HT2026 Groups & Group Actions Lecture Notes

Remaining TODOs: 1

1. Groups

Definition 1.1: Let $𝑆$ be a set. A binary operation on $𝑆$ is a function $* : 𝑆 \times 𝑆 \to 𝑆$ .

We define $𝑎 * 𝑏 ≔ * (𝑎, 𝑏)$ .

Remark: If

*

is understood implicitly, we often write

𝑎 𝑏

for

𝑎 * 𝑏

Definition 1.2:

*

admits an identity element if

\exists 𝑒 \in 𝑆

s.t.

𝑎 * 𝑒 = 𝑎 = 𝑒 * 𝑎

\forall 𝑎 \in 𝑆

Definition 1.3: Suppose that

𝑒

is an identity element, then inverses exist for

(𝑆, *)

\forall 𝑎 \in 𝑆

\exists 𝑏 \in 𝑆

s.t.

𝑎 * 𝑏 = 𝑒 = 𝑏 * 𝑎

Definition 1.4: A group is a pair

(𝐺, *)

where

* : 𝐺 \times 𝐺 \to 𝐺

is a binary operation which is associative, admits an identity, and for which inverses exist.

Definition 1.5:

𝐺

is an abelian group if

*

is commutative.

Lemma 1.6: Let $*$ be a binary operation on $𝑆$ .

If $𝑒_{1}, 𝑒_{2}$ are identity elements, then $𝑒_{1} = 𝑒_{2}$ .
If $𝑒 \in 𝑆$ is an identity element, $𝑎 \in 𝑆$ , $*$ is associative, and $𝑏_{1}, 𝑏_{2}$ are inverses of $𝑎$ under $*$ , then $𝑏_{1} = 𝑏_{2}$ .

Proof:

$𝑒_{1} = 𝑒_{1} * 𝑒_{2} = 𝑒_{2}$
$\begin{aligned} 𝑏_{1} & \overset{(id)}{=} & 𝑏_{1} * 𝑒 \\ \overset{(inv)}{=} & 𝑏_{1} * (𝑎 * 𝑏_{2}) \\ \overset{(assoc)}{=} & (𝑏_{1} * 𝑎) * 𝑏_{2} \\ \overset{(inv)}{=} & 𝑒 * 𝑏_{2} \\ \overset{(id)}{=} & 𝑏_{2} \end{aligned}$

Definition 1.7: Let

(𝑆, *)

be a set with a binary operation, which admits inverses, then

\forall 𝑎 \in 𝑆

, we write

𝑎^{- 1}

to mean the inverse of

𝑎

Remark:

- 𝑎

is also commonly used to denote the inverse of

𝑎

(𝑆, +)

is an abelian group.

Example (symmetry group):

Definition 1.8: Let

Sym (𝑋)

≔{𝑓:𝑋→𝑋|𝑓bijective}.

Proposition 1.9: $(Sym (𝑋), \circ)$ is a group.

Proof: Suppose $𝑓, 𝑔$ are bijections. Then:

$𝑓 \circ 𝑔$ is injective:
$\begin{aligned} (𝑓 \circ 𝑔) (𝑥) = (𝑓 \circ 𝑔) (𝑦) \\ ⟹ & 𝑓 (𝑔 (𝑥)) = 𝑓 (𝑔 (𝑦)) \\ ⟹ & 𝑔 (𝑥) = 𝑔 (𝑦) \\ ⟹ 𝑥 = 𝑦 \end{aligned}$
$𝑓 \circ 𝑔$ is surjective:
$\begin{aligned} \forall 𝑥 \in 𝑋, & \exists 𝑦 \in 𝑋 s.t. 𝑧 = 𝑓 (𝑦) \\ \exists 𝑥 \in 𝑋 s.t. 𝑦 = 𝑔 (𝑥) \\ ⟹ & 𝑧 = 𝑓 (𝑔 (𝑥)) = (𝑓 \circ 𝑔) (𝑥) . \end{aligned}$

So $𝑓 \circ 𝑔$ is a bijection, and $𝑓 \circ 𝑔$ is a binary operation on $Sym (𝑥)$ .

