Difference between revisions of "Normal subgroup"
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Let {{M|(G,\times)}} be a [[Group|group]] and {{M|H}} a [[Subgroup|subgroup]] of {{M|G}}, we say {{M|H}} is a '''normal subgroup'''<ref name="LangUA">Undergraduate Algebra - Serge Lang</ref> of {{M|G}} if: | Let {{M|(G,\times)}} be a [[Group|group]] and {{M|H}} a [[Subgroup|subgroup]] of {{M|G}}, we say {{M|H}} is a '''normal subgroup'''<ref name="LangUA">Undergraduate Algebra - Serge Lang</ref> of {{M|G}} if: | ||
* <math>\forall x\in G[xH=Hx]</math> where the {{M|xH}} and {{M|Hx}} are left and right [[Coset|cosets]] | * <math>\forall x\in G[xH=Hx]</math> where the {{M|xH}} and {{M|Hx}} are left and right [[Coset|cosets]] | ||
| + | ** This is the sameas saying: {{M|1=\forall x\in G[xHx^{-1}=H]}} | ||
| + | According to Serge Lang<ref name="LangUA"/> this is equivalent (that is say '''''if and only if''''' or {{M|\iff}}) | ||
| + | * {{M|H}} is the kerel of some [[Homomorphism|homomorphism]] of {{M|G}} into some other group | ||
| + | |||
| + | ==Proof of claims== | ||
| + | {{Begin Theorem}} | ||
| + | Claim 1: <math>\forall x\in G[xH=Hx]\iff\forall x\in G[xHx^{-1}=H]</math> | ||
| + | {{Begin Proof}} | ||
| + | '''Proof of: <math>\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H]</math>''' | ||
| + | : Suppose that for whatever {{M|g\in G}} we have that {{M|1=gH=Hg}} - we wish to show that for any {{M|1=x\in G[xHx^{-1}=H]}} | ||
| + | :: Let {{M|x\in G}} be given. | ||
| + | :: Recall that <math>X=Y\iff[X\subseteq Y\wedge X\supseteq Y]</math> so we need to show: | ||
| + | ::# <math>xHx^{-1}\subseteq H</math> | ||
| + | ::#* By the [[Implies and subset relation|implies-subset relation]] this is the same as <math>y\in xHx^{-1}\implies y\in H</math> | ||
| + | ::# <math>xHx^{-1}\supseteq H</math> | ||
| + | ::#* By the [[Implies and subset relation|implies-subset relation]] this is the same as <math>y\in H\implies y\in xHx^{-1}</math> | ||
| + | :: Let us show 1: | ||
| + | ::: Suppose <math>y\in xHx^{-1}</math> we wish to show <math>\implies y\in H</math> (that is <math>xHx^{-1}\subseteq H</math>) | ||
| + | :::: <math>y\in xHx^{-1}\implies \exists h_1\in H:y=xh_1x^{-1}</math> | ||
| + | :::: <math>\implies yx=xh_1</math> - note that {{M|xh_1\in xH}} | ||
| + | ::::: By hypothesis, <math>\forall g\in G[gH=Hg]</math> | ||
| + | :::: So, as {{M|1=yx=xh_1\in xH}} we see {{M|1=yx\in Hx}} | ||
| + | :::: This means <math>\exists h_2\in H</math> such that <math>yx=h_2x</math> | ||
| + | ::::: Using [[Group#Cancellation laws|the cancellation laws for groups]] we see that | ||
| + | :::: <math>y=h_2</math> as {{M|h_2\in H}} we must have {{M|y\in H}} | ||
| + | ::: We have now shown that <math>[y\in xHx^{-1}\implies y\in H]\iff[xHx^{-1}\subseteq H]</math> | ||
| + | :: Now to show 2: | ||
| + | ::: Suppose that {{M|y\in H}} we wish to show that <math>\implies y\in xHx^{-1}</math> (that is <math>H\subseteq xHx^{-1}</math>) | ||
| + | :::: Note that {{M|yx\in Hx}} by definition of {{M|Hx}} | ||
| + | ::::: By hypothesis {{M|1=Hx=xH}} so | ||
| + | :::: we see that {{M|yx\in xH}} | ||
| + | :::: this means {{M|1=\exists h_1\in H[yx=xh_1]}} | ||
| + | :::: and this means {{M|1=y=xh_1x^{-1} }} | ||
| + | :::: such a {{M|h_1}} existing is the very definition of {{M|xh_1x^{-1}\in xHx^{-1} }} | ||
| + | :::: thus {{M|y\in xHx^{-1} }} | ||
| + | ::: We have now shown that <math>[y\in H\implies y\in xHx^{-1}]\implies[H\subseteq xHx^{-1}]</math> | ||
| + | : Combining this we hve shown that {{M|1=\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H]}} | ||
| + | |||
