Krdytkyu

How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?

Why isn't there other possibility?

Thanks :)

As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.

subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.

I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:

Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.

Keeping that in mind, we first prove the two easy facts about subgroups of index $2$

Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.

Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.

Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.

Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.

From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.

Other Way :

$A_n$ is generated by all 3 cycles in $S_n$

If $Hneq A_n$ and $|S_n:H|=2$ then atleast one 3 cycle not in H.

WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distant cosets which is contradiction to fact that H hads index 2