Composition is associative.

We have an identity, ${id}_{𝑋} : 𝑋 \to 𝑋; 𝑥 \mapsto 𝑥$ , which satisfies $𝑓 \circ {id}_{𝑥} = 𝑓 = {id}_{𝑋} \circ 𝑓$ , $\forall 𝑓 \in Sym (𝑓)$ .

We have inverses: $𝑓 \circ 𝑓^{- 1} = 𝑓^{- 1} \circ 𝑓 = {id}_{𝑋}$ .

Definition 1.10:If

𝑋 = {1, 2, \dots, 𝑛}

𝑆_{𝑛}

≔Sym(𝑋) is the

𝑛

th symmetric group.

Remark:

Sym (𝑋)

can also be interpreted as the permutations of

𝑋

Example (fields):

Let $(𝐹, +, \times)$ be a field.

Then $(𝐹, +)$ is an abelian group, with identity element $0_{𝐹}$ , whereas $(𝐹 \ {0_{𝐹}}, \times)$ (sometimes denoted $𝐹^{*}$ ) is also an abelian group with identity element $1_{𝐹}$ .

For example, $ℚ \subset ℝ \subset ℂ$ all form groups under addition, or under multiplication when zero is excluded.

Also $ℤ_{𝑝}^{*} ≔$ non-zero integers mod p, for prime $𝑝$ , is a group under multiplication.

Example (matrices):

$(𝑀_{𝑛} (𝐹), +)$ is an abelian group (where $𝑀_{𝑛} (𝐹)$ is the set of $𝑛 \times 𝑛$ matrices over $𝐹$ ).

Definition 1.11:

𝐺 𝐿_{𝑛} (𝐹)

≔{𝐴∈𝑀𝑛(𝐹)|det𝐴≠0}.

$(𝐺 𝐿_{𝑛} (𝐹), \cdot)$ is the $𝑛$ th general linear group over $𝐹$ .

Let $𝑆 𝐿_{𝑛} (𝐹)$ ≔{𝐴∈𝐺𝐿𝑛(𝐹)|det𝐴=1}; this is the $𝑛$ th standard linear group over $𝐹$ under matrix multiplication.

Example:

Definition 1.12: Let $𝑋$ be a set, let $𝑌 \subseteq 𝑋$ .

Let $Stab (𝑌) ≔ {𝑓 \in Sym (𝑋) : 𝑓 (𝑌) = 𝑌}$ .

Proposition 1.13: $(Stab (𝑌), \circ)$ is a group.

Proof: Necessary to prove: $\forall 𝑓, 𝑔 \in Stab (𝑌), 𝑓 \circ 𝑔 \in Stab (𝑌)$ : $(𝑓 \circ 𝑔) (𝑌) = 𝑓 (𝑔 (𝑌)) = 𝑓 (𝑌) = 𝑌$ .

Associativity is inherited from the symmetric group.

${id}_{𝑋} (𝑌) = 𝑌 ⟹ {id}_{𝑋} \in Stab (𝑌)$ .

Let $𝑓 \in Stab (𝑌)$ , then $𝑓 (𝑌) = 𝑌$ . Hence $𝑓^{- 1} (𝑓 (𝑌)) = 𝑌$ , and $𝑓^{- 1} (𝑓 (𝑌)) = 𝑓^{- 1} (𝑌)$ , so $𝑓^{- 1} = 𝑌$ , so $𝑓^{- 1} \in Stab (𝑌)$ .

Definition 1.14:

(Stab (𝑌), \circ)

is the stabiliser subgroup.

