Alternating Group is Set of Even Permutations

Theorem

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

Let $A_n$ be the alternating group on $n$ letters.

Then $A_n$ consists of the set of even permutations of $S_n$.

Proof

Let $\sgn$ denote the sign of permutation on n Letters.

We have that $\map \sgn {S_n}$ is onto $C_2$.

Thus from the First Isomorphism Theorem, $A_n$ consists of the set of even permutations of $S_n$.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 81$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $9$: Permutations: Definition $9.19$: Remark