Definition
Let (G,×) 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
- xHx^{-1}\supseteq H
- 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
References
- ↑ Jump up to: 1.0 1.1 Undergraduate Algebra - Serge Lang