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Math4302Modern Algebra (Lecture 13)

Math4302 Modern Algebra (Lecture 13)

Groups

Cosets

Last time we see that (left coset) aba\sim b (to differentiate from right coset, we may denote it as aLba\sim_L b) by a1bHa^{-1}b\in H defines an equivalence relation.

Definition of Equivalence Class

Let aHa\in H, and the equivalence class containing aa is defined as:

aH={xax}={xa1xH}={xx=ah for some hH}aH=\{x|a\simeq x\}=\{x|a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}

Properties of Equivalence Class

aH=bHaH=bH if and only if aba\sim b.

Proof

If aH=bHaH=bH, then since aaH,abHa\in aH, a\in bH, then for some hh, a=bha=bh, since b1aHb^{-1}a\in H, so a1bHa^{-1}b\in H, therefore aba\simeq b.

If aba\sim b, then aHbHaH\subseteq bH, since anything in aHaH is related to aa, therefore it is related to bb so abHa\in bH.

bHaHbH\subseteq aH, apply the reflexive property for equivalence relation, therefore baHb\in aH.

So aH=bHaH=bH.

If aHbHaH\cap bH\neq \emptyset, then aH=bHaH=bH.

Proof

If $x\in aH\cap bH$, then $x\sim a$ and $x\sim b$, so $a\sim b$, so $aH=bH$.

aH=HaH=H if and only if aHa\in H.

Proof

$aH=eH$ if and only if $a\sim e$, if and only if $a\in H$.

aHaH is called left coset of aa in HH.

Examples

Consider G=S3={e,ρ,ρ2,τ1,τ2,τ3}G=S_3=\{e,\rho,\rho^2,\tau_1,\tau_2,\tau_3\}.

where ρ=(123),ρ2=(132),τ1=(12),τ2=(23),τ3=(13)\rho=(123),\rho^2=(132),\tau_1=(12),\tau_2=(23),\tau_3=(13).

H={e,ρ,ρ2}H=\{e,\rho,\rho^2\}.

All the left coset for HH is H=eH=ρH=ρ2HH=eH=\rho H=\rho^2H.

τ1ρ=(23)=τ2τ1ρ2=(13)=τ3τ2ρ=(31)=τ3τ2ρ2=(12)=τ1τ3ρ=(12)=τ1τ3ρ2=(23)=τ2\tau_1\rho=(23)=\tau_2\\ \tau_1\rho^2=(13)=\tau_3\\ \tau_2\rho=(31)=\tau_3\\ \tau_2\rho^2=(12)=\tau_1 \tau_3\rho=(12)=\tau_1\\ \tau_3\rho^2=(23)=\tau_2 τ1H={τ1,τ2,τ3}=τ2H=τ3H\tau_1H=\{\tau_1,\tau_2,\tau_3\}=\tau_2H=\tau_3H\\

Consider G=ZG=\mathbb{Z} with H=5ZH=5\mathbb{Z}.

We have 5 cosets, H,1+H,2+H,3+H,4+HH,1+H,2+H,3+H,4+H.

Lemma for size of cosets

Any coset of HH has the same cardinality as HH.

Define ϕ:HaH\phi:H\to aH by ϕ(h)=ah\phi(h)=ah.

ϕ\phi is an bijection, if ah=ah    h=hah=ah'\implies h=h', it is onto by definition of aHaH.

Corollary: Lagrange’s Theorem

If GG is a finite group, and HGH\leq G, then HG|H|\big\vert |G|. (size of HH divides size of GG)

Proof

Suppose HH has rr distinct cosets, then G=rH|G|=r|H|, so H|H| divides G|G|.

Corollary for Lagrange’s Theorem

If G=p|G|=p, where pp is a prime number, then GG is cyclic.

Proof

Prick eaGe\neq a\in G, let H=aGH=\langle a\rangle \leq G, then H|H| divides G|G|, since pp is prime, then H=G|H|=|G|, so G=aG=\langle a \rangle.

If GG is finite and aGa\in G, then ord(a)G\operatorname{ord}(a)\big\vert|G|.

Proof

Since ord(a)=a\operatorname{ord}(a)=|\langle a\rangle|, and a\langle a\rangle is a subgroup, so ord(a)G\operatorname{ord}(a)\big\vert|G|.

Definition of index

Suppose HGH\leq G, the number of distinct left cosets of HH is called the index of HH in GG. Notation is (G:H)(G:H).

Definition of right coset

Suppose HGH\leq G, define the equivalence relation by aba\sim 'b (or aRba\sim_R b in some textbook) if ab1Ha b^{-1}\in H. (note the in left coset, we use a1bHa^{-1}b \in H, or equivalently b1aHb^{-1}a \in H, these are different equivalence relations)

The equivalent class is defined

Ha={xGxa}={xGxa1H}={xx=ha for some hH}Ha=\{x\in G|x\sim'a\}=\{x\in G|xa^{-1}\in H\}=\{x|x=ha\text{ for some }h\in H\}

Some properties are the same as the left coset

  • Ha=H    aHHa=H\iff a\in H
  • Ha=HbHa=Hb if and only if ab    ab1Ha\sim'b\iff a b^{-1}\in H.
  • HaHb    Ha=HbHa\cap Hb\neq \emptyset\iff Ha=Hb.

Some exercises: Find all the left and right cosets of G=S3G=S_3, there should be 2 left cosets and 2 right cosets (giving different partition of GG).

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