Definition 1.15: Let

𝐺

be a group, with binary operation

*

. A subset

𝐻 \subseteq 𝐺

is a subgroup if

*

restricts to a map

* : 𝐻 \times 𝐻 \to 𝐻

which then makes

(𝐻, *_{/ 𝐻})

into a group. We write

𝐻 \leq 𝐺

Example:

(ℝ, +) \leq (ℂ, +)

Proposition 1.16 (Subgroup Test): Let $𝐺$ be a group. A subset $𝐻$ of $𝐺$ is a subgroup iff :

$𝑒 \in 𝐻$
$\forall 𝑥, 𝑦 \in 𝐻, 𝑥 𝑦^{- 1} \in 𝐻$

Proof:

“ $⟹$ ”:

Assume $𝐻$ is a subgroup of $𝐺$ . Then $𝐻$ is closed under multiplication and inversion. Therefore $\forall 𝑥, 𝑦 \in 𝐻, 𝑥 𝑦^{- 1} \in 𝐻$ and also $𝑒 \in 𝐻$ .

“ $⟸$ ”: Associativity on $𝐻$ is inherited from $𝐺$ .

(i) $⟹ 𝑒 \in 𝐻$ .

Apply (ii) with $𝑥 = 𝑒$ , giving $\forall 𝑦 \in 𝐻, 𝑦^{- 1} \in 𝐻$ .

So $(𝐻, *_{/ 𝐻})$ is a group.

Example: Let $𝐺 = 𝐺 𝐿_{𝑛} (ℝ)$ .

Let $𝑂 (𝑛) = {𝐀 \in 𝐺 | 𝐀^{⊤} 𝐀 = 𝐈}$ , the orthogonal matrices.

$𝐈^{⊤} 𝐈 = 𝐈 ⟹ 𝐈 \in 𝑂 (𝑛)$ .
$\begin{aligned} 𝐀, 𝐁 \in 𝑂 (𝑛) ⟹ {(𝐀 𝐁^{- 1})}^{⊤} (𝐀 𝐁^{- 1}) & = {(𝐁^{- 1})}^{⊤} 𝐀^{⊤} 𝐀 𝐁^{- 1} \\ = {(𝐁^{- 1})}^{⊤} 𝐁^{- 1} \\ = {(𝐁^{⊤})}^{- 1} 𝐁^{- 1} \\ = {(𝐁 𝐁^{⊤})}^{- 1} \\ = 𝐈^{- 1} (𝐁^{⊤} 𝐁 = 𝐈 ⟹ 𝐁 𝐁^{⊤} = 𝐈 for square 𝐁) \\ = 𝐈 \end{aligned}$

Hence by the subgroup test, $𝑂 (𝑛) \leq 𝐺 𝐿_{𝑛} (ℝ)$ .

Also $𝑆 𝐿_{𝑛} (𝐹) \leq 𝐺 𝐿_{𝑛} (𝐹)$ can be shown by the subgroup test.

Lemma 1.17: If

𝐻 \leq 𝐺

and

𝐾 \leq 𝐺

, then

𝐻 \cap 𝐾 \leq 𝐺

Corollary 1.18:

𝑆 𝑂 (𝑛) = 𝑆 𝐿_{𝑛} (ℝ) \cap 𝑂 (𝑛) \leq 𝐺 𝐿_{𝑛} (ℝ)

Definition 1.19: Let $(𝐺, *_{𝐺}), (𝐻, *_{𝐻})$ be groups.

The product group is $(𝐺 \times 𝐻, *_{𝐺 \times 𝐻})$ , where $(𝑔_{1}, ℎ_{1}) *_{𝐺 \times 𝐻} (𝑔_{2}, ℎ_{2}) = (𝑔_{1} *_{𝐺} 𝑔_{2}, ℎ_{1} *_{𝐻} ℎ_{2})$ .

Lemma 1.20: The product of two groups is a group.

Proof: Associativity is easy.

