# Showing that the alternating group is normal

## 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##).

## 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.

Last edited:

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member 587159
Do you know the fact that any subgroup of index 2 is normal?

fresh_42
Mentor
And you only need a transposition as ##\sigma ##, so you could e.g. apply a signature argument. Both facts, @Math_QED's index argument as well as the property of the signature function are of general interest. Once you know how to prove them, you could take that proof one by one for the situation you have given.

Btw., instead of the technical ##\sigma A \sigma^{-1} \subseteq A##, one can also say: left cosets coincide with right cosets.

Do you know the fact that any subgroup of index 2 is normal?
And you only need a transposition as ##\sigma ##, so you could e.g. apply a signature argument. Both facts, @Math_QED's index argument as well as the property of the signature function are of general interest. Once you know how to prove them, you could take that proof one by one for the situation you have given.

Btw., instead of the technical ##\sigma A \sigma^{-1} \subseteq A##, one can also say: left cosets coincide with right cosets.
The thing is, in my class we haven't learned much at all about normal subgroups or indices. So I tried to stay as close to what we have covered as possible. Does the proof I have given work in this case?

fresh_42
Mentor
The thing is, in my class we haven't learned much at all about normal subgroups or indices. So I tried to stay as close to what we have covered as possible. Does the proof I have given work in this case?
I will have a look later on, just as a start: ##A_2## Abelian is irrelevant, ##S_2## Abelian is needed. But to explain the coset issue, as it is really important, let me say the following.

For any subgroup ##U \subseteq G## we can define an equivalence relation on ##G## by ##x \sim y \Longleftrightarrow xy^{-1} \in U \Longleftrightarrow Ux=Uy## (ex.!) Every equivalence relation on a set allows to write that set as a union of disjoint subsets, namely equivalence classes, i.e. two elements belong to the same class, if they are equivalent (ex.!). In our case we even have equally many elements in each class, because of the group properties, i.e. ##|Ux|=|Uy|=|Ue|=|U|## (ex.!). The set of equivalence classes (i.e. each class considered as an element of this set) is denoted by ##G/\sim \,=\, G/U##.

Now the question is obvious, whether ##G/U## carries a group structure again or not. It turns out that this is the case if and only if ##U## is a normal subgroup. Otherwise we can't get a well-defined operation, aka multiplication on the set of classes (ex.!). A subgroup is normal if ##Ux=xU \Longleftrightarrow xUx^{-1} \subseteq U##.

Our suggestions now read: Either consider ##S_n = 1 \cdot A_n \cup \sigma \cdot A_n = A_n \cdot 1 \cup A_n \cdot \sigma ## as disjoint union of sets and ask yourself what this means for ##S_n - A_n## (@Math_QED) or consider ##\operatorname{sgn}(\sigma \tau \sigma^{-1})## where ##\sigma## is a transposition and ##\tau \in A_n = \{\,\sigma \in S_n \,|\,\operatorname{sgn}(\sigma )=+1\,\}## and ##\operatorname{sgn}()## the signature of a permutation, i.e. ##+1## for an even and ##-1## for an odd permutation (my idea).

In any case, I recommend to do these little exercises I noted by (ex!) above. It's worth to do them at least once to understand the concept of factor (aka quotient) groups, rings, algebras, vector spaces. It occurs in all these areas and is always the same idea.

member 587159
fresh_42
Mentor
I don't think your proof is correct. I saw that you read the permutations from left to right as you wrote ##(ijk)=(ij)(ik)##, which shows ##W_n \subseteq A_n##. This is true as can be seen by ##\operatorname{sgn}(ijk)=+1##.

However, I don't see the converse. You wrote ##(ij)(kl)=(ijk)(jkl)## which is not true from left to right: ##i## lands at ##j## on the LHS and at ##k## on the RHS. If you want to have it read from right to left, then the previous formula is wrong. I would have been nice of you, if you had mentioned the convention you use. I don't think it is right at all, but I'm not sure. I also don't see the length preservation in your second part of the proof. However, if you want to argue about the length, then better use the signature formula. ##\operatorname{sgn}(\tau)## counts the number of wrong positions with respect to the natural ordering and sets it to ##+1## if even and ##-1## if odd. ##A_n## contains all even permutations.

I don't think your proof is correct. I saw that you read the permutations from left to right as you wrote ##(ijk)=(ij)(ik)##, which shows ##W_n \subseteq A_n##. This is true as can be seen by ##\operatorname{sgn}(ijk)=+1##.

However, I don't see the converse. You wrote ##(ij)(kl)=(ijk)(jkl)## which is not true from left to right: ##i## lands at ##j## on the LHS and at ##k## on the RHS. If you want to have it read from right to left, then the previous formula is wrong. I would have been nice of you, if you had mentioned the convention you use. I don't think it is right at all, but I'm not sure. I also don't see the length preservation in your second part of the proof. However, if you want to argue about the length, then better use the signature formula. ##\operatorname{sgn}(\tau)## counts the number of wrong positions with respect to the natural ordering and sets it to ##+1## if even and ##-1## if odd. ##A_n## contains all even permutations.
https://math.stackexchange.com/a/914352/207516

I based part of my proof off of this. Is this wrong?

fresh_42
Mentor
In any case your multiplications are wrong: either you read them from left to right or the other way, but not both ways in the same proof. So ##A_2## Abelian is a wrong argument, although easy to fix, and it cannot be ##(ijk)=(ij)(ik) \; , \; \rightarrow \; , \;## and ##(ij)(kl)=(ijk)(jkl)\; , \;\leftarrow \; , \;## at the same time. I don't know whether ##A_n## can be generated by ##3-##cycles, but your argument doesn't match because of this discrepancy on how to read the multiplication and therefore I do not see that ##W_n## generates ##A_n##.

Nevertheless, why don't you use the same argument for products of ##2-##cycles? Then ##W_n## wouldn't be needed at all. You will only have to check, whether you get an even number of them.

I simply think, that it is far too complicated. @Math_QED 's argument is easier and of more general value, and counting false positions for ##\operatorname{sgn}## as well. I have chosen the multiplicative setup for ##\operatorname{sgn}## so all you need is ##\operatorname{sgn}(\sigma \tau)=\operatorname{sgn}(\sigma) \operatorname{sgn}(\tau)## which is also a formula worth to keep in mind.

Mr Davis 97