Math 360, Fall 2013, Assignment 9 Proof 1
Contents
Theorem[edit]
The alternating group on n letters ($A_n$) is a normal subgroup of the symmetric group on n letters ($S_n$).
Remark[edit]
We've already shown that $A_n$ is a subgroup, since it contains the identity permutation, which is even, it is closed under the operation since composition is a closed operation, and since inverting a permutation doesn't change its parity, it is also closed under inversion. Additionally, it has already been shown via the first of the criteria for normality that the above statement is true, so we will instead use the third of the criteria, which is just as valid for the purposes of this proof.
Proof[edit]
Choose any $\sigma \in S_n$ and any $\tau \in A_n$. Look at $\sigma\tau\sigma^{-1}$. We know that $\tau$ has an even parity, since it resides in $A_n$. Additionally, we know that inverting a permutation does not alter its parity. Thus $\sigma$ and $\sigma^{-1}$ have the same parity. This leads to a proof by cases:
CASE I: $\sigma$ is of even parity
Suppose that $\sigma$ is of even parity. Then we have three even permutations composed with one another, and by the properties of permutation parity, this means that the resulting product is even and therefore $\sigma\tau\sigma^{-1} \in A_n$
CASE II: $\sigma$ is of odd parity
Suppose that $\sigma$ is of odd parity. Then we have an odd permutation composed with an even permutation composed with an odd permutation. An even permutation composed with an odd permutation is odd, and an odd result composed with an odd permutation is again even, thus $\sigma\tau\sigma^{-1}\in A_n$
We can say for certain that these cases exhaust all possibilities for the parity of $\sigma$. Therefore, we conclude that $\sigma\tau\sigma^{-1} \in A_n$. Since we took $\sigma$ and $\tau$ to be arbitrary, we conclude that the third of the criteria for normality is met.
Therefore, the alternating group on n letters is a normal subgroup of the symmetric group on n letters.
Q.E.D.
