Identity Mapping on Symmetric Group is Even Permutation

Theorem

Let $S_n$ denote the symmetric group on $n$ letters.

Let $e \in S_n$ be the identity permutation on $S_n$.

Then $e$ is an even permutation on $S_n$.

Proof

By definition of the identity permutation:

$\forall i \in \N_{>0}: \map e i = i$

Thus $e$ fixes all elements of $S_n$.

We have that a Transposition is of Odd Parity.

Hence a permutation is odd if and only if if is the composition of an odd number of transpositions.

The identity permutation is the composition of $0$ (zero) transpositions.

Hence the result.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 82$