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 :)

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7 Answers 7


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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    $\begingroup$ @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. $\endgroup$
    – jstnchng
    Nov 30, 2014 at 18:49
  • $\begingroup$ @jstnchng Suppose $0 = f(t) \neq f(t') = 1$ for transpositions $t, t' \in S_n$. Then $1 = f(t) + f(t') = f(tt') = f(tgtg^{-1})$ for some $g \in S_n$. But then $1 = 0 + 0$ is a contradiction. $\endgroup$
    – jskattt797
    Apr 12, 2021 at 22:07

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 $2$.

  • $\begingroup$ How do we know $(132)\notin H$? $\endgroup$
    – Lotte
    Jun 8, 2019 at 23:40
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    $\begingroup$ @Lotte If $(132)\in H$ as H is group $(132)^{-1}=(123)\in H$ But which is not . $\endgroup$ Aug 22, 2019 at 16:00
  • $\begingroup$ How do you know these are in fact distinct sets? $\endgroup$ Jan 11, 2022 at 1:59
  • $\begingroup$ Coset are partition of group. So by definition they are distinct. Let me know if you have any doubts $\endgroup$ Jan 11, 2022 at 5:06
  • $\begingroup$ why would at least one 3-cycle not be in $H$ ? because if we assume all 3-cycles are in $H$, we can only know that $A_n \subseteq H$ right ?but I don't see why this would imply that $A_n = H$. $\endgroup$
    – Wicowan
    Dec 20, 2022 at 9:59

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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    $\begingroup$ 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 $S_n$...... $\endgroup$
    – Vladhagen
    Oct 30, 2013 at 19:42
  • $\begingroup$ @Vladhagen you define the normal subgroup $G$ as being non-trivial. If $G \preccurlyeq S_n$ such that $[G:S_n]=2$ and $G \ne A_n$ then $[A_n \cap G:S_n]=2$, but $A_n$ is simple and obviously $A_n \cap G \preccurlyeq A_n$ so $A_n \cap G = A_n$ or identity. If identity, then $G$ contains identity and a transposition, but then $G$ isn't normal. So the intersection must be $A_n$, and by the order divisor theorem $\# A_n = n!/2 \mid \# G$ and $\# G \mid \# S_n = n!$ so $G = S_n$. $\endgroup$
    – Stephanie
    Oct 8, 2019 at 13:14

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.

  • $\begingroup$ How do 3-cycles have an odd order? Or were you referring to something else? $\endgroup$ Oct 30, 2018 at 20:14
  • $\begingroup$ @Maximalista 3-cycles have order 3. On the other hand, a 3-cycle is an even permutation. $\endgroup$ Jan 1, 2021 at 10:48

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.


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 $H=A_n$.


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 $H=A_n$.


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