$A_n$ is the only subgroup of $S_n$ of index $2$.

Question

How to prove that the only subgroup of the symmetric group $S_n$ of index 2 is $A_n$?

Why isn't there other possibility?

Thanks :)

Answer 1

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 $f=A_n$.

Answer 2

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 $2$.

Answer 3

• 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.

[To see that $A_n\cap H\neq\{1\}$ as brought up in a comment, note that if this were the case, then $S_n\backslash(A_n\cup H)=\{x\}$ for some single element $x\in S_n$. Both $A_n$ and $H$ are normal, so $x\in Z(S_n)=\{1\}$, impossible.]

The better answer is probably the accepted one: The sign homomorphism is the only non-trivial homomorphism $\phi:S_n\to \{\pm1\}$. This uniquely identifies $A_n$ as the only index 2 subgroup of $S_n$.

Answer 4

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.

Answer 5

The quotient map for $A_n$ is a surjective homomorphism to $C_2$.

Any other index two subgroup $H$ of $A_n$ gives you a distinct surjective homomorphism to $C_2$.

Therefore taking the product of these we obtain a surjective homomorphism $S_n$ to $C_2 \times C_2$. But then this has kernel of order $\frac{n!}{4}$, by the first iso theorem, and thus the image of $A_n$ cannot be 1 or $C_2 \times C_2$, so must be of order 2.

But this implies that $A_n$ has an index two subgroup, a contradiction as $A_n$ is simple for $n \geq 5$, and when $n = 4$, $A_4$ has no order 6 subgroup.

Answer 6

Let $n\geq 2$ and $H \leq S_n$ be such that $\mid S_n:H \mid=2$. Then $H \trianglelefteq S_n$ and $S_n/H$ being isomorphic to $\Bbb Z/2\Bbb Z$, is cyclic. Consider the natural surjection $\pi :S_n\to S_n/H$. Since $S_n/H$ is abelian, the commutator subgroup $[S_n,S_n]\subseteq \ker(\pi)=H$. We know $[S_n,S_n]=A_n$, hence $A_n \subseteq H$. Clealy $\mid A_n \mid=\mid H \mid$, as both of them are subgroups of a finite group of equal index. Therefore we have $H=A_n$.

Answer 7

Let $H\le S_n$ such that $[S_n:H]=2$ and $H\ne A_n$. Denoted $d:=|H\cap A_n|$, we get: $$|HA_n|=\frac{(n!/2)^2}{d} \tag1$$ By Lagrange, $d\mid n!/2$ and, since $HA_n\le S_n$ (as $A_n$ is normal in $S_n$) also $\bigl((n!/2)^2/d\bigr)\mid n!$. Therefore: $$\frac{n!}{2}=kd \tag2$$ and $$1=l\frac{n!}{4d} \tag3$$ for some positive integers $k$ and $l$. From $(2)$ and $(3)$ follows $kl=2$, and hence $k=1$ or $k=2$. But since $H\ne A_n$, it must be $d\ne n!/2$ and hence, by $(2)$, $k\ne 1$. So $k=2$ and hence, again by $(2)$, $d=n!/4$, whence by $(1)$ $|HA_n|=n!$, and finally $HA_n=S_n$: contradiction, by the argument in this post. Therefore, necessarily $H=A_n$.