What is the order of the alternating group $A_4$?

Question

When I write out all the elements of $S_4$, I count only 11 transpositions. But in my text, the order of $A_4$ is $12$. What am I missing?

$A_4=\{(12)(34),(13)(24),(14)(23),(123),(124),(132),(134),(142),(143),(234),(243)$

$|A_4|=11$

Answer 1

You are missing the identity element. It can be written as an "even" permutation: $$(12)(12)$$

Answer 2

The order of $A_n$ is always half the order of $S_n$, consider the bijective map from the even permutations to the odd permutations where $\varphi(\pi)=(12)\pi $. This is a bijection since the inverse is the map from the odd permutations to the even permutations $\varphi^{-1}(\pi)=(12)\pi$.

Answer 3

As $A_n$ is a subgroup of $S_n$, there should be identity and identity is a even permutation.

identity $e=(13)(13)$

Answer 4

Consider the sign homomorphism $\varphi:S_n\rightarrow \{\pm 1\}$, which maps even permutations to 1 and odd permutations to -1. Notice that $\ker \varphi = A_n$, and since $\varphi$ is surjective, \begin{align*} \lvert \ker \varphi \lvert = \frac{\lvert S_n \lvert}{\lvert \{\pm 1\}\lvert}=\frac{n!}{2}. \end{align*}