Is An a Normal Subgroup of Sn? A Proof and Explanation

[kimkibun]

good day! i need to prove that the alternating group A_n is a normal subgroup of symmetric group, S_n, and i just want to know if my proving is correct.

we know that normal subgroup is subgroup where the right and left cosets coincides. but i got this equivalent definition of normal group from fraleigh's book, which states that for all gεG and hεH, a subgroup H of G is normal iff gHg^-1=H.


now here's my proof using the definition i got,

I. I need to show that for all τεS_n, τA_nτ^-1 is subset of A_n.

Let λετA_nτ^-1, then λ=τστ^-1, for all σεA_n. but since multiplication of transpositions are commutative and therefore,

λ=τστ^-1=σττ^-1=σ, thus, λεA_n, and therefore τA_nτ^-1 is a subset of A_n.

II. I need to prove that A_n is a subset of τA_nτ^-1.

Let σεA_n and τεS_n, since σ is an even transposition, τσ must be an odd transposition since no permutation is a product of both even or odd transposition. Also, since multiplication of transposition is commutative I now have,

σ=σe=σττ^-1=τστ^-1

thus, A_n is a subset of τA_nτ^-1.

since I've shown that the two sets are subsets of each other, I therefore conclude that τA_nτ^-1=A_n and A_n is normal.

thanks and God bless!

 

[algebrat]

looks hard to read. your grader isn't looking forward to digging through that either.

maybe there's a slicker approach.

isn't tA the same size as At? and there's only two cosets, since A is half of S. so either tA=At, or they are disjoint. but e is in A, so tA intersects At.

i used facts about size of cosets being same, A being half of S, and cosets are always disjoint or equal, so if you don't have these facts or close to them, then maybe you're on the right track.

 

[micromass]

since multiplication of transpositions are commutative

Huh, since when? And why can you limit your proof to transpositions??

Maybe another approach will help. Can you prove that A_n is the kernel of a suitable homomorphism??

 

[haruspex]

Kimkibun,
As micromass says, you cannot assume $\sigma$ and $\tau$ commute.
I also note that you have nowhere used any property of An other than the fact that it is a subgroup. You need to start with some property of An that subgroups in general don't have. algebrat's approach of using the fact that An is half of Sn does work, though I would word it differently. Left cosets (tH) are not necessarily disjoint from or identical to right cosets (Hu) but they are disjoint from (or identical to) each other. Since all cosets have the same size, either tA = A or tA = S-A. Likewise At = A or At = S-A. Therefore either tA = At or tA = S-At. Since t is an element of both, tA = At.
The 'disjoint or equal' property of left cosets is easily proved. Suppose xH and yH have a common element, xh = ym. Let xk ε xH. So ymh^-1k = xk. mh^-1k ε H, therefore xk ε yH.
That they all have the same size is also easy.

Btw, this wording is a bit sloppy:

for all gεG and hεH, a subgroup H of G is normal iff gHg^-1=H.

Better is
a subgroup H of G is normal in G iff (for all gεG gHg^-1=H).

 

[blue_raver22]

Your proof is correct. You have shown that for any permutation τ in Sn, τAnτ-1 is a subset of An and that An is a subset of τAnτ-1. This satisfies the definition of a normal subgroup, which states that for any element g in G, gHg-1 is a subset of H. In this case, τ is the element g and An is the subgroup H. Therefore, An is a normal subgroup of Sn. Well done!