Difference between revisions of "Normal subgroup"
From Maths
m |
m (Completed proof of second claim) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 5: | Line 5: | ||
According to Serge Lang<ref name="LangUA"/> this is equivalent (that is say '''''if and only if''''' or {{M|\iff}}) | 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 | * {{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== | ==Proof of claims== | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
| Line 47: | Line 49: | ||
{{Todo|Simple proof}} | {{Todo|Simple proof}} | ||
{{End Proof}}{{End Theorem}} | {{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
