Normal subgroup

Definition

Let $(G,\times)$ be a group and $H$ a subgroup of $G$, we say $H$ is a normal subgroup 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 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

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

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

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TODO: Simple proof

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