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## Homework Statement

For each natural number ##n##, let ##V_n## be the subset of the symmetric group ##S_n## defined by

$$V_n = \{(i j)(k l) | i,j,k,l \in \{1,\ldots, n\}, i \neq j, \text{ and } k \neq l\},$$

that is, ##V_n## is the set of all products of two 2-cycles. Let ##A_n## be the subgroup of ##S_n## generated by ##V_n##; the group ##A_n## is called the alternating group of degree ##n##. For any ##\sigma\in S_n## define the set $$\sigma A_n\sigma^{-1}=\{\sigma \tau \sigma^{-1} \mid \sigma\in S_n, a\in A_n\}.$$

Prove that ##\sigma A_n\sigma^{-1}=A_n## for any ##\sigma\in S_n## (that is, ##A_n## is a normal subgroup of ##S_n##).

## Homework Equations

## The Attempt at a Solution

If ##n \le 2## then the fact that ##\sigma A_n \sigma^{-1} = A_n## is immediate because in that case ##A_n## is abelian. So suppose ##n \ge 3##. Let ##W_n = \{(i j k) | i,j,k \in \{1,\ldots, n\}, i \neq j \neq k\}##. We want to show that ##\langle V_n \rangle = \langle W_n \rangle##. First, we show that ##\langle W_n \rangle \subseteq \langle V_n \rangle ##. Since ##(ijk) = (ij)(ik)##, we see that every product of ##3##-cycles is also a product of an even number of transpositions, so every element in ##\langle W_n \rangle## is in ##\langle V_n \rangle##. Second, we show that ##\langle V_n \rangle \subseteq \langle W_n \rangle ##. Note that there are two cases for a pair of transpositions. Case 1 is that the two transpositions share a number: ##(ij)(ik) = (ijk)##, so this ends up being a ##3##-cycle. Case 2 is if all the numbers in the two transpositions are distinct: ##(ij)(kl) = (ijk)(jkl)##, and so this ends up being a pair of ##3##-cycles. Hence, any product of an even number of transpositions is also a product of ##3##-cycles. So ##\langle V_n \rangle \subseteq \langle W_n \rangle ##. Hence ##\langle V_n \rangle = \langle W_n \rangle##.

We will now show that conjugation preserves the length of a cycle.

If ##\mu## is the ##m## cycle ##(x_1 x_2 \dots x_m)##, then

##\tau \mu \tau^{-1} = (\tau(x_1) \tau(x_2) \dots \tau(x_m))##

since ##\tau\mu\tau^{-1}(\tau(x_i)) = \tau(\mu(x_i)) = \tau(x_{i+1})##. So ##\tau \mu \tau^{-1}## is again an ##m## cycle.

Therefore, conjugation by ##\tau \in S_n## preserves the length of ##\mu##. Now, ##A_n## is generated by the ##3##-cycles. Hence, if ##\sigma \in A_n## then ##\sigma = \mu_1\dots \mu_k## where ##\mu_i## are ##3##-cycles. Then

##\tau \sigma \tau^{-1} = \tau(\mu_1\dots \mu_k)\tau^{-1} = \tau \mu_1 \tau^{-1} \dots \tau\mu_k\tau^{-1}##

which is a product of ##3##-cycles.

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