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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 :)

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    T$h$ere should be a link below the [abstract al$g$ebra] tag that says "edit". Click it, a$n$d you can edit.2011-03-14

4 Answers 4

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As mentioned by yoyo: if $H\subset S_n$ 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 $f:S_n\to C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)\in C_2$ is the same element for every transposition $t\in S_n$ (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 $t\in S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.

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    @user8268 Can you explain how $f(t)\in C_2$ is the same element for every transposition $t\in S_n$, and how $S_n$ is generated by transpositions? Thank you.2014-11-30
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subgroups of index two are normal (exercise). $A_n$ is simple, $n\geq 5$ (exercise). if there were another subgroup $H$ of index two, then $H\cap A_n$ would be normal in $A_n$, contradiction.

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    How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......2013-10-30
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Other Way :

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

If $H\neq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$

WLOG assume say $(123)\notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.

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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.

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    How do 3-cycles have an odd order? Or were you referring to something else?2018-10-30