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 | ||
| + | *: This can be summed up as the following two statements: | ||
| + | *:# The [[Kernel|kernel]] of a [[Homomorphism|homomorphism]] is a normal subgroup | ||
| + | *:# Every normal subgroup is the kernel of some homomorphism | ||
| + | ==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]}} | ||
| − | {{Todo| | + | |
| + | |||
| + | 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}} | ||
| + | {{Begin Theorem}} | ||
| + | Claim 2: The kernel of a homomorphism is a normal subgroup | ||
| + | {{Begin Proof}} | ||
| + | We wish to show that given a homomorphism {{M|f:G\rightarrow X}} (where {{M|X}} is some group) that the kernel of {{M|f}}, {{M|H}} is normal. Which is to say that: | ||
| + | * <math>\forall x\in G[xHx^{-1}=H]</math> (which is <math>\forall x\in G[x\text{Ker}(f)x^{-1}=\text{Ker}(f)]</math> ) | ||
| + | |||
| + | '''Proof that <math>\forall x\in G[xHx^{-1}\subseteq H]</math>''' | ||
| + | : Let {{M|x\in G}} be given | ||
| + | :: Let {{M|y\in xHx^{-1} }} be given | ||
| + | ::: Then <math>\exists h_1\in H:y=xh_1x^{-1}</math> | ||
| + | ::: <math>f(y)=f(xh_1x^{-1})=f(x)f(h_1)f(x^{-1})</math> | ||
| + | :::: But {{M|H}} is the kernel of {{M|f}} so {{M|1=f(h_1)=e}} where {{M|e}} is the identity of {{M|X}} | ||
| + | ::: <math>f(y)=f(x)ef(x^{-1})</math> | ||
| + | :::: It is a property of homomorphisms that {{M|1=f(x^{-1})=(f(x))^{-1} }} | ||
| + | ::: <math>f(y)=f(x)f(x^{-1})=f(x)f(x)^{-1}=e</math> | ||
| + | :: Thus {{M|1=y\in\text{Ker}(f)=H}} | ||
| + | : So we see that {{M|1=xHx^{-1}\subseteq H}} | ||
| + | |||
| + | |||
| + | '''Proof that <math>\forall x\in G[H\subseteq xHx^{-1}]</math>''' | ||
| + | : As before, let {{M|x\in G}} be given. | ||
| + | :: From above we know that <math>\forall x\in G[xHx^{-1}\subseteq H]</math>, using this we see that <math>x^{-1}Hx\subseteq H</math> | ||
| + | :: Let {{M|y\in H}} be given (we will show that then {{M|y\in xHx^{-1} }}) | ||
| + | ::: We know that {{M|x^{-1}Hx\subseteq H}}, as {{M|y\in H}} we know that {{M|x^{-1}yx\in H}} | ||
| + | ::: This means <math>\exists h\in H[x^{-1}yx=h]</math> | ||
| + | ::: <math>\implies yx=xh</math> | ||
| + | ::: <math>\implies y=xhx^{-1}</math> | ||
| + | ::: Thus <math>y\in xHx^{-1}</math> | ||
| + | : We have shown that <math>[y\in H\implies y\in xHx^{-1}]\iff[H\subseteq xHx^{-1}]</math> | ||
| + | |||
| + | |||
| + | We have shown that <math>\forall x\in G[xHx^{-1}\subseteq H\wedge H\subseteq xHx^{-1}]</math>, which is exactly: | ||
| + | * Given a group homomorphism {{M|f:G\rightarrow X}} where {{M|1=\text{Ker}(f)=H}}, we have shown that {{M|1=\forall x\in G[H=xHx^{-1}]}} which is the first definition of a normal subgroup | ||
| + | {{End Proof}}{{End Theorem}} | ||
| + | {{Begin Theorem}} | ||
| + | Claim 3: Every normal subgroup is the kernel of some homomorphism | ||
| + | {{Begin Proof}} | ||
| + | {{End Proof}}{{End Theorem}} | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Definition|Abstract Algebra}} | {{Definition|Abstract Algebra}} | ||
Latest revision as of 18:07, 17 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
- This can be summed up as the following two statements:
- The kernel of a homomorphism is a normal subgroup
- Every normal subgroup is the kernel of some homomorphism
- This can be summed up as the following two statements:
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
Claim 2: The kernel of a homomorphism is a normal subgroup
We wish to show that given a homomorphism [ilmath]f:G\rightarrow X[/ilmath] (where [ilmath]X[/ilmath] is some group) that the kernel of [ilmath]f[/ilmath], [ilmath]H[/ilmath] is normal. Which is to say that:
- [math]\forall x\in G[xHx^{-1}=H][/math] (which is [math]\forall x\in G[x\text{Ker}(f)x^{-1}=\text{Ker}(f)][/math] )
Proof that [math]\forall x\in G[xHx^{-1}\subseteq H][/math]
- Let [ilmath]x\in G[/ilmath] be given
- Let [ilmath]y\in xHx^{-1} [/ilmath] be given
- Then [math]\exists h_1\in H:y=xh_1x^{-1}[/math]
- [math]f(y)=f(xh_1x^{-1})=f(x)f(h_1)f(x^{-1})[/math]
- But [ilmath]H[/ilmath] is the kernel of [ilmath]f[/ilmath] so [ilmath]f(h_1)=e[/ilmath] where [ilmath]e[/ilmath] is the identity of [ilmath]X[/ilmath]
- [math]f(y)=f(x)ef(x^{-1})[/math]
- It is a property of homomorphisms that [ilmath]f(x^{-1})=(f(x))^{-1}[/ilmath]
- [math]f(y)=f(x)f(x^{-1})=f(x)f(x)^{-1}=e[/math]
- Thus [ilmath]y\in\text{Ker}(f)=H[/ilmath]
- Let [ilmath]y\in xHx^{-1} [/ilmath] be given
- So we see that [ilmath]xHx^{-1}\subseteq H[/ilmath]
Proof that [math]\forall x\in G[H\subseteq xHx^{-1}][/math]
- As before, let [ilmath]x\in G[/ilmath] be given.
- From above we know that [math]\forall x\in G[xHx^{-1}\subseteq H][/math], using this we see that [math]x^{-1}Hx\subseteq H[/math]
- Let [ilmath]y\in H[/ilmath] be given (we will show that then [ilmath]y\in xHx^{-1} [/ilmath])
- We know that [ilmath]x^{-1}Hx\subseteq H[/ilmath], as [ilmath]y\in H[/ilmath] we know that [ilmath]x^{-1}yx\in H[/ilmath]
- This means [math]\exists h\in H[x^{-1}yx=h][/math]
- [math]\implies yx=xh[/math]
- [math]\implies y=xhx^{-1}[/math]
- Thus [math]y\in xHx^{-1}[/math]
- We have shown that [math][y\in H\implies y\in xHx^{-1}]\iff[H\subseteq xHx^{-1}][/math]
We have shown that [math]\forall x\in G[xHx^{-1}\subseteq H\wedge H\subseteq xHx^{-1}][/math], which is exactly:
- Given a group homomorphism [ilmath]f:G\rightarrow X[/ilmath] where [ilmath]\text{Ker}(f)=H[/ilmath], we have shown that [ilmath]\forall x\in G[H=xHx^{-1}][/ilmath] which is the first definition of a normal subgroup
Claim 3: Every normal subgroup is the kernel of some homomorphism
