Alternating groups, specifically $A_6$

Question

My question is what are the possible order of $A_6$? And how would I show I get $\frac{6!}{2}=360$. Any tips? I know that $A_6$ is the group of even permutations on six elements. I also know that $360=2^3*3^2*5^1$. How do I got about doing this?

Answer 1

Consider a homomorphism $S_n \rightarrow \{-1, 1\}$ defined such that $\phi(\pi) \mapsto \operatorname{sgn}(\pi)$.

What is the kernel of this homomorphism? From here, apply the fact that $S_n/\ker(\phi) \cong \operatorname{Im}(\phi)$. What must be the orders of each of these?

Alternately, if you've proven that the only normal subgroup of $S_n$ is $A_n$, then what do you know about $\ker(\phi)$?

Answer 2

Here's an alternative method (which is equivalent, but avoids the language of homomorphisms and kernels). Let $O\subset S_n$ be the set of odd permutations and $E\subset S_n$ be the set of even permutations. Choose any odd permutation $\pi$. Then since the product of odd permutations is even, and the product of an odd and an even permutation is odd, we have $\pi O\subseteq E$ and $\pi E\subseteq O$. But $\pi E \cup \pi O = \pi S_n = S_n$, so that $$S_n = O\cup E = \pi E\cup \pi O.$$ It follows that $\pi E = O$; but the map $S_n\to S_n:\sigma\mapsto\pi\sigma$ is one-to-one, so that $|E| = |O|$. From this you can easily deduce $|A_n|$.

Answer 3

Facts that you might already know, or might need to prove:

• The symmetric group $S_n$ has order $n!$
• If I take an odd permutation $\sigma$ in $S_n$, then left-multiplication by $\sigma$ sends the set of even elements of $S_n$ to the set of odd elements, and vice-versa. So there are an equal number of even and odd permutations.
• Since $A_n$ is the set of even permutations, $|A_n| = \frac{1}{2}|S_n| = \frac{n!}{2}$.