chapter 5 summary

# Definitions

Cosets
Let $G$ be a group and $H\leq G$ . Define a left (respect to right) coset of $H$ with representative $g$ to be

(1)
\begin{align} $gH = \{gh: h \in H \}$ (Left) \end{align}
(2)
\begin{align} $Hg = \{hg: h \in H \}$ (Right) \end{align}

.

Note: If $G$ is abelian and $g \in G$ and $H \leq G$ then $gH = Hg$.

Index: Let $H \leq G$. The index of $H$ in $G$ is the number of left cosets of $H$, denoted $[G:H]$. Example in "Examples and Notes."

# Theorems

Lemma 5.1:

Let $H$ be a subgroup of $G$ and $g_{1}$, $g_{2}$ $\epsilon$ $G$. The following are equivalent conditions

1. $g_{1}H=g_{2}H$

2. $Hg_{1}^{-1}=Hg_{2}^{-1}$

3. $g_{1}H\subseteq\ g_{2}H$

4. $g_{2}\epsilon\ g_{1}H$

5. $g_{1}^{-1}g_{2}\epsilon\ H$

Theorem 5.2: Let $H \leq G$. Then the left (right) cosets of $H$ in $G$, partition $G$.

Theorem 5.3: Let $H \leq G$. Then the number of left cosets equals the number of right cosets. (Note: this is not saying the elements are the same, just that the total number of left cosets is the same as the total number of right cosets.)

Proposition 5.4: Let $H\leq G$ with $g\in G$. Define $\phi: H\rightarrow gH$ via $\phi(h) = gh$. The $\phi$ is a bijection (both onto and one-to-one), So that $H$ and $gh$ have the same cardinality (same number of elements in the sets).

Theorem 5.5 (LaGrange) (book) Let $G$ be a finite group and let $H$ be a subgroup of $G$. Then $|G|/|H|$=$[G:H]$ is the number of distinct left cosets of $H$ in $G$. In particular, the number of elements in $H$ must divide the number of elements in $G$.
(notes) Let $G$ be a finite group and let $H \leq G$. Then $[G:H]$=$|G|/|H|$, and in particular the order of $H$ divides the order of $G$.

Corollary 5.6 Suppose that $G$ is a finite group and $g \in G$. Then the order of $g$ must divide the number of elements in $G$.

Corollary 5.7 Let $|G|=p$ with $p$ a prime number. Then $G$ is cyclic for any $g \in G$, such that $g \neq e$ is a generator.

Corollary 5.8 Let $H$ and $K$ be subgroups of a finite group $G$ such that $G \supset H \supset K$. Then $[G:K]=[G:H][H:K]$.

# Examples & Notes

$D_{4} = <r,s> , H = <s,r^{2}>$

$H = {e, s, r^{2}, sr^{2}}$

$rH = {re, rs, rr^2, rsr^{2}}$

$= {r, sr^{3}, r^{3}, sr}$

Example of index. $G = S_3$, $H = <(12)>$.

(3)
\begin{align} \begin{align} S_3 &= \{ (1), (12), (13), (23), (123), (132) \} \\ H= <(12)> &= \{ (12), (12)^2 \} = \{ (12), (12)(12) \} = \{ (12), (1) \} \\ (123)H = (13)H &= \{(13)(12), (13)(1) \} = \{ (123), (13) \} \\ (132)H = (23)H &= \{ (23)(12), (23)(1) \} = \{ (132), (23) \} \\ \end{align}

Therefore the number of distinct left cosets of H is 3, so $[S_3: H] = 3$

Example of Lagrange's Theorem:

$D_4$ in $S_4$.

$\dfrac{|S_4|}{|D_4|} = \dfrac{24}{8} = 3$

24 = the number of elements in the entire group.
8 = the number of elements in each coset, including $D_4$.
3 = the number of cosets.

Note: Converse of Lagrange's Theorem is false: if $k/|G|$ then there may or may not be a subgroup size k
Ex.
$|A_4| = 12$
there is no subgroup size 6 even though 6|12

Example from homework 17
1) Suppose that G is a finite group with an element of g of order 5 and and element of h of order 7. Why must $35\leq|G|$?
|g|:(5) and |h|:(7) must both divide |G|, and the smallest number that they both divide is 35. LCM(5,7)=35.

General way of thinking about dihedral groups

(4)
$$r^ks = sr^{n-k}$$

Most Useful way of thinking about dihedral groups

(5)
$$r^{-1}s = sr^{n-1}$$
page revision: 20, last edited: 20 Apr 2010 22:15