Difference between revisions of "Notes:Coset stuff"

Revision as of 00:15, 24 October 2016

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

For x,y∈G we can define an equivalence relation on G: x∼y⟺x−1y∈H^{[Note 1]}. 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.
$\forall x,g\in G[x\in[g]\iff x\in gH]$ . Done - Alec (talk) 21:23, 23 October 2016 (UTC)
For x,y∈G we can define another equivalence relation on G: x∼′y⟺xy−1∈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.
$\forall x,g\in G[x\in[g]'\iff x\in Hg]$ . 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!

∀(g,h),(g′,h′)∈G[π(g,h)=π(g′,h′)⟹×(g,h)=×(g′,h′)]^{[Note 2]}
- Then $([g],[h])=([g'],[h'])$ so $[g]=[g']$ and $[h]=[h']$
- So g∼g′ and h∼h′
  - Thus ∃h1,h2∈H such that g−1g′=h1 and h−1h′=h2. 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

x∼y⟺x−1y∈H is an equivalence relation
1. Reflexive: x∼x holds
  - Trivial, $x^{-1}x=e\in H$ as $H$ a subgroup
2. Symmetric: x∼y⟹y∼x
  - Suppose x∼y, then x−1y∈H, i.e. ∃h∈H such that x−1y=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: ∀x,y,z∈G[(x∼y∧y∼z)⟹x∼z]
  - Suppose x,y,z∈G given such that x∼y and y∼z then:
    - ∃h1,h2∈H such that x−1y=h1 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.
- Let gH be a coset. I claim this is [g]. That is:
  - x∈gH⟺x∈[g] (by the implies-subset relation and from gH⊆[g] and [g]⊆gH)
    1. ⟹
      - 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. ⟸
      - 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.
Follows by doing 1 again but with $xy^{-1}$ instead
Follows by doing 2 again, but slightly differently. I should copy and paste it and make the alterations.

Notes

Jump up ↑ It must be this way as we will require $x\sim x$ , then we get $x^{-1}x=e\in H$ as $H$ is a subgroup.
Jump up ↑
We're informal about how we use π here, we really mean:
- $\pi_1:G\times G\rightarrow\frac{G}{\sim}\times\frac{G}{\sim}$ given by $\pi':(g,h)\mapsto([g],[h])=(\pi(g),\pi(h))$

[1] Jump up ↑ It must be this way as we will require $x\sim x$ , then we get $x^{-1}x=e\in H$ as $H$ is a subgroup.

[2] Jump up ↑ We're informal about how we use $\pi$ here, we really mean:
$\pi_1:G\times G\rightarrow\frac{G}{\sim}\times\frac{G}{\sim}$ given by $\pi':(g,h)\mapsto([g],[h])=(\pi(g),\pi(h))$

[3] $\pi_1:G\times G\rightarrow\frac{G}{\sim}\times\frac{G}{\sim}$ given by $\pi':(g,h)\mapsto([g],[h])=(\pi(g),\pi(h))$

[Note 1]

[Note 2]

@@ Line 12: / Line 12: @@
 # {{M|1=\forall x,g\in G[x\in[g]'\iff x\in Hg]}}. Done - [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 21:23, 23 October 2016 (UTC)
 ==Going forward==
-I have shown we get two [[equivalence relations]]. {{M|\sim}} and {{M|\sim'}} Thus we get two {{plural|partition|s}} of {{M|G}}.
+I have shown we get two [[equivalence relations]]. {{M|\sim}} and {{M|\sim'}} Thus we get two {{plural|partition|s}} of {{M|G}}, and:
+* {{M|\pi:G\rightarrow\frac{G}{\sim} }} given by {{M|\pi:g\rightarrow[g] }} and {{M|\pi':G\rightarrow\frac{G}{\sim'} }} given by {{M|\pi':g\rightarrow[g]'}}
+Can we [[factor (function)|factor]] anything through {{M|\frac{G}{\sim} }} or {{M|\frac{G}{\sim'} }}?
+We can factor a map, {{M|f:G\rightarrow X}} (for some other thing {{M|X}}) if:
+* {{M|1=\forall g,h\in G[\pi(g)=\pi(h)\implies f(g)=f(h)]}}
+Actually lets try factoring the group operation through this!
+* {{M|1=\forall (g,h),(g',h')\in G[\pi(g,h)=\pi(g',h')\implies \times(g,h)=\times(g',h')]}}<ref group="Note">We're informal about how we use {{M|\pi}} here, we really mean:
+* {{M|\pi_1:G\times G\rightarrow\frac{G}{\sim}\times\frac{G}{\sim} }} given by {{M|1=\pi':(g,h)\mapsto([g],[h])=(\pi(g),\pi(h))}}</ref>
+** Then {{M|1=([g],[h])=([g'],[h'])}} so {{M|1=[g]=[g']}} and {{M|1=[h]=[h']}}
+** So {{M|g\sim g'}} and {{M|h\sim h'}}
+*** Thus {{M|1=\exists h_1,h_2\in H}} such that {{M|1=g^{-1}g'=h_1}} and {{M|1=h^{-1}h'=h_2}}. We want to show {{M|1=gh=g'h'}}
+**** Well {{M|1=h_1h_2=g^{-1}g'h^{-1}h'}}
+Here we get stuck. If we had {{M|1=gH=Hg}} then we could go further and factor the {{M|\times}} through, then we can proceed to:
+* {{M|1=g'h_4h'=gh}} for some {{M|h_4\in H}} (I did {{M|1=h_4:=h_1h_3}} on paper but I don't know if I used the same {{M|h_1}} here. {{M|h_3}} comes from turning either {{M|gH}} into {{M|Hg}} or {{M|hH}} into {{M|Hh}} with {{M|h_1}} or {{M|h_2}} what it was "before")
+The result is, if {{M|1=h_4}} is known to be {{M|e}} then we can factor. So if {{M|H}} is the [[trivial group]], we can factor. We must have {{M|1=\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:
+: <span style="font-size:1.5em;"><m>\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} }</m></span>
+If we want {{M|\pi}} to be a [[group homomorphism]] we require this in fact.
 ==Proof of claims==
+{{Begin Inline Theorem}}
+Proofs here
+{{Begin Inline Proof}}
 # {{M|1=x\sim y\iff x^{-1}y\in H}} is an equivalence relation
 ## Reflexive: {{M|x\sim x}} holds
@@ Line 42: / Line 68: @@
 # Follows by doing 1 again but with {{M|xy^{-1} }} instead
 # Follows by doing 2 again, but slightly differently. I should copy and paste it and make the alterations.
+{{End Proof}}{{End Theorem}}
 ==Notes==
 <references group="Note"/>

Difference between revisions of "Notes:Coset stuff"

Revision as of 00:15, 24 October 2016

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