Chapter 2 Groups and Vector Spaces

2.2 Groups

2.2.4 Lagrange’s Theorem

Lemma 2.26

Every coset of $H$ in a group $G$ has the same number of elements.

Proof:

Let the coset be $a * H$. Then we can make an one-to-one mapping between $a * H$ and $H$:

$$f : a * H \longrightarrow H, a * h \longrightarrow h$$

So $a * H$ has the same number of elements as $H$.

$\square$

Lemma 2.28

The distinct cosets of $H$ in a group $G$ are disjoint.

Proof:

Assume $a * H$ and $b * H$ are 2 distinct cosets of $H$, assume $c \in a * H \cap b * H$, then we can find $h_1, h_2$ such that

$a h_1 = c = b h_2$, so we have $a h_1 h_2^{-1} = b$, so $b \in a * H$. That means $a * H = b * H$.

Then we reach a contradiction.

$\square$

Theorem 2.29 (Lagrange’s theorem)

Let $G$ be a group of finite order and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$. That is, $|H|$ divides $|G|$.

Proof:

Just apply Lemma 2.26 and Lemma 2.28.

$\square$

2.2.5 Induced Operations; Isomorphism

Definition 2.36

Let $⟨G,*⟩$ be a group, $H$ a subgroup and let $S = \{H_0 = H,H_1,H_2, ...,H_M \}$ be the set of cosets of $H$ in $G$. Then the induced operation between cosets $A$ and $B$ in $S$ is defined by

$$A * B = C \text{ if and only if } a * b= c$$

for any $a ∈ A, b ∈ B$ and some $c ∈ C$, provided that this operation is well defined. The operation is well defined if for every $a ∈ A$ and $b ∈ B, a * b ∈ C$; there is thus no ambiguity in the induced operation.

Theorem 2.38

If $H$ is a subgroup of a commutative group $⟨G,*⟩$, the induced operation $*$ on the set of cosets of $H$ satisfies

$$(a * b) * H = (a * H) * (b * H).$$

Furthermore, the operation is well defined.

Proof:

We will prove it's well defined, Let $a_1 \in a * H, b_1 \in b * H$, we want to show

$$(a * b) * H = (a_1 * b_1) * H$$

Since $a_1 \in a * H$, we have $a_1 = a h_1, h_1 \in H$. Similarly, we have $b_1 = b h_2, h_2 \in H$.

Since $G$ is commutative, $a_1 b_1 = a h_1 b h_2 = (ab) (h_1h_2)$. Since $H$ is a subgroup, $h_1 h_2 \in H$.

So we have

$$(a * b) * H = (a_1 * b_1) * H$$

Then the operation is well defined.

$\square$

2.4 Review of Linear Algebra

Definition 2.62

Let $W$ be a $k$-dimensional subspace of a vector space $V$. The set of all vectors $𝐮 ∈ V$ which are orthogonal to all the vectors of $W$ is called the dual space of $W$ (sometimes called the orthogonal complement of $W$ or dual space), denoted $W_⟂$. (The symbol $W_⟂$ is sometimes pronounced $“W$ perp,” for “perpendicular.”) That is,

$$W_⟂ = \{𝐮 ∈ V ∶ 𝐮 \cdot 𝐰 = 0 \text{ for all } 𝐰 ∈ W\}.$$

$\square$

Theorem 2.64

Let $V$ be a finite-dimensional vector space of $n$-tuples, $\mathbb{F}^n$, with a subspace $W$ of dimension $k$. Let $U= W_⟂$ be the dual space of $W$. Then

$$\dim(W_⟂) = \dim(V)−\dim(W) = n−k.$$

Proof:

We will use a different approach compare to the textbook.

Let $e_1, \cdots, e_n$ be the standard basis of $\mathbb{F}^n$. And $\psi_1, \cdots, \psi_n$ be the dual basis of $e_1, \cdots, e_n$.

Then we can construct an isomorphism from

$$T: \mathbb{F}^n \longrightarrow (\mathbb{F}^{n})', (a_1, \cdots, a_n) \longrightarrow a_1 \psi_1 + \cdots + a_n \psi_n$$

Then in fact, $W_⟂$ is isomorphic to the annihilator of $W$, i.e. $W^0$. From Linear Algebra Done Right 4th edition, 3.125 dimension of the annihilator, we know

$$\dim W^0 = \dim 𝑉 − \dim W.$$

So

$$\dim(W_⟂) = \dim W^0 = \dim 𝑉 − \dim W$$

$\square$