Notes:Coset stuff

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

Stuff

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

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

[ilmath]H[/ilmath] itself is a coset as [ilmath]eH=H[/ilmath] clearly (for [ilmath]e[/ilmath] the identity of [ilmath]G[/ilmath])

Claims

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

Going forward

I have shown we get two equivalence relations. [ilmath]\sim[/ilmath] and [ilmath]\sim'[/ilmath] Thus we get two partitions of [ilmath]G[/ilmath], and:

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

Can we factor anything through [ilmath]\frac{G}{\sim} [/ilmath] or [ilmath]\frac{G}{\sim'} [/ilmath]?

We can factor a map, [ilmath]f:G\rightarrow X[/ilmath] (for some other thing [ilmath]X[/ilmath]) if:

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

Actually lets try factoring the group operation through this!

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

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

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

The result is, if [ilmath]h_4[/ilmath] is known to be [ilmath]e[/ilmath] then we can factor. So if [ilmath]H[/ilmath] is the trivial group, we can factor. We must have [ilmath]\times=\overline{\times}\circ\pi[/ilmath] 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:

[ilmath]\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} }[/ilmath]

If we want [ilmath]\pi[/ilmath] to be a group homomorphism we require this in fact.

Proof of claims

Proofs here

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