$(𝑔, ℎ) (𝑒, 𝑒) = (𝑔 𝑒, ℎ 𝑒) = (𝑔, ℎ) = (𝑒, 𝑒) (𝑔, ℎ)$ , so $(𝑒, 𝑒)$ is an identity.

The inverse of $(𝑔, ℎ)$ is $(𝑔^{- 1}, ℎ^{- 1})$ .

Definition 1.21: Let $𝐺$ be a group. $𝐺$ is cyclic if $\exists 𝑔 \in 𝐺$ s.t. $𝐺 = {𝑔^{𝑛} | 𝑛 \in ℤ}$ , where

𝑔^{𝑛} = {\begin{cases} 𝑔 𝑔^{𝑛 - 1} & if 𝑛 > 0 \\ 𝑒 & if 𝑛 = 0 \\ {(𝑔^{𝑛})}^{- 1} & if 𝑛 < 0 . \end{cases}

$𝑔$ is known as a generator of $𝐺$ .

Example:

Definition 1.22: $𝐺 = (ℤ, +)$ is a cyclic group with $𝑔 = 1$ .

This is the infinite cyclic group.

Definition 1.23:For $𝑛 \in ℤ_{> 0}$ , $𝐶_{𝑛} = {𝑒, 𝑔, 𝑔^{2}, \dots, 𝑔^{𝑛 - 1}}$ is a cyclic group with binary operation

𝑔^{𝑎} 𝑔^{𝑏} = {\begin{cases} 𝑔^{𝑎 + 𝑏} & if 𝑎 + 𝑏 \leq 𝑛 - 1 \\ 𝑔^{𝑎 + 𝑏 - 𝑛} & if 𝑛 \leq 𝑎 + 𝑏 \leq 2 𝑛 - 2 \end{cases}

and inverse

{(𝑔^{𝑎})}^{- 1} = 𝑔^{𝑛 - 𝑎}

for

0 \leq 𝑎 < 𝑛

𝐶_{𝑛}

is the cyclic group of order

𝑛

Remark: A cyclic group can have more than one generator; for instance,

(ℤ, +)

also has

- 1

as a generator.

Remark:

𝐶_{5} = ⟨ 𝑔^{2} ⟩

, where, for

ℎ \in 𝐻

, we write

⟨ ℎ ⟩ ≔ {ℎ^{𝑛} | 𝑛 \in ℤ}

. This is because 2 and 5 are coprime.

Definition 1.24: An isometry

𝑇 : ℝ^{𝑛} \to ℝ^{𝑛}

is a distance-preserving function, i.e.

| 𝑇 (𝒗) - 𝑇 (𝒘) | = | 𝒗 - 𝒘 |

for all

𝒗, 𝒘 \in ℝ^{𝑛}

Remark: By the subgroup test,

Isom (ℝ^{𝑛}) \leq Sym (ℝ^{𝑛})

Remark:

\forall 𝑇 \in Isom (ℝ^{𝑛}), \exists! 𝐀 \in 𝑂 (𝑛), 𝒃 \in ℝ^{𝑛}

s.t.

𝑇 (𝒗) = 𝐀 𝒗 + 𝒃

for all

𝒗 \in ℝ^{𝑛}

Definition 1.25: Let $𝑛 \in ℤ_{> 0}$ . Let $𝑃_{𝑛}$ be the regular $𝑛$ -sided polygon centred at the origin $𝑂$ .

The dihedral group $𝐷_{2 𝑛}$ is

𝐷_{2 𝑛} ≔ Isom (ℝ^{2}) \cap Stab (𝑃_{𝑛});

that is, it is the group of distance-preserving transformations that stabilise the regular $𝑛$ -sided polygon centred at the origin.

We denote the rotation by $2 𝜋 / 𝑛$ anticlockwise by $𝑟$ , and the reflection in an axis (which one depends on which way we choose to draw the polygons) by $𝑠$ . Then the elements of $𝐷_{2 𝑛}$ are $𝑒, 𝑟, 𝑟^{2}, \dots, 𝑟^{𝑛 - 1}, 𝑠, 𝑟 𝑠, 𝑟^{2} 𝑠, \dots, 𝑟^{𝑛 - 1} 𝑠$ . Note that this is $2 𝑛$ elements.

