Cosets and Theorem of Lagrange

 

Def 1

$G$ : group,  $H \leq G$

Let $g \in G$

1.  The left coset of $H$ containing $g$ is

$gH = \left\{gh \; | \; h \in H \right\}$  $(\subset G)$

2.  The right coset of $H$ containing $g$ is

$Hg = \left\{hg \; | \; h \in H \right\}$  $(\subset G)$

 

 

 

Thm 1

Let $G$ be a group  and  $H \leq G$. Then for $a, b, c \in G$,

1.  (a)  $aH = bH$  $\Leftrightarrow$  $a \in bH$     ($\Leftrightarrow$  $b \in aH$)

                                $\Leftrightarrow$  $a = bh$  for some $h \in H$     ($\Leftrightarrow$  $b = ah$  for some $h \in H$)

 

     (b)  $Ha = Hb$  $\Leftrightarrow$  $a \in Hb$

 

2.  (a)  $aH = bH$  $\Leftrightarrow$  $(ca)H = (cb)H$

 

     (b)  $Ha = Hb$  $\Leftrightarrow$  $H(ac) = H(bc)$

 

pf)

(a)

$\Rightarrow)$

  $a = ae \in aH = bH$

 

$\Leftarrow)$

  $\forall x \in aH$

  So there exists $h \in H$  s.t.

  $x = ah$

  Since $a \in bH$, there exists $h' \in H$  s.t.

  $a = bh'$

 

  $\therefore$  $x = (bh')h = b(h'h) \in bH$

  $\therefore$  $aH \subset bH$

 

  Similarly we can show $bH \subset aH$

  $\therefore$  $aH = bH$

 

(b)

  Similarly

 

(a)

$\Rightarrow)$

  By 1, there exists $h \in H$  s.t.

  $a = bh$

  $\therefore$  $ca = cbh$

  $\therefore$  $caH = cbH$

 

$\Leftarrow)$

  By 1, there exists $h \in H$  s.t.

  $ca = cbh$

  $\therefore$  $a = bh$

  $\therefore$  $aH = bH$

 

(b)

  Similarly

 

 

 

Cor 1

1.  $aH = bH$  $\Leftrightarrow$  $b^{-1}a H = H$

                          $\Leftrightarrow$  $(a^{-1}b)H = H$

2.  $Ha = Hb$  $\Leftrightarrow$  $H = H(ba^{-1})$

                          $\Leftrightarrow$  $H = H(ab^{-1})$

 

 

 

Thm 2

Let $H \leq G$. Then $\forall a, b \in G$, 

either  $aH = bH$     or     $aH \cap bH = \varnothing$

 

  Case 1 : $aH \cap bH = \varnothing$

  It is clear.

 

  end

 

 

  Case 2 : $aH \cap bH \neq \varnothing$

  Then there exists $c$  s.t.

  $c \in aH \cap bH$

  Since $c \in aH$  and  $c \in bH$, by Thm 1

  $cH = aH$ and $cH = bH$

  $\therefore$  $aH = bH$

 

  end

 

 

 

Cor 2

$G$ is a disjoint union of left cosets

 

 

 

Thm 3

Let $H \leq G$

For $a, b \in G$, define $a \sim b$ if $aH = bH$. Then

$\sim$ is an equivalence relation  and  $[a] = aH$

 

  Show 1 : $a \sim a$

  It is clear.

 

  end

 

 

  Show 2 : $a \sim b$  $\Rightarrow$  $b \sim a$

  $a \sim b$  $\Rightarrow$  $aH = bH$  $\Rightarrow$  $bH = aH$  $\Rightarrow$  $b \sim a$

 

  end

 

 

  Show 3 : $a \sim b$, $b \sim c$  $\Rightarrow$  $a \sim c$

  $a \sim b$  $\Rightarrow$  $aH = bH$

  $b \sim c$  $\Rightarrow$  $bH = cH$

  $\therefore$  $aH = cH$

  $\therefore$  $a \sim c$

 

  end

 

 

  Show 4 : $[a] = aH$

  $[a] = \left\{x \in G \; | \; x \sim a \right\}$
         $= \left\{x \in G \; | \; xH = aH \right\}$
         $= \left\{x \in G \; | \; x \in aH \right\}$
         $= aH$

 

  end

 

 

 

Thm 4

Let $H \leq G$. For $a, b \in G$,

$\left|aH \right| = \left|bH \right| = \left|H \right|$

($\left|Ha \right| = \left|Hb \right| = \left|H \right|$)

 

# So the number of left cosets of $H$ and right cosets of $H$ is the same.

 

  Define $\phi : H \to aH$ by

  $\phi(h) = ah$

 

 

  Show : $\phi$ is 1-1

  $\forall h, h' \in H$

  $\phi(h) = \phi(h')$  $\Rightarrow$  $ah = ah'$  $\Rightarrow$  $h = h'$

 

  end

 

 

 

Thm 5 Theorem of Lagrange

Let $G$ be a finite group. Then

$H \leq G$  $\Rightarrow$  $\left|H \right| \mid \left|G \right|$

 

  By Cor 2,  

  $G$ is a disjoint union of left cosets of $H$

 

  Since $G$ is finite, there exists $a_{1}, a_{2}, \cdots, a_{n} \in G$  s.t.

