Notes:Coset stuff

See /Quotient group for this applied to the quotient group (which is after all the point)

Stuff

Let $(G,\times)$ be a group and let $H\subseteq G$ be a subgroup. Proper or not. Then

• Any set of the form $gH$ is called a left coset, where $gH:=\{g\times h\ \vert\ h\in H\}$
• Any set of the form $Hg$ is called a right coset, where $Hg:=\{h\times g\ \vert\ h\in H\}$

$H$ itself is a coset as $eH=H$ clearly (for $e$ the identity of $G$)

Claims

1. For $x,y\in G$ we can define an equivalence relation on $G$: $x\sim y\iff x^{-1}y\in H$. Done - Alec (talk) 21:23, 23 October 2016 (UTC)
   • By symmetry this is/must be the same as $y^{-1}x\in H$ - this is true as $H$ is a subgroup.
2. $\forall x,g\in G[x\in[g]\iff x\in gH]$. Done - Alec (talk) 21:23, 23 October 2016 (UTC)
3. For $x,y\in G$ we can define another equivalence relation on $G$: $x\sim'y\iff xy^{-1}\in H$. Done - Alec (talk) 21:23, 23 October 2016 (UTC)
   • By symmetry this is/must be the same as $yx^{-1}\in H$ - this is true as $H$ is a subgroup.
4. $\forall x,g\in G[x\in[g]'\iff x\in Hg]$. Done - Alec (talk) 21:23, 23 October 2016 (UTC)

Going forward

I have shown we get two equivalence relations. $\sim$ and $\sim'$ Thus we get two partitions of $G$, and:

• $\pi:G\rightarrow\frac{G}{\sim}$ given by $\pi:g\rightarrow[g]$ and $\pi':G\rightarrow\frac{G}{\sim'}$ given by $\pi':g\rightarrow[g]'$

Can we factor anything through $\frac{G}{\sim}$ or $\frac{G}{\sim'}$?

We can factor a map, $f:G\rightarrow X$ (for some other thing $X$) if:

• $\forall g,h\in G[\pi(g)=\pi(h)\implies f(g)=f(h)]$

Actually lets try factoring the group operation through this!

• $\forall (g,h),(g',h')\in G[\pi(g,h)=\pi(g',h')\implies \times(g,h)=\times(g',h')]$
  • Then $([g],[h])=([g'],[h'])$ so $[g]=[g']$ and $[h]=[h']$
  • So $g\sim g'$ and $h\sim h'$
    • Thus $\exists h_1,h_2\in H$ such that $g^{-1}g'=h_1$ and $h^{-1}h'=h_2$. We want to show $gh=g'h'$
      • Well $h_1h_2=g^{-1}g'h^{-1}h'$

Here we get stuck. If we had $gH=Hg$ then we could go further and factor the $\times$ through, then we can proceed to:

• $g'h_4h'=gh$ for some $h_4\in H$ (I did $h_4:=h_1h_3$ on paper but I don't know if I used the same $h_1$ here. $h_3$ comes from turning either $gH$ into $Hg$ or $hH$ into $Hh$ with $h_1$ or $h_2$ what it was "before")

The result is, if $h_4$ is known to be $e$ then we can factor. So if $H$ is the trivial group, we can factor. We must have $\times=\overline{\times}\circ\pi$ so this is another way of saying a group "over" the trivial group is isomorphic to the group.

Really stupid and pointless way of saying it.

We can get multiplication through:

$\xymatrix { G\times G \ar[dr]^{\pi\circ\times} \ar[r]^\times \ar[d]_{(\pi,\pi)} & G \ar[d]^{\pi} \\ \frac{G}{\sim}\times\frac{G}{\sim} \ar@{.>}[r]_{\overline{x} } & \frac{G}{\sim} }$

If we want $\pi$ to be a group homomorphism we require this in fact.

Proof of claims

[Expand]

Proofs here

---

1. $x\sim y\iff x^{-1}y\in H$ is an equivalence relation
   1. Reflexive: $x\sim x$ holds
      • Trivial, $x^{-1}x=e\in H$ as $H$ a subgroup
   2. Symmetric: $x\sim y\implies y\sim x$
      • Suppose $x\sim y$, then $x^{-1}y\in H$, i.e. $\exists h\in H$ such that $x^{-1}y=h]$
        • Thus $y=xh$ so $e=y^{-1}xh$ and lastly $h^{-1}=y^{-1}x$. Notice $h^{-1}\in H$ as $H$ is a subgroup.
      • So $y^{-1}x\in H$ too! And $y^{-1}x\in H\iff y\sim x$. As required.
   3. Transitive: $\forall x,y,z\in G[(x\sim y\wedge y\sim z)\implies x\sim z]$
      • Suppose $x,y,z\in G$ given such that $x\sim y$ and $y\sim z$ then:
        • $\exists h_1,h_2\in H$ such that $x^{-1}y=h_1$ and
          • As $H$ is a subgroup $h_1h_2\in H$. So we see:
          • $h_1h_2=x^{-1}yy^{-1}z\in H$, tidying up: $x^{-1}z\in H$
        • But $x^{-1}z\in H\iff x\sim z$
      • Since the $x,y,z$ were arbitrary we have shown this for all. As required.
2. • Let $gH$ be a coset. I claim this is $[g]$. That is:
     • $x\in gH\iff x\in [g]$ (by the implies-subset relation and from $gH\subseteq[g]$ and $[g]\subseteq gH$)
       1. $\implies$
          • Let $x\in gH$ be given. Then $\exists h_1\in H[x=gh_1]$
            • But then $g^{-1}x=h_1$ so $g^{-1}x\in H$ so $g\sim x$ or $x\sim g$, Thus $x\in [g]$ as $[g]:=\{k\in G\ \vert\ k\sim g\}$
       2. $\impliedby$
          • Let $x,g\in G$ be given, then we claim $x\in[g]\implies x\in gH$
            • If $x\in[g]$ then $x\sim g$ so $x^{-1}g\in H$ so $\exists h_2\in H[x^{-1}g=h]$
              • Thus $g=xh$ so $gh^{-1}=x$
            • As $H$ is a subgroup $h^{-1}\in H$. So $gh^{-1}\in gH$ so $x=gh^{-1}\in gH$ or just:
          • $x\in gH$ as required.
3. Follows by doing 1 again but with $xy^{-1}$ instead
4. Follows by doing 2 again, but slightly differently. I should copy and paste it and make the alterations.