Proposition 1.26: $𝑆_{2 𝑛} = {𝑟^{𝑖} 𝑠^{𝑗} | 0 \leq 𝑖 < 𝑛, 𝑗 \in {0, 1}}$ .

Proof: Suppose $𝑟^{𝑖} 𝑠^{𝑗} = 𝑟^{𝑎} 𝑠^{𝑏}$ w.l.o.g. $𝑖 \geq 𝑎$ . Then $𝑟^{- 𝑎} 𝑟^{𝑖} 𝑠^{𝑗} = 𝑟^{- 𝑎} 𝑟^{𝑎} 𝑠^{𝑏} = 𝑠^{𝑏}$ , so w.l.o.g. $𝑎 = 0$ .

If $𝑗 = 𝑏$ , then $𝑟^{𝑖} = 𝑟^{0} = 𝑒 ⟹ 𝑟^{𝑖} (𝑝_{0}) = 𝑒 (𝑝_{0})$ for some point $𝑝_{0}$ , so $𝑖 = 0$ .

If $𝑗 \neq 𝑏$ , $𝑟^{𝑖} = 𝑠$ . Take dets to get $1 = - 1$ , contradiction.

Now let $𝑃 \in 𝐷_{2 𝑛}$ . Then $𝑃 (𝑝_{0}) = 𝑝_{𝑖} = 𝑟^{𝑖} (𝑝_{0})$ for some $𝑖 \in {0, \dots, 𝑛 - 1}$ .

Hence $𝑟^{- 𝑖} 𝑃 (𝑝_{0}) = 𝑝_{0}$ .

Now $| 𝑟^{- 𝑖} 𝑃 (𝑝_{1}) - 𝑟^{- 𝑖} 𝑃 (𝑝_{0}) | = | 𝑝_{1} - 𝑝_{0} |$ , hence $𝑟^{- 𝑖} 𝑃 (𝑝_{1}) \in {𝑝_{1}, 𝑝_{𝑛 - 1}}$ .

TODO finish this off

Definition 1.27: The order of a group

𝐺

| 𝐺 |

, the number of elements in

𝐺

Definition 1.28: If

𝐺 = {𝑔_{1}, \dots, 𝑔_{𝑛}}

, its Cayley table is an

𝑛 \times 𝑛

array where with

𝑔_{𝑖} 𝑔_{𝑗}

in the

𝑖

th row and

𝑗

th column.

Definition 1.29: Let $𝐺, 𝐻$ be groups. An isomorphism between $𝐺$ and $𝐻$ is a bijection $𝜑 : 𝐺 \to 𝐻$ s.t.

𝜑 (𝑔_{1} 𝑔_{2}) = 𝜑 (𝑔_{1}) 𝜑 (𝑔_{2}) .

If there is an isomorphism between $𝐺$ and $𝐻$ , then $𝐺$ and $𝐻$ are isomorphic, written $𝐺 ≅ 𝐻$ .

Remark:Isomorphic finite groups will have the “same” Cayley tables (up to row/column orders).

Definition 1.30: For a group $𝐺$ , the order of an element $𝑔 \in 𝐺$ is

𝑜 (𝑔) ≔ min {𝑟 \in ℤ_{> 0} | 𝑔^{𝑟} = 𝑒} .

If $𝑔^{𝑟} \neq 𝑒$ for all $𝑟 \in ℤ_{> 0}$ , we say that $𝑔$ has infinite order.

Lemma 1.31: If $𝜑 : 𝐺 \to 𝐻$ is an isomorphism, then $\forall 𝑔 \in 𝐺$ ,

𝑜 (𝜑 (𝑔)) = 𝑜 (𝑔) .

