Group theory definitions & proofs (1)

18 Mar, 2026

The following exercises are from Appendix 2 of Quantum Computation and Quantum Information by Isaac Chuang & Michael Nielsen. I am currently reading this book. I hope I can finish this book by Friday — will work hard to accomplish that.

A group $(G, \cdot)$ is a non-empty set $G$ with a binary group multiplication operation “ $\cdot$ ”, with the following properties:

Closure: For all $a, b \in G$ , the product $a \cdot b \in G$ .
Associativity: For all $a, b, c \in G$ , $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ .
Identity: There exists $e \in G$ such that for all $a \in G$ , $e \cdot a = a \cdot e = a$ .
Inverse: For every $a \in G$ , there exists $a^{- 1} \in G$ such that $a \cdot a^{- 1} = a^{- 1} \cdot a = e$ .

The order of a finite group $G$ is the number of elements it contains, denoted as $| G |$ .

The order of an element $g \in G$ is the smallest positive integer $r$ such that $g^{r} = e$ which is the identity element.

A group $G$ is Abelian if $g_{1} g_{2} = g_{2} g_{1}$ for all $g_{1}, g_{2} \in G$ .

A subgroup $H$ of $G$ is a subset of $G$ which forms a group under the same group multiplication operation as $G$ .

The conjugate of $g_{2}$ with respect to $g_{1}$ is $g_{1}^{- 1} g_{2} g_{1}$ .

A normal subgroup is a subgroup $H$ of $G$ such that $g^{- 1} H g = H$ for all $g \in G$ .

The conjugacy class $G_{x}$ of an element $x \in G$ is defined by $G_{x} \equiv {g^{- 1} x g ∣ g \in G}$ .

Exercise A2.1: Prove that for any element $g$ of a finite group, there always exists a positive integer $r$ such that $g^{r} = e$ . That is, every element of such a group has an order.

Proof: Let $G$ be a finite group and $g \in G$ . $G$ has $| G |$ elements.

According to closure property, $g, g^{2}, \dots, g^{| G |}$ a total of $| G |$ elements are $\in G$ .

According to identity property, there exist an identity $e \in G$ .

Now we have $| G | + 1$ elements in the discussion. For a finite $G$ of only $| G |$ elements, there must exist $g^{i}, g^{j}$ ( $0 \leq i < j \leq | G |$ ) such that $g^{i} = g^{j}$ .

$⟹ g^{j} \cdot g^{- i} = g^{i} \cdot g^{- i} = e ⟹ g^{j - i} = e$ , where $j - i > 0$ . QED.

Lagrange’s theorem: If $H$ is a subgroup of a finite group $G$ then $| H |$ divides $| G |$ .

Proof: For each $g \in G$ , we define its left coset $g H = {g \cdot h ∣ h \in H}$ which has an order of $| H |$ .

For any two left cosets $g_{1} H, g_{2} H$ , suppose there exist $x \in g_{1} H \cap g_{2} H$ , i.e., $x = g_{1} h_{1} = g_{2} h_{2}$ for some $h_{1}, h_{2} \in H$ .

Then $g_{1} = g_{2} h_{2} h_{1}^{- 1} \in g_{2} H$ .

Therefore every $g_{1} h \in g_{2} H$ , i.e., $g_{1} H \subseteq g_{2} H$ .

By a symmetric argument, $g_{2} H \subseteq g_{1} H$ .

Therefore, $g_{1} H = g_{2} H$ . That means, $g_{1} H$ and $g_{2} H$ are either identical or disjoint.

By their identical cosets, we can partition elements $g$ in $G$ into several subsets, each has an order of $| H |$ . Let $k$ be the number of distinct cosets, then $| G | = k | H |$ . QED.

Exercise A2.3: Show that the order of an element $g \in G$ divides $| G |$ .

Proof: $g^{r} = e$ . Consider a subgroup ${e, g, g^{2}, \dots, g^{r - 1}} \subseteq G$ which has $r$ distinct elements.

Then according to Lagrange’s theorem, $r$ divides $| G |$ . QED.

Exercise A2.4: Show that if $y \in G_{x}$ then $G_{y} = G_{x}$ .

Proof: $G_{x} \equiv {g^{- 1} x g ∣ g \in G}$ , $G_{y} \equiv {g^{- 1} y g ∣ g \in G}$ .

Suppose $y = g_{1}^{- 1} x g_{1} \in G_{x}$ where $g_{1} \in G$ .

$G_{y} = {g^{- 1} g_{1}^{- 1} x g_{1} g ∣ g \in G} = {g_{2}^{- 1} x g_{2} ∣ g_{2} \equiv g_{1} g \in G} = G_{x}$ . QED.

Exercise A2.5: Show that if $x$ is an element of an Abelian group $G$ then $G_{x} = {x}$ .

Proof: $G_{x} \equiv {g^{- 1} x g ∣ g \in G}$ .

In an Abelian group, $g^{- 1} x g = g^{- 1} g x = e x = x$ .

Therefore, $G_{x} = {x}$ . QED.

#Math notes