NoteNextra-origin/content/Math4302/Math4302_L14.md

Math4302 Modern Algebra (Lecture 14)

Group

Cosets

Left cosets:

aH=\{x|a\sim x\}=\{x\in G|a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}

Right cosets:

Ha=\{x|x\sim'a\}=\{x\in G|xa^{-1}\in H\}=\{x|x=ha\text{ for some }h\in H\}

And G=\sqcup_{a\in G}aH=\sqcup_{a\in G}Ha (all sets are disjoint)

And H is both a left and right coset of G

Example of left and right cosets

G=S_3=\{e,\rho,\rho^2,\tau_1,\tau_2,\tau_3\} with H=\{e,\rho, \rho^2\}, \tau_1=(12), \tau_2=(23), \tau_3=(13).

Number of distinct coset is |G|/|H|=2.

The (left and right) cosets are:

\tau_1 H=\tau_2 H=\tau_3 H=\{\tau_1,\tau_2,\tau_3\}\\
H=\rho H=\rho^2 H=\{e,\rho,\rho^2\}

For this case, left and right cosets are the same (gives the same partition of G).

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H=\{e,\tau\}

Left cosets:

e H=H=\tau_1 H\\
\rho H=\{\tau_3,\rho\}=\tau_3 H\\
\rho^2 H=\{\tau_2,\rho^2\}=\tau_2 H

Right cosets:

H=H e=H\tau_1\\
H\tau_2=\{\tau_2,\rho\}=H\rho \\
H\tau_3=\{\tau_3,\rho^2\}=H\rho^2

Definition of Normal Subgroup

A subgroup H\leq G is called a normal subgroup if aH=Ha for all a\in G. We denote it by H\trianglelefteq G

Example of normal subgroup

Every subgroup of an abelian group is a normal subgroup.

Prove using direct product of cyclic groups.

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If G is finite, and |H|=\frac{|G|}{2}, then H\trianglelefteq G.

there are exactly two cosets, and one of them must be H, then the left coset G\setminus H will always be the same as the right G\setminus H.

A_n\trianglelefteq S_n

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If \phi:G\to G' is a homomorphism, then \ker(\phi)\trianglelefteq G

We will use the equivalent definition of normal subgroup. (aha^{-1}\in H for all a\in G, h\in H)

\phi(aha^{-1})=phi(a)\phi(h)\phi(a)^{-1}=\phi(a)e'\phi(a)^{-1}=e', so aha^{-1}\in \ker(\phi)

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Consider \operatorname{GL}(n,\mathbb{R}) be all the invertible matrices of size n\times n

Let H=\{A\in \operatorname{GL}(n,\mathbb{R})|\det(A)=1\}.

H\trianglelefteq \operatorname{GL}(n,\mathbb{R})

\phi:\operatorname{GL}(n,\mathbb{R})\to (\mathbb{R}-\{0\},\cdot) where \phi(A)=\det(A)

Then H=\ker(\phi)

Lemma for equivalent definition of normal subgroup

The following are equivalent:

1. H\trianglelefteq G
2. aHa^{-1}=H for all a\in G
3. aHa^{-1}\subseteq H for all a\in G, that is aha^{-1}\in H for all a\in G

Proof

We first show that 1\implies 2.

aHa^{-1}\subseteq H:

If aH=Ha, for every h\in H, ah=h'a for some h', so aha^{-1}=h'\in H.

H\subseteq aHa^{-1}:

we have Ha=aH, so for every h\in H, ha=ah' for some h', so h=ah'a^{-1}\in aHa^{-1}.

2\implies 3: clear

3\implies 1:

aH\subseteq Ha. for any h\in H, \forall aha^{-1}\in H, so aha^{-1}=h'\in H, so ah=h'a\in Ha so aH\subseteq Ha.

Ha\subseteq aH: apply previous part to a^{-1}., and a^{-1}H\subseteq Ha^{-1}, so \forall h\in H a^{-1}h=h'a^{-1}\in Ha^{-1}, so ha=ah'.