Proof:

First we show that $𝜑 (𝑒) = 𝑒$ :

\begin{aligned} 𝜑 (𝑒) & = 𝜑 (𝑒^{2}) \\ = {𝜑 (𝑒)}^{2} \\ ⟹ 𝑒 & = {𝜑 (𝑒)}^{- 1} 𝜑 (𝑒) \\ = {𝜑 (𝑒)}^{- 1} {𝜑 (𝑒)}^{2} \\ = 𝜑 (𝑒) . \end{aligned}

Then

\begin{aligned} 𝑔^{𝑟} = 𝑒 & ⟺ 𝜑 (𝑔^{𝑟}) = 𝜑 (𝑒) = 𝑒 \\ ⟺ {𝜑 (𝑔)}^{𝑟} = 𝑒 . \end{aligned}

Hence

\begin{aligned} 𝑜 (𝑔) & = min {𝑟 > 0 | 𝑔^{𝑟} = 𝑒} \\ = min {𝑟 > 0 | {𝜑 (𝑔)}^{𝑟} = 𝑒} & = 𝑜 (𝜑 (𝑔)) . \end{aligned}

2. Permutation groups

Remark:

| 𝑆_{𝑛} | = 𝑛!

Remark: We write permutations on the right, i.e. if $𝑖 \in {1, \dots, 𝑛}, 𝜎 \in 𝑆_{𝑛}$ , then $𝑖 𝜎 ≔ 𝜎 (𝑖)$ .

This means that composition is backwards from the usual order.

Definition 2.1: A cycle is a permutation $𝜎 \in 𝑆_{𝑛}$ such that $\exists {𝑎_{1}, \dots, 𝑎_{𝑚}} \subseteq {1, \dots, 𝑛}$ s.t. $𝑎_{𝑘} 𝜎 = 𝑎_{𝑘 + 1}$ for $𝑘 \in {1, \dots, 𝑚 - 1}$ and $𝑎_{𝑚} 𝜎 = 𝑎_{1}$ , and for all $𝑥 \in {1, \dots, 𝑛} \ {𝑎_{1}, \dots, 𝑎_{𝑚}}$ , $𝑥 𝜎 = 𝑥$ .

We denote this cycle as $(𝑎_{1} 𝑎_{2} \dots 𝑎_{𝑚})$ .

Remark:We could denote

𝑒 = (1) (2) \dots (𝑛)

Lemma 2.2: Disjoint cycles commute.

Proof: Let $𝜎 = (𝑎_{1} \dots 𝑎_{𝑚})$ and $𝜏 = (𝑏_{1} \dots 𝑏_{𝑘})$ be disjoint cycles, i.e. ${𝑎_{1}, \dots, 𝑎_{𝑚}} \cap {𝑏_{1}, \dots, 𝑏_{𝑘}} = \emptyset$ .

Let $𝑖 \in {1, \dots, 𝑛}$ . Then

𝑖 𝜎 𝜏 = {\begin{cases} 𝑎_{𝑗 + 1} 𝜏 = 𝑎_{𝑗 + 1} & if 𝑖 = 𝑎_{𝑗} \\ 𝑖 𝜏 = 𝑏_{𝑗 + 1} & if 𝑖 = 𝑏_{𝑗} \\ 𝑖 & otherwise; \end{cases}

𝑖 𝜏 𝜎 = {\begin{cases} 𝑖 𝜎 = 𝑎_{𝑗 + 1} & if 𝑖 = 𝑎_{𝑗} \\ 𝑏_{𝑗 + 1} 𝜎 = 𝑏_{𝑗 + 1} & if 𝑖 = 𝑏_{𝑗} \\ 𝑖 & otherwise . \end{cases}

Theorem 2.3: Every $𝜎 \in 𝑆_{𝑛}$ can be written as a product of disjoint cycles.

Moreover, this factorisation is unique up to cycling of elements within cycles, and order of (commuting) factors.