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

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?

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

$|A_4|=11$

• What do transpositions have to do with the order of $\;A_4\;$ here? And BTW: none of the elements of $\;A_4\;$ is a transposition... – Timbuc Nov 4 '14 at 5:28
• It's still an element of $A_4$ – Eoin Nov 4 '14 at 5:29
• I think you may be confusing transpositions = permutations of the form $\;(i_1\;i_2)\;$ with permutations in general. – Timbuc Nov 4 '14 at 5:30
• @Eoin, ah yes, since $A_4$ is group, must have a identity. Clicked – atherton Nov 4 '14 at 5:31
• How many times do you have to swap elements in order to "implement" the identity permutation? Is this number of times even? – David Nov 4 '14 at 5:35

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

• It's also an empty product of zero transpositions, but that is strangely more difficult to write! – Derek Holt Nov 4 '14 at 5:41
• $$\ \ \ \ \ \ \ \$$ – David Nov 4 '14 at 5:44
• Quite easy to write, see my previous comment ;-) – David Nov 4 '14 at 5:44

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 $\varphi^{-1}(\pi)=(12)\pi$.

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*}

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

identity $e=(13)(13)$