Normal subgroup

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 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 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]

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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:

1. [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]
2. [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]

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]

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]

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 [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]

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TODO: Simple proof

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