6 - Symmetric groups

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Symmetric group

Let $S$ be a set, then the set of all bijection on $S$ , with function composition as the operator, is a group, called the symmetric group on $S$ , denoted by $S y m (S)$ .

Here, the identity map $i d : S \to S$ acts as the group identity, with any $f \in S y m (S)$ $i d f = f i d = f$ .

The inverse function $f^{- 1}$ acts as the inverse element of each $f \in S y m (S)$ .

Associativity and closure follow as properties of composition of bijections.

Set cardinalities categorize symmetric groups

Let $S$ and $T$ be sets, such that the cardinalities of $S$ and $T$ are equal. Then $S y m (S)$ is isomorphic to $S y m (T)$

Commutative Diagram

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$* p r o o f * :$

Let $S$ , $T$ be sets with equal cardinality, and $ϕ : S \to T$ be a bijection. Moreover, let $f_{S}$ denote a bijection on $S$ , and $f_{T}$ denote a bijection on $T$ .

By observing the commutative diagram above, we see that for every $f_{T} \in S Y M (T)$ there exists an $f_{S}$ such that $f_{S} = ϕ^{- 1} f_{T} ϕ$ . Note that this indeed a bijection on $S$ , because the composition of bijections is a bijection.

Hence we claim that the map $f_{T} \mapsto ϕ^{- 1} f_{T} ϕ$ is an isomorphism from $S Y M (T)$ to $S Y M (S)$ .

Homomorphicity: $F_{T 1} F_{T 2} \mapsto ϕ F_{T 1} F_{T 2} ϕ^{- 1} = (ϕ^{- 1} f_{T_{1}} f_{T_{2}} ϕ) (ϕ^{- 1} f_{T_{1}} f_{T_{2}} ϕ)$

Surjectivity:
Let $f_{S}$ be an arbitrary element of $S Y M (S)$ , then we have $f_{T} = ϕ f_{S} ϕ^{- 1} \in S Y M (T)$ That maps to $f_{S}$ .

Injectivity:
let $ϕ^{- 1} f_{T_{1}} ϕ = ϕ^{- 1} f_{T_{2}} ϕ$ Then due to the properties of bijections, $f_{T_{1}} = f_{T_{2}}$ .

$◻$

All groups are isomorphic to a subgroup of a symmetric group.

Let $G$ be a group, then $G$ is isomorphic to a subgroup of symmetric group on the set $G$ , $S y m (G)$ .

$* p r o o f * :$
Because left coset maps of a group are bijective They are a subset of $S y m (G)$ . Because left coset maps on a group are isomorphic to the group itself, $G$ is isomorphic to a subgroup of $S y m (G)$ .

$◻$