Difference between revisions of "Normal subgroup"

Revision as of 16:33, 16 May 2015

Definition

Let $(G,\times)$ be a group and $H$ a subgroup of $G$ , we say $H$ is a normal subgroup^[1] of $G$ if:

$\forall x\in G[xH=Hx]$ where the xH and Hx are left and right cosets
- This is the sameas saying: $\forall x\in G[xHx^{-1}=H]$

According to Serge Lang^[1] this is equivalent (that is say if and only if or $\iff$ )

$H$ is the kerel of some homomorphism of $G$ into some other group

Proof of claims

[Expand]
Claim 1:
$\forall x\in G[xH=Hx]\iff\forall x\in G[xHx^{-1}=H]$

Proof of:

$\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H]$

Suppose that for whatever

$g\in G$ we have that

$gH=Hg$ - we wish to show that for any

$x\in G[xHx^{-1}=H]$

Let

$x\in G$ be given.

Recall that

$X=Y\iff[X\subseteq Y\wedge X\supseteq Y]$ so we need to show:

$xHx^{-1}\subseteq H$
- By the implies-subset relation this is the same as $y\in xHx^{-1}\implies y\in H$
$xHx^{-1}\supseteq H$
- By the implies-subset relation this is the same as $y\in H\implies y\in xHx^{-1}$

Let us show 1:

Suppose

$y\in xHx^{-1}$ we wish to show

$\implies y\in H$ (that is

$xHx^{-1}\subseteq H$ )

$y\in xHx^{-1}\implies \exists h_1\in H:y=xh_1x^{-1}$

$\implies yx=xh_1$ - note that

$xh_1\in xH$

By hypothesis,

$\forall g\in G[gH=Hg]$

So, as

$yx=xh_1\in xH$ we see

$yx\in Hx$

This means

$\exists h_2\in H$ such that

$yx=h_2x$

Using the cancellation laws for groups we see that

$y=h_2$ as

$h_2\in H$ we must have

$y\in H$

We have now shown that

$[y\in xHx^{-1}\implies y\in H]\iff[xHx^{-1}\subseteq H]$

Now to show 2:

Suppose that

$y\in H$ we wish to show that

$\implies y\in xHx^{-1}$ (that is

$H\subseteq xHx^{-1}$ )

Note that

$yx\in Hx$ by definition of

$Hx$

By hypothesis

$Hx=xH$ so

we see that

$yx\in xH$

this means

$\exists h_1\in H[yx=xh_1]$

and this means

$y=xh_1x^{-1}$

such a

$h_1$ existing is the very definition of

$xh_1x^{-1}\in xHx^{-1}$

thus

$y\in xHx^{-1}$

We have now shown that

$[y\in H\implies y\in xHx^{-1}]\implies[H\subseteq xHx^{-1}]$

Combining this we hve shown that

$\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H]$

$\forall x\in G[xHx^{-1}=H]\implies\forall x\in G[xH=Hx]$

TODO: Simple proof

References

↑ ^{Jump up to: 1.0} ^1.1 Undergraduate Algebra - Serge Lang

[LangUA-1] {Jump up to: 1.0} ^1.1 Undergraduate Algebra - Serge Lang

[1]

@@ Line 2: / Line 2: @@
 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]]
+** 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}}
-{{Todo|Flesh out page}}
 ==References==
 <references/>
 {{Definition|Abstract Algebra}}

Difference between revisions of "Normal subgroup"

Revision as of 16:33, 16 May 2015

Definition

Proof of claims

References

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