Difference between revisions of "Normal subgroup"

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:

1. $$xHx^{-1}\subseteq H$$
   • By the implies-subset relation this is the same as $$y\in xHx^{-1}\implies y\in H$$
2. $$xHx^{-1}\supseteq H$$
   • 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 $\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H]$

Next:

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

TODO: Simple proof

We wish to show that given a homomorphism $f:G\rightarrow X$ (where $X$ is some group) that the kernel of $f$, $H$ is normal. Which is to say that:

• $$\forall x\in G[xHx^{-1}=H]$$ (which is $$\forall x\in G[x\text{Ker}(f)x^{-1}=\text{Ker}(f)]$$ )

Proof that $$\forall x\in G[xHx^{-1}\subseteq H]$$

Let $x\in G$ be given

Let $y\in xHx^{-1}$ be given

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

$$f(y)=f(xh_1x^{-1})=f(x)f(h_1)f(x^{-1})$$

But $H$ is the kernel of $f$ so $f(h_1)=e$ where $e$ is the identity of $X$

$$f(y)=f(x)ef(x^{-1})$$

It is a property of homomorphisms that $f(x^{-1})=(f(x))^{-1}$

$$f(y)=f(x)f(x^{-1})=f(x)f(x)^{-1}=e$$

Thus $y\in\text{Ker}(f)=H$

So we see that $xHx^{-1}\subseteq H$

Proof that $$\forall x\in G[H\subseteq xHx^{-1}]$$

As before, let $x\in G$ be given.

From above we know that $$\forall x\in G[xHx^{-1}\subseteq H]$$ , using this we see that $$x^{-1}Hx\subseteq H$$

Let $y\in H$ be given (we will show that then $y\in xHx^{-1}$)

We know that $x^{-1}Hx\subseteq H$, as $y\in H$ we know that $x^{-1}yx\in H$

This means $$\exists h\in H[x^{-1}yx=h]$$

$$\implies yx=xh$$

$$\implies y=xhx^{-1}$$

Thus $$y\in xHx^{-1}$$

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

We have shown that $$\forall x\in G[xHx^{-1}\subseteq H\wedge H\subseteq xHx^{-1}]$$ , which is exactly:

• Given a group homomorphism $f:G\rightarrow X$ where $\text{Ker}(f)=H$, we have shown that $\forall x\in G[H=xHx^{-1}]$ which is the first definition of a normal subgroup