It's also symmetric $$x^{-1}y {,} \forall x,y\in H$$ and since the identity element exist we know that an element $j$ exist such that $$x^{-1}yj=1 <=> j=y^{-1}x \in H$$ Therefore its symmetric. The relation is also transitive: $$x∼y .AND. y∼z. THEN.x∼z \forall x,y,z \in H$$ $$(x^{-1}y)(y^{-1}z)=x^{-1}z \in H$$ Since the relation have all the properties listed above, the relation is an equialance relation. Now we are going to show that the equivalence classes of this relation is the left cosets. Since the distinct left cosests form a partition of $G$, its equal to the equivalence classes because the equivalence classes are the parts of the partition of $G$, which means the equivalence classes are also forming the partition.

The equivalence classes are the set

$$[x]=\{y \in G | y∼x\}$$ and by the reflexivitive we get $$[x]=\{y \in G | x∼y\}=\{y\in G|x^{-1}y\in H\}$$ $x^{-1}y\in H$ gives us that $$y=x(x^{-1}y)=y\in xH$$ This means that $$y=xh$$ for some $h \in H$ and this gives us $$h=x^{-1}y$$ which is the relation $x∼y$. Therefore $$[x]=xH$$

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2 Answers 2

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You start wrong. The property you have to prove are

  • $\sim$ is reflexive, that is, $x\sim x$ for all $x\in G$,

  • $\sim$ is symmetric, that is, for all $x,y\in G$, $x\sim y$ implies $y\sim x$,

  • $\sim$ is transitive, that is, for all $x,y,z\in G$, $x\sim y$ and $y\sim z$ implies $x\sim z$.

Note that the relation is on $G$, not on $H$.

With this correction, your proof is good, apart from the symmetry.

Suppose $x\sim y$; then $x^{-1}y\in H$ and therefore $(x^{-1}y)^{-1}\in H$. Since $(x^{-1}y)^{-1}=y^{-1}(x^{-1})^{-1}=y^{-1}x\in H$, we conclude that $y\sim x$.

The proof that $[x]=xH$ is good, but not clearly written down.

We have $[x]=\{y\in G:x\sim y\}$ by definition.

  • Suppose $y\in[x]$. Then $x^{-1}y\in H$, so $y=x(x^{-1}y)\in xH$. Therefore $[x]\subseteq xH$.

  • Suppose $y\in xH$. Then $y=xh$ for some $h\in H$, so $x^{-1}y=h\in H$. Therefore $xH\subseteq[x]$.

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  • $\begingroup$ How could i write the proof that [x]=xH ? And how could i explain it better (with words and symbols). @egreg $\endgroup$
    – Bonnie
    Commented May 23, 2014 at 10:02
  • $\begingroup$ @Bonnie What you write up to “This means that“ is the proof of $[x]\subseteq xH$. The rest is the proof of the converse inclusion: you shouldn't start it with “This means that”, but rather with “Suppose $y=xh$ with $h\in H$”. $\endgroup$
    – egreg
    Commented May 23, 2014 at 10:06
  • $\begingroup$ Ok thanks! But everything else is ok ? To prove this with equivalence classes was abit harder then the "normal" proof. $\endgroup$
    – Bonnie
    Commented May 23, 2014 at 10:19
  • $\begingroup$ @Bonnie Your argument for symmetry is convoluted and not very clear. Note that in your edit you have changed too many $H$'s into $G$'s (and you shouldn't have done it anyway). $\endgroup$
    – egreg
    Commented May 23, 2014 at 10:32
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You have the right idea for the first part of the proof, but certain things could be proved more clearly/succinctly. The main issue I have, though, comes with your attempt to prove that $[x] = xH$ for all $x \in G$. What you currently have written down shows (twice over) that $[x] \subseteq xH$ for all $xH$. How do you know that $xH \subseteq [x]$? (It might help you to instead prove (equivalently) that for all $x,y \in G$ you have $x \sim y \iff y \in xH$.)

Edit: When I said that something could be proved more clearly, I was referring to the proof that $\sim$ is symmetric (along with what I assume amounts to problems you're having with typesetting in $\LaTeX$ elsewhere). Since I'm not sure if that was due to you just rushing through typing out your solution here or you being new to writing proofs, I've provided an example below of how you might clearly write up the argument:

Assume that $x \sim y$. Then $x^{-1}y \in H$ by the definition of $\sim$. So, since $H$ is a group, we also know that $(x^{-1}y)^{-1} \in H$. But $$ (x^{-1}y)^{-1} = y^{-1}x \in H, $$ and therefore $y \sim x$.

[You should show that $(x^{-1}y)^{-1} = y^{-1}x$ directly (just like you did in your original solution) if you're not far enough along in the subject to take this for granted.]

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