  $G = \displaystyle \bigcup_{i=1}^{n}a_{i}H$     ( if $i \neq j$,  $aH_{i} \cap a_{j}H = \varnothing$ )

 

  By Thm 4,

  $\left|G \right| = \displaystyle \sum_{i=1}^{n} \left|a_{i}H \right| = \sum_{i=1}^{n} \left|H \right| = n \left|H \right|$

  $\therefore$  $\left|H \right| \mid \left|G \right|$

 

 

 

Cor 3

1.  If $G$ is a finite group and $a \in G$, then $O(a) \mid \left|G \right|$

2.  If $G$ is a group with $\left|G \right| = p$  ( prime ), then $G \simeq \mathbb{Z}_{p}$

 

 

 

Def 2

Let $H \leq G$

The index of $H$ in $G$ is

$(G : H)$ $=$ number of left cosets of $H$ ( $=$ number of right cosets of $H$ )

 

 

 

Prop 1

1.  If $\left|G \right| < \infty$, then $(G : H) = \dfrac{\left|G \right|}{\left|H \right|}$

2.  If $K \leq H \leq G$, then $(G : K) = (G : H)(H : K)$

 

# 1 is clearly

 

pf)

  Since $H \leq G$, there exists $a_{i} \in G$  for $i \in I$  s.t.

  $G = \displaystyle \bigcup_{i \in I} a_{i}H$

  &     If $i_{1} \neq i_{2}$,  $a_{i_{1}}H \cap a_{i_{2}}H = \varnothing$

  Since $K \leq H$, there exists $b_{j} \in H$  for $j \in J$  s.t.

  $H = \displaystyle \bigcup_{j \in J} b_{j}K$

  &     If $j_{1} \neq j_{2}$,  $b_{j_{1}}K \cap b_{j_{2}}K = \varnothing$

 

  It is sufficient to show

  $\left\{(a_{i}b_{j})K \; | \; i \in I, \; j \in J \right\}$ is the collection of distinct left cosets of $K$ in $H$

 

 

  Claim 1 : Fix $i \in I$.  $\forall j \in J$,  $(a_{i}b_{j})K \subset a_{i}H$

  $\forall g \in (a_{i}b_{j})K$

  So there exists $k \in K$  s.t.

  $g = a_{i}b_{j}k$

 

  Since $K \subset H$,

  $k \in H$

  $\therefore$  $b_{j}k \in H$

  $\therefore$  $g \in a_{i}H$

  $\therefore$  $(a_{i}b_{j})K \subset a_{i}H$

 

  end

 

 

  Claim 2 : For $i \in I$,  $a_{i}H \subset \displaystyle \bigcup_{i \in J} (a_{i}b_{j}) K$

  $\forall g \in a_{i}H$

  So there exists $h \in H$  s.t.

  $g = a_{i}h$

 

  Since $h \in H = \displaystyle \bigcup_{j \in J} b_{j}K$, there exists $j_{1} \in J$  s.t.

  $h \in b_{j_{1}}K$

  So there exists $k \in K$  s.t.

  $h = b_{j_{1}}k$

  $\therefore$  $g = a_{i}h = a_{i}b_{j_{1}}k \in a_{i}b_{j_{1}}K$

  $\therefore$  $g \in \displaystyle \bigcup_{j \in J} (a_{i}b_{j})K$

  $\therefore$  $a_{i}H \subset \displaystyle \bigcup_{j \in J} (a_{i}b_{j})K$

 

  end

 

 

  Show 1 : $\displaystyle \bigcup_{(i, j) \in I \times J} (a_{i}b_{j}) K = G$

  By Claim 1,

  $\displaystyle \bigcup_{j \in J} (a_{i}b_{j}) K \subset a_{i}H$

  By Claim 2,

  $a_{i}H = \displaystyle \bigcup_{j \in J} (a_{i}b_{j}) K $

 

  Since $G = \displaystyle \bigcup_{i \in I} a_{i} H$,

  $\displaystyle \bigcup_{(i, j) \in I \times J} (a_{i}b_{j}) K = G$

 

  end

 

 

  Show 2 : If $(i, j) \neq (p, q)$,  $(a_{i}b_{j}) K \cap (a_{p}b_{q}) K = \varnothing$

  - case 1 : $i \neq p$

  By Claim 1,

  $(a_{i}b_{j})K \subset a_{i}H$

  $(a_{p}b_{q})K \subset a_{p}H$

 

  Since $a_{i}H \cap a_{p}H = \varnothing$,

  $(a_{i}b_{j}) K \cap (a_{p}b_{q}) K = \varnothing$

 

  - case 2 : $i = p$, $j \neq q$

  So $b_{j}K \cap b_{q}K = \varnothing$

 

  By Thm 2,

  $b_{j}K \neq b_{q}K$

  By Thm 1 - 2,

  $(a_{i}b_{j})K \neq (a_{i}b_{q})K$

  By Thm 2,

  $(a_{i}b_{j}) K \cap (a_{p}b_{q}) K = \varnothing$

 

  end