| + | |||
| + | |||
| + | Next: | ||
| + | |||
| + | '''Proof of: <math>\forall x\in G[xHx^{-1}=H]\implies\forall x\in G[xH=Hx]</math>''' | ||
| + | {{Todo|Simple proof}} | ||
| + | {{End Proof}}{{End Theorem}} | ||
| − | |||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Definition|Abstract Algebra}} | {{Definition|Abstract Algebra}} | ||
Revision as of 16:33, 16 May 2015
Definition
Let [ilmath](G,\times)[/ilmath] be a group and [ilmath]H[/ilmath] a subgroup of [ilmath]G[/ilmath], we say [ilmath]H[/ilmath] is a normal subgroup[1] of [ilmath]G[/ilmath] if:
- [math]\forall x\in G[xH=Hx][/math] where the [ilmath]xH[/ilmath] and [ilmath]Hx[/ilmath] are left and right cosets
- This is the sameas saying: [ilmath]\forall x\in G[xHx^{-1}=H][/ilmath]
According to Serge Lang[1] this is equivalent (that is say if and only if or [ilmath]\iff[/ilmath])
- [ilmath]H[/ilmath] is the kerel of some homomorphism of [ilmath]G[/ilmath] into some other group
Proof of claims
Claim 1: [math]\forall x\in G[xH=Hx]\iff\forall x\in G[xHx^{-1}=H][/math]
Proof of: [math]\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H][/math]
- Suppose that for whatever [ilmath]g\in G[/ilmath] we have that [ilmath]gH=Hg[/ilmath] - we wish to show that for any [ilmath]x\in G[xHx^{-1}=H][/ilmath]
- Let [ilmath]x\in G[/ilmath] be given.
- Recall that [math]X=Y\iff[X\subseteq Y\wedge X\supseteq Y][/math] so we need to show:
- [math]xHx^{-1}\subseteq H[/math]
- By the implies-subset relation this is the same as [math]y\in xHx^{-1}\implies y\in H[/math]
- [math]xHx^{-1}\supseteq H[/math]
- By the implies-subset relation this is the same as [math]y\in H\implies y\in xHx^{-1}[/math]
- [math]xHx^{-1}\subseteq H[/math]
- Let us show 1:
- Suppose [math]y\in xHx^{-1}[/math] we wish to show [math]\implies y\in H[/math] (that is [math]xHx^{-1}\subseteq H[/math])
- [math]y\in xHx^{-1}\implies \exists h_1\in H:y=xh_1x^{-1}[/math]
- [math]\implies yx=xh_1[/math] - note that [ilmath]xh_1\in xH[/ilmath]
- By hypothesis, [math]\forall g\in G[gH=Hg][/math]
- So, as [ilmath]yx=xh_1\in xH[/ilmath] we see [ilmath]yx\in Hx[/ilmath]
- This means [math]\exists h_2\in H[/math] such that [math]yx=h_2x[/math]
- Using the cancellation laws for groups we see that
- [math]y=h_2[/math] as [ilmath]h_2\in H[/ilmath] we must have [ilmath]y\in H[/ilmath]
- We have now shown that [math][y\in xHx^{-1}\implies y\in H]\iff[xHx^{-1}\subseteq H][/math]
- Suppose [math]y\in xHx^{-1}[/math] we wish to show [math]\implies y\in H[/math] (that is [math]xHx^{-1}\subseteq H[/math])
- Now to show 2:
- Suppose that [ilmath]y\in H[/ilmath] we wish to show that [math]\implies y\in xHx^{-1}[/math] (that is [math]H\subseteq xHx^{-1}[/math])
- Note that [ilmath]yx\in Hx[/ilmath] by definition of [ilmath]Hx[/ilmath]
- By hypothesis [ilmath]Hx=xH[/ilmath] so
- we see that [ilmath]yx\in xH[/ilmath]
- this means [ilmath]\exists h_1\in H[yx=xh_1][/ilmath]
- and this means [ilmath]y=xh_1x^{-1}[/ilmath]
- such a [ilmath]h_1[/ilmath] existing is the very definition of [ilmath]xh_1x^{-1}\in xHx^{-1} [/ilmath]
- thus [ilmath]y\in xHx^{-1} [/ilmath]
- Note that [ilmath]yx\in Hx[/ilmath] by definition of [ilmath]Hx[/ilmath]
- We have now shown that [math][y\in H\implies y\in xHx^{-1}]\implies[H\subseteq xHx^{-1}][/math]
- Suppose that [ilmath]y\in H[/ilmath] we wish to show that [math]\implies y\in xHx^{-1}[/math] (that is [math]H\subseteq xHx^{-1}[/math])
- Combining this we hve shown that [ilmath]\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H][/ilmath]
Next:
Proof of: [math]\forall x\in G[xHx^{-1}=H]\implies\forall x\in G[xH=Hx][/math]
TODO: Simple proof
