Difference between revisions of "Normal subgroup"

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]

TODO: Simple proof

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]

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