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 :)

  • 6
    $\begingroup$ Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening. $\endgroup$ Mar 14, 2011 at 21:00
  • $\begingroup$ I am terribly sorry. But I don't know how to reedit it. $\endgroup$ Mar 14, 2011 at 21:07
  • 5
    $\begingroup$ There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit. $\endgroup$ Mar 14, 2011 at 21:11

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 $f=A_n$.

  • 3
    $\begingroup$ @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
  • 1
    $\begingroup$ @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 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 at 5:06

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.

  • 13
    $\begingroup$ 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 $n\geq 2$ 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 $H=A_n$.


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 